🧠:Okay, let's see. The problem is to find the maximum value of 'a' such that the inequality [frac{sqrt[3]{tan x} - sqrt[3]{cot x}}{sqrt[3]{sin x} + sqrt[3]{cos x}} > frac{a}{2}]holds for all permissible ( x ) in the interval ( left(frac{3pi}{2}, 2piright) ). Hmm, first I need to understand the domain where this inequality is defined. So, the expressions inside the cube roots must be valid. Since cube roots can handle negative numbers, but (tan x) and (cot x) have their own domains. Let me check the interval ( left(frac{3pi}{2}, 2piright) ). In that interval, which is the fourth quadrant, right? So, from ( 270^circ ) to ( 360^circ ). In this quadrant, (sin x) is negative, (cos x) is positive. So, (tan x = frac{sin x}{cos x}) will be negative, and (cot x = frac{cos x}{sin x}) will also be negative. Therefore, both (tan x) and (cot x) are negative here. But cube roots of negative numbers are allowed, so the numerator and denominator are both real numbers. However, we need to make sure that (tan x) and (cot x) are defined. So, (tan x) is defined when (cos x neq 0), which is true in the open interval ( left(frac{3pi}{2}, 2piright) ), and (cot x) is defined when (sin x neq 0), which is also true in the open interval. So, the expression is defined for all ( x ) in that interval.Next, let's simplify the given inequality. Maybe substituting variables or using trigonometric identities could help. Let me write down the inequality again:[frac{sqrt[3]{tan x} - sqrt[3]{cot x}}{sqrt[3]{sin x} + sqrt[3]{cos x}} > frac{a}{2}]Since (tan x = frac{sin x}{cos x}) and (cot x = frac{cos x}{sin x}), maybe expressing everything in terms of (sin x) and (cos x) would be useful. Let's denote ( s = sin x ) and ( c = cos x ). Then, the expression becomes:Numerator: ( sqrt[3]{frac{s}{c}} - sqrt[3]{frac{c}{s}} )Denominator: ( sqrt[3]{s} + sqrt[3]{c} )So the entire fraction is:[frac{sqrt[3]{frac{s}{c}} - sqrt[3]{frac{c}{s}}}{sqrt[3]{s} + sqrt[3]{c}}]Hmm, maybe we can factor this expression or simplify it somehow. Let's see. Let's rewrite the numerator:First, let's note that ( sqrt[3]{frac{s}{c}} = frac{sqrt[3]{s}}{sqrt[3]{c}} ), and similarly ( sqrt[3]{frac{c}{s}} = frac{sqrt[3]{c}}{sqrt[3]{s}} ). So the numerator is:[frac{sqrt[3]{s}}{sqrt[3]{c}} - frac{sqrt[3]{c}}{sqrt[3]{s}} = frac{(sqrt[3]{s})^2 - (sqrt[3]{c})^2}{sqrt[3]{s}sqrt[3]{c}} = frac{s^{2/3} - c^{2/3}}{(sc)^{1/3}}]So numerator simplifies to ( frac{s^{2/3} - c^{2/3}}{(sc)^{1/3}} ). The denominator is ( sqrt[3]{s} + sqrt[3]{c} = s^{1/3} + c^{1/3} ). So the entire expression becomes:[frac{s^{2/3} - c^{2/3}}{(sc)^{1/3} cdot (s^{1/3} + c^{1/3})}]Hmm, notice that the numerator is a difference of squares: ( s^{2/3} - c^{2/3} = (s^{1/3} - c^{1/3})(s^{1/3} + c^{1/3}) ). So substituting that in, we get:Numerator: ( (s^{1/3} - c^{1/3})(s^{1/3} + c^{1/3}) )Denominator: ( (sc)^{1/3} cdot (s^{1/3} + c^{1/3}) )We can cancel out the ( (s^{1/3} + c^{1/3}) ) terms, assuming they are not zero. But in the interval ( left( frac{3pi}{2}, 2pi right) ), ( s = sin x ) is negative and ( c = cos x ) is positive. So ( s^{1/3} ) is the cube root of a negative number, which is negative, and ( c^{1/3} ) is the cube root of a positive number, which is positive. Therefore, ( s^{1/3} + c^{1/3} ) could be positive or negative depending on their magnitudes. However, since ( sin x ) is between -1 and 0 in this interval, and ( cos x ) is between 0 and 1. Therefore, ( |s^{1/3}| = | sqrt[3]{sin x} | = sqrt[3]{ | sin x | } ), and ( |c^{1/3}| = sqrt[3]{ cos x } ). Since ( | sin x | ) is between 0 and 1, and ( cos x ) is between 0 and 1. Therefore, ( s^{1/3} + c^{1/3} ) is the sum of a negative number and a positive number. Whether this sum is positive or negative depends on which term is larger in magnitude.Wait, but since ( x ) is in ( left( frac{3pi}{2}, 2pi right) ), let's take specific points to test. For example, near ( 2pi ), ( sin x ) is close to 0 (negative), and ( cos x ) is close to 1. So ( s^{1/3} ) is a small negative number, and ( c^{1/3} ) is close to 1. So ( s^{1/3} + c^{1/3} ) is close to 1 - a small negative number, so it's positive. Near ( frac{3pi}{2} ), ( sin x ) is -1, and ( cos x ) is 0. So ( s^{1/3} = sqrt[3]{-1} = -1 ), and ( c^{1/3} = 0 ). Therefore, ( s^{1/3} + c^{1/3} = -1 + 0 = -1 ). Therefore, the denominator can be both positive and negative in this interval. Therefore, when we cancel ( s^{1/3} + c^{1/3} ), we need to consider the sign. Wait, but if the denominator ( s^{1/3} + c^{1/3} ) is negative, then when we cancel it from numerator and denominator, the inequality sign would flip. Wait, but in the original inequality, the left-hand side (LHS) is compared to ( frac{a}{2} ). So maybe we need to handle the sign carefully. Hmm, perhaps my initial approach is not the best. Let me try another method.Alternatively, maybe substituting ( t = sqrt[3]{tan x} ). Let's try that. Let ( t = sqrt[3]{tan x} ). Then ( tan x = t^3 ), and since ( cot x = 1/tan x ), then ( sqrt[3]{cot x} = sqrt[3]{1/t^3} = 1/t ). Therefore, the numerator becomes ( t - 1/t ).The denominator is ( sqrt[3]{sin x} + sqrt[3]{cos x} ). Let's express this in terms of ( t ). Since ( tan x = frac{sin x}{cos x} ), so ( t^3 = frac{sin x}{cos x} ), which implies ( sin x = t^3 cos x ). Let's substitute this into the denominator.Denominator: ( sqrt[3]{t^3 cos x} + sqrt[3]{cos x} = sqrt[3]{t^3} cdot sqrt[3]{cos x} + sqrt[3]{cos x} = t cdot sqrt[3]{cos x} + sqrt[3]{cos x} = sqrt[3]{cos x} (t + 1) )Therefore, the entire expression becomes:Numerator: ( t - 1/t )Denominator: ( sqrt[3]{cos x} (t + 1) )So the fraction is:[frac{t - frac{1}{t}}{sqrt[3]{cos x} (t + 1)} = frac{t^2 - 1}{t cdot sqrt[3]{cos x} (t + 1)} = frac{(t - 1)(t + 1)}{t cdot sqrt[3]{cos x} (t + 1)} = frac{(t - 1)}{t cdot sqrt[3]{cos x}}]So we simplified the expression to ( frac{t - 1}{t cdot sqrt[3]{cos x}} ). Interesting. Now, let's recall that ( t = sqrt[3]{tan x} ). Since ( x ) is in the fourth quadrant, ( tan x ) is negative, so ( t ) is the cube root of a negative number, which is negative. Therefore, ( t ) is negative. Also, ( cos x ) is positive in the fourth quadrant, so ( sqrt[3]{cos x} ) is positive. Therefore, the denominator ( t cdot sqrt[3]{cos x} ) is negative (since ( t ) is negative and ( sqrt[3]{cos x} ) is positive). The numerator ( t - 1 ): since ( t ) is negative, ( t - 1 ) is negative minus 1, which is more negative. So numerator is negative, denominator is negative. Therefore, the entire fraction is positive.Therefore, the original fraction simplifies to a positive value. Therefore, the inequality is equivalent to:[frac{t - 1}{t cdot sqrt[3]{cos x}} > frac{a}{2}]But since both numerator and denominator are negative, their ratio is positive. So this simplifies to:[frac{1 - t}{ |t| cdot sqrt[3]{cos x} } > frac{a}{2}]Wait, maybe not. Let me check again. Since ( t ) is negative, ( t = -|t| ). So numerator: ( t - 1 = -|t| - 1 ), denominator: ( t cdot sqrt[3]{cos x} = -|t| cdot sqrt[3]{cos x} ). Therefore, the fraction becomes:[frac{ - (|t| + 1) }{ - |t| cdot sqrt[3]{cos x} } = frac{ |t| + 1 }{ |t| cdot sqrt[3]{cos x} }]Ah, that's better! So the expression simplifies to ( frac{ |t| + 1 }{ |t| cdot sqrt[3]{cos x} } ). So the original inequality becomes:[frac{ |t| + 1 }{ |t| cdot sqrt[3]{cos x} } > frac{a}{2}]Now, since ( t = sqrt[3]{tan x} ), and ( tan x = frac{sin x}{cos x} ). Since ( x ) is in ( (frac{3pi}{2}, 2pi) ), ( sin x ) is negative, ( cos x ) is positive, so ( tan x ) is negative. Therefore, ( t = sqrt[3]{tan x} ) is negative, so ( |t| = | sqrt[3]{tan x} | = sqrt[3]{ | tan x | } ). Let me denote ( |t| = sqrt[3]{ | tan x | } = sqrt[3]{ frac{ | sin x | }{ cos x } } ). Let me denote ( u = |t| = sqrt[3]{ frac{ | sin x | }{ cos x } } ). Then, since ( | sin x | = - sin x ) (because ( sin x ) is negative in this interval), we can write ( u = sqrt[3]{ frac{ - sin x }{ cos x } } = sqrt[3]{ - tan x } ). But since ( tan x ) is negative, ( - tan x ) is positive, so ( u ) is the cube root of a positive number, hence positive. Therefore, ( u > 0 ).Therefore, the expression simplifies to:[frac{ u + 1 }{ u cdot sqrt[3]{cos x} } > frac{a}{2}]So, we need to find the minimal value of the left-hand side (LHS) over ( x in (frac{3pi}{2}, 2pi) ), because the inequality must hold for all permissible x. Therefore, the minimal value of LHS should be greater than ( frac{a}{2} ). Hence, the maximum a is twice the minimal value of LHS.So, the problem reduces to finding the minimum of ( frac{ u + 1 }{ u cdot sqrt[3]{cos x} } ), where ( u = sqrt[3]{ frac{ - sin x }{ cos x } } ).Let me express everything in terms of a single variable. Let's set ( y = cos x ). Since ( x in (frac{3pi}{2}, 2pi) ), ( cos x ) ranges from 0 to 1. Specifically, when ( x ) approaches ( frac{3pi}{2} ), ( cos x ) approaches 0, and as ( x ) approaches ( 2pi ), ( cos x ) approaches 1. Also, ( sin x = - sqrt{1 - y^2} ) because ( sin x ) is negative in this interval.Therefore, ( u = sqrt[3]{ frac{ - sin x }{ cos x } } = sqrt[3]{ frac{ sqrt{1 - y^2} }{ y } } ). So ( u = left( frac{ sqrt{1 - y^2} }{ y } right)^{1/3} ). Therefore, substituting back into the expression:[frac{ u + 1 }{ u cdot y^{1/3} } = frac{ left( frac{ sqrt{1 - y^2} }{ y } right)^{1/3} + 1 }{ left( frac{ sqrt{1 - y^2} }{ y } right)^{1/3} cdot y^{1/3} } ]Simplify the denominator:[left( frac{ sqrt{1 - y^2} }{ y } right)^{1/3} cdot y^{1/3} = left( frac{ sqrt{1 - y^2} }{ y } cdot y right)^{1/3} = ( sqrt{1 - y^2} )^{1/3 } = (1 - y^2)^{1/6}]So denominator simplifies to ( (1 - y^2)^{1/6} ).Numerator:[left( frac{ sqrt{1 - y^2} }{ y } right)^{1/3} + 1 = u + 1]Therefore, the entire expression becomes:[frac{ u + 1 }{ (1 - y^2)^{1/6} } ]But ( u = left( frac{ sqrt{1 - y^2} }{ y } right)^{1/3} ), so substituting back:[frac{ left( frac{ sqrt{1 - y^2} }{ y } right)^{1/3} + 1 }{ (1 - y^2)^{1/6} } ]Let me simplify this expression. Let's write ( z = y ), where ( z in (0, 1) ). Then:First, ( sqrt{1 - z^2} = (1 - z^2)^{1/2} ), so:Numerator term 1: ( left( frac{(1 - z^2)^{1/2}}{ z } right)^{1/3} = frac{(1 - z^2)^{1/6}}{ z^{1/3} } )Numerator term 2: 1Denominator: ( (1 - z^2)^{1/6} )Therefore, the expression is:[frac{ frac{(1 - z^2)^{1/6}}{ z^{1/3} } + 1 }{ (1 - z^2)^{1/6} } = frac{1}{ z^{1/3} } + frac{1}{(1 - z^2)^{1/6} }]Wait, let's check that step. Let me rewrite:Expression = [ ( (1 - z²)^{1/6} / z^{1/3} ) + 1 ] / (1 - z²)^{1/6}So split the fraction:= [ (1 - z²)^{1/6} / ( z^{1/3} (1 - z²)^{1/6} ) ) + 1 / (1 - z²)^{1/6} ]Wait, no, perhaps more straightforward:Numerator: A + B, where A = (1 - z²)^{1/6} / z^{1/3}, B = 1.Then, dividing by denominator (1 - z²)^{1/6}:= A / denominator + B / denominator = [ (1 - z²)^{1/6} / z^{1/3} ] / (1 - z²)^{1/6} ) + 1 / (1 - z²)^{1/6}Simplify:= 1 / z^{1/3} + 1 / (1 - z²)^{1/6}So the entire expression simplifies to:[frac{1}{ z^{1/3} } + frac{1}{ (1 - z^2)^{1/6} }]Where ( z = cos x ), and ( z in (0, 1) ). Therefore, our problem reduces to finding the minimum of the function:[f(z) = frac{1}{ z^{1/3} } + frac{1}{ (1 - z^2)^{1/6} }]over ( z in (0, 1) ).Now, to find the minimum of ( f(z) ), we can take the derivative and set it to zero. Let's compute ( f(z) ):( f(z) = z^{-1/3} + (1 - z^2)^{-1/6} )Compute the derivative ( f'(z) ):First term derivative: ( d/dz [ z^{-1/3} ] = (-1/3) z^{-4/3} )Second term derivative: ( d/dz [ (1 - z^2)^{-1/6} ] = (-1/6)(1 - z^2)^{-7/6} (-2z) = ( z / 3 ) (1 - z^2)^{-7/6} )Therefore:( f'(z) = - frac{1}{3} z^{-4/3} + frac{z}{3} (1 - z^2)^{-7/6} )Set ( f'(z) = 0 ):[- frac{1}{3} z^{-4/3} + frac{z}{3} (1 - z^2)^{-7/6} = 0]Multiply both sides by 3 to eliminate denominators:[- z^{-4/3} + z (1 - z^2)^{-7/6} = 0]Bring the first term to the other side:[z (1 - z^2)^{-7/6} = z^{-4/3}]Multiply both sides by ( z^{4/3} ):[z^{1 + 4/3} (1 - z^2)^{-7/6} = 1]Simplify exponent of z: ( 1 + 4/3 = 7/3 ), so:[z^{7/3} (1 - z^2)^{-7/6} = 1]Express ( (1 - z^2)^{-7/6} = (1 - z^2)^{-7/6} ). Let's write both terms with exponents:[z^{7/3} cdot (1 - z^2)^{-7/6} = 1]Take natural logarithm on both sides:[frac{7}{3} ln z - frac{7}{6} ln(1 - z^2) = 0]Factor out ( frac{7}{6} ):[frac{7}{6} left( 2 ln z - ln(1 - z^2) right ) = 0]Since ( frac{7}{6} neq 0 ), we have:[2 ln z - ln(1 - z^2) = 0]Which simplifies to:[2 ln z = ln(1 - z^2)]Exponentiate both sides:[z^2 = 1 - z^2]Solve for ( z^2 ):[2 z^2 = 1 implies z^2 = frac{1}{2} implies z = frac{1}{sqrt{2}} approx 0.7071]Since ( z in (0,1) ), we take the positive root. Therefore, critical point at ( z = frac{1}{sqrt{2}} ). Now, we need to check if this is a minimum.Compute the second derivative or test intervals around the critical point. Let's take test values around ( z = 1/sqrt{2} approx 0.707 ).Take z = 0.6:Compute f'(0.6):First term: - (0.6)^{-4/3} / 3 ≈ - (0.6)^{-1.333} / 3 ≈ - (1 / 0.6^{1.333}) /3 ≈ - (1 / (0.6 * 0.6^{0.333})) /3 ≈ 0.6^{0.333} ≈ cube root of 0.6 ≈ 0.843, so 0.6^{1.333} ≈ 0.6 * 0.843 ≈ 0.5058, so 1 / 0.5058 ≈ 1.977, so -1.977 /3 ≈ -0.659Second term: 0.6 /3 * (1 - 0.6^2)^{-7/6} = 0.2 * (1 - 0.36)^{-7/6} = 0.2 * (0.64)^{-7/6} ≈ 0.2 * (0.64)^{-1.1667} ≈ 0.64^-1 = 1.5625, and 0.64^-0.1667 ≈ e^{ -0.1667 ln 0.64 } ≈ e^{ -0.1667*(-0.4463) } ≈ e^{0.0744} ≈ 1.077. So total ≈ 0.2 * 1.5625 * 1.077 ≈ 0.2 * 1.683 ≈ 0.3366So total f'(0.6) ≈ -0.659 + 0.3366 ≈ -0.322, which is negative.Take z = 0.8:First term: - (0.8)^{-4/3} /3 ≈ - (0.8)^{-1.333} /3 ≈ - (1 / 0.8^{1.333}) /3. 0.8^1 = 0.8, 0.8^0.333 ≈ cube root of 0.8 ≈ 0.928, so 0.8^{1.333} ≈ 0.8 * 0.928 ≈ 0.742, so 1 / 0.742 ≈ 1.347, so -1.347 /3 ≈ -0.449Second term: 0.8 /3 * (1 - 0.8^2)^{-7/6} = 0.8 /3 * (1 - 0.64)^{-7/6} = 0.2667 * (0.36)^{-7/6} ≈ 0.2667 * (0.36)^{-1.1667} ≈ 0.36^-1 = 2.7778, and 0.36^-0.1667 ≈ e^{ -0.1667 ln 0.36 } ≈ e^{ -0.1667*(-1.0217) } ≈ e^{0.170} ≈ 1.185. So total ≈ 0.2667 * 2.7778 * 1.185 ≈ 0.2667 * 3.298 ≈ 0.879.Thus, f'(0.8) ≈ -0.449 + 0.879 ≈ 0.43, which is positive.Therefore, around z = 1/√2 ≈ 0.707, the derivative goes from negative to positive, indicating a minimum at z = 1/√2. Therefore, the minimal value occurs at z = 1/√2.Therefore, the minimal value of f(z) is f(1/√2). Let's compute that.First, compute ( z = 1/sqrt{2} approx 0.7071 )Compute ( f(z) = z^{-1/3} + (1 - z^2)^{-1/6} )First term: ( (1/sqrt{2})^{-1/3} = (sqrt{2})^{1/3} = 2^{1/6} approx 1.1225 )Second term: ( (1 - (1/2))^{-1/6} = (1/2)^{-1/6} = 2^{1/6} approx 1.1225 )Therefore, ( f(1/sqrt{2}) = 2^{1/6} + 2^{1/6} = 2 cdot 2^{1/6} = 2^{7/6} approx 2^{1 + 1/6} = 2 cdot 1.1225 ≈ 2.245 )Wait, 2^{7/6} is 2^(1 + 1/6) = 2 * 2^{1/6} ≈ 2 * 1.1225 ≈ 2.245. So approximately 2.245.But we need to verify this calculation. Let me compute 2^{1/6}:2^{1/3} ≈ 1.26, so 2^{1/6} = sqrt(2^{1/3}) ≈ sqrt(1.26) ≈ 1.1225. So yes, 2^{1/6} ≈ 1.1225.So f(z) at z=1/√2 is 1.1225 + 1.1225 = 2.245, which is 2^{7/6}. Since 2^{1/6} is approximately 1.1225, then 2^{7/6} = 2^{1 + 1/6} = 2 * 2^{1/6} ≈ 2.245.Therefore, the minimal value of f(z) is 2^{7/6} ≈ 2.245. Therefore, the minimal value of the original expression is 2^{7/6}, so the maximum a is 2 * 2^{7/6} = 2^{1 + 7/6} = 2^{13/6} ≈ ?Wait, wait. Wait, the original inequality is:LHS > a/2We found that the minimal value of LHS is 2^{7/6}. Therefore, to have 2^{7/6} > a/2, so a/2 < 2^{7/6}, hence a < 2 * 2^{7/6} = 2^{1 + 7/6} = 2^{13/6}But 2^{13/6} = 2^{2 + 1/6} = 4 * 2^{1/6} ≈ 4 * 1.1225 ≈ 4.49Wait, but hold on. Let's make sure.Wait, the minimal value of LHS is 2^{7/6}, so to have that minimal value greater than a/2, we have:2^{7/6} > a/2 => a < 2 * 2^{7/6} = 2^{1 + 7/6} = 2^{13/6}But 13/6 is equal to 2 and 1/6, so 2^{13/6} = (2^{1/6})^13 ≈ ?Wait, alternatively, 2^{13/6} = e^{(13/6) ln 2} ≈ e^{(13/6)(0.693)} ≈ e^{(1.5015)} ≈ 4.48.Yes, approximately 4.48. So the maximum a is approximately 4.48, which would round to 4.48, but let's check if that's precise.Alternatively, 2^{1/6} is the sixth root of 2, which is approximately 1.122462048. Therefore, 2^{7/6} = 2^{1 + 1/6} = 2 * 2^{1/6} ≈ 2 * 1.122462 ≈ 2.244924. Then, 2.244924 * 2 ≈ 4.489848, which is approximately 4.49 when rounded to the nearest hundredth.But wait, let's confirm that this critical point is indeed the minimum. Since we checked the derivative before and after and found that the function decreases before z=1/√2 and increases after, so yes, it's the minimum.Therefore, the minimal value of the LHS is 2^{7/6} ≈ 2.2449, hence a < 2 * 2.2449 ≈ 4.4898. Therefore, the maximum a is approximately 4.4898, which rounds to 4.49.But let me verify this calculation again to ensure there's no error in substitution.Wait, we had earlier that the expression simplifies to f(z) = z^{-1/3} + (1 - z²)^{-1/6}, and at z = 1/√2, f(z) = (1/√2)^{-1/3} + (1 - 1/2)^{-1/6} = (2^{1/2})^{1/3} + (1/2)^{-1/6} = 2^{1/6} + 2^{1/6} = 2 * 2^{1/6} = 2^{7/6}Yes, that's correct. So 2^{7/6} is indeed the minimal value. Therefore, the maximum a is 2 * 2^{7/6} = 2^{13/6} ≈ 4.49.But to confirm, let's check the original expression at z = 1/√2 (i.e., x where cos x = 1/√2, which is x = 7π/4, which is 315 degrees, in the interval (3π/2, 2π)).At x = 7π/4, sin x = -√2/2, cos x = √2/2.Compute tan x = sin x / cos x = -1, cot x = -1. Therefore, numerator: cube root(tan x) - cube root(cot x) = cube root(-1) - cube root(-1) = (-1) - (-1) = 0. Wait, but according to our previous analysis, the expression at z = 1/√2 should be 2^{7/6} ≈ 2.245. But here, if we substitute x = 7π/4, the numerator becomes 0. What's happening here?Wait, this is a contradiction. That means there is a mistake in the reasoning.Wait, when x = 7π/4, sin x = -√2/2, cos x = √2/2, so tan x = -1, cot x = -1. Therefore, the original expression:Numerator: cube root(tan x) - cube root(cot x) = cube root(-1) - cube root(-1) = (-1) - (-1) = 0.Denominator: cube root(sin x) + cube root(cos x) = cube root(-√2/2) + cube root(√2/2) = - cube root(√2/2) + cube root(√2/2) = 0.So the original expression becomes 0/0, which is undefined. But according to our substitution, we obtained a minimal value at z = 1/√2, which corresponds to x = 7π/4. But at that point, the original expression is undefined. Therefore, there's an error in the analysis.This suggests that the critical point at z = 1/√2 is actually a point where the function is undefined, so the minimal value must occur elsewhere.Wait, where is the mistake here? Let's track back.When we substituted variables, we set z = cos x, and found a critical point at z = 1/√2. However, at x = 7π/4, which corresponds to z = cos x = √2/2 ≈ 0.7071, but at this x, sin x = -√2/2, so tan x = -1. But then, in the original expression, both numerator and denominator become zero, leading to an indeterminate form 0/0. Therefore, even though mathematically we found a critical point at z = 1/√2, in reality, the function has a discontinuity there.Therefore, our previous approach is flawed. We need to reconsider.Therefore, there must be a mistake in the substitution or simplification steps.Wait, let's retrace the steps. Initially, we set t = cube root(tan x), leading to substitution. Then, we expressed the expression in terms of z = cos x. However, the critical point found corresponds to x = 7π/4, where the original function is undefined. Therefore, the minimum cannot occur there. Therefore, there must be a mistake in the calculus approach.Alternatively, perhaps the function approaches a limit as x approaches 7π/4 from either side. Let's consider the limit as z approaches 1/√2 from below and above.Wait, but z = cos x is in (0,1) for x in (3π/2, 2π). So approaching z = 1/√2 from either side is possible. However, at z = 1/√2, the original expression is undefined. Therefore, we need to check the behavior near z = 1/√2.But perhaps the minimal value occurs at the limits of the interval, i.e., as z approaches 0 or 1.Let me analyze the behavior as z approaches 0+ and 1-.First, as z approaches 0+ (x approaches 3π/2+):In this case, cos x approaches 0, so z approaches 0. Then, sin x approaches -1, so |sin x| = 1. Therefore, u = cube root( |tan x| ) = cube root( |sin x / cos x| ) ≈ cube root(1 / z ) → infinity as z approaches 0. Therefore, f(z) = 1/z^{1/3} + 1/(1 - z^2)^{1/6} ≈ infinity + 1/1^{1/6} = infinity. Therefore, f(z) approaches infinity as z approaches 0+.As z approaches 1- (x approaches 2π-):Here, cos x approaches 1, sin x approaches 0 (negative). So |tan x| = |sin x / cos x| ≈ | -0 / 1 | = 0. Therefore, u = cube root(0) = 0. Therefore, f(z) = 1/z^{1/3} + 1/(1 - z^2)^{1/6} ≈ 1/1 + 1/(0)^{1/6}. But as z approaches 1-, 1 - z^2 approaches 0+, so (1 - z^2)^{-1/6} approaches infinity. Therefore, f(z) approaches 1 + infinity = infinity.Therefore, f(z) tends to infinity at both ends. However, there is a critical point at z = 1/√2, but at this point, the original function is undefined. Therefore, the minimum must occur near the critical point but avoiding the undefined point.But how is this possible?Wait, the function f(z) was derived under the substitution that t = cube root(tan x), but when x approaches 7π/4, tan x approaches -1, so cube root(tan x) approaches -1, but in the substitution steps, we had to express things in terms of z = cos x. However, the error occurs when assuming the critical point is at z = 1/√2, but in reality, the original function is undefined there.Therefore, there must be a mistake in the substitution process.Alternatively, perhaps the expression simplifies differently.Let me revisit the original expression:After substitutions, the expression simplifies to:f(z) = z^{-1/3} + (1 - z^2)^{-1/6}But this was derived under the assumption that the original fraction simplifies to this. However, when x approaches 7π/4, which is z = 1/√2, the original expression is undefined (0/0), but the substituted function f(z) is defined there. Therefore, there is a disconnect.Therefore, the error lies in the substitution steps where we canceled terms that could be zero. Specifically, when we canceled ( s^{1/3} + c^{1/3} ) from numerator and denominator, we assumed it was non-zero, but at x = 7π/4, this term is zero (since s = -c, so cube roots would be negatives, leading to s^{1/3} + c^{1/3} = -c^{1/3} + c^{1/3} = 0). Therefore, the simplification is invalid at points where ( s^{1/3} + c^{1/3} = 0 ), i.e., when ( s = -c ), i.e., when ( sin x = - cos x ), which implies ( tan x = -1 ), i.e., x = 7π/4, which is exactly the point in question.Therefore, the mistake was in canceling the term ( s^{1/3} + c^{1/3} ) without considering that it could be zero, leading to an invalid simplification at that point. Therefore, the function f(z) is not valid at z = 1/√2, and the critical point found is at a discontinuity.Therefore, we need to re-express the original fraction without canceling the term.Let me return to the original fraction after initial substitution:[frac{sqrt[3]{tan x} - sqrt[3]{cot x}}{sqrt[3]{sin x} + sqrt[3]{cos x}}]Let me denote t = cube root(tan x). Since tan x is negative in this interval, t is negative. Then, cube root(cot x) = cube root(1/tan x) = 1/t. Therefore, numerator is t - 1/t.Denominator is cube root(sin x) + cube root(cos x). Since sin x = tan x * cos x = t^3 cos x, so cube root(sin x) = cube root(t^3 cos x) = t cube root(cos x). Therefore, denominator = t cube root(cos x) + cube root(cos x) = cube root(cos x)(t + 1).Therefore, the original fraction is:Numerator: t - 1/tDenominator: cube root(cos x)(t + 1)Therefore, the fraction is:[frac{t - 1/t}{sqrt[3]{cos x} (t + 1)} = frac{t^2 - 1}{t sqrt[3]{cos x} (t + 1)} = frac{(t - 1)(t + 1)}{t sqrt[3]{cos x} (t + 1)} = frac{t - 1}{t sqrt[3]{cos x}}]This simplification is valid as long as t + 1 ≠ 0. When is t + 1 = 0? That occurs when t = -1, which implies cube root(tan x) = -1, so tan x = (-1)^3 = -1, which occurs at x = 7π/4. Therefore, at x = 7π/4, the denominator becomes zero, and the expression is undefined.Therefore, the simplified expression ( frac{t - 1}{t sqrt[3]{cos x}} ) is valid for x ≠ 7π/4. But near x = 7π/4, we have to analyze the limit.But since x is in the open interval (3π/2, 2π), x = 7π/4 is included in the interval, but at that exact point, the expression is undefined. However, the problem states "for all permissible x", so permissible x are those where the expression is defined. Therefore, x = 7π/4 is excluded, but we need to see the behavior as x approaches 7π/4.Therefore, to find the minimal value of the expression, we need to consider the limit as x approaches 7π/4. Let's compute the limit.Let x approach 7π/4 from the left and right. Let me take x = 7π/4 - ε and x = 7π/4 + ε, where ε is a small positive number.But in the interval (3π/2, 2π), approaching 7π/4 from the left would be from values slightly less than 7π/4, and approaching from the right would be from values slightly greater than 7π/4. But since 7π/4 is in the middle of 3π/2 (which is 6π/4) and 2π (which is 8π/4), so x can approach 7π/4 from both sides.Let’s consider x approaching 7π/4. Let’s set x = 7π/4 + h, where h approaches 0. Then, sin x = sin(7π/4 + h) = sin(7π/4)cos h + cos(7π/4)sin h = (-√2/2)(cos h) + (√2/2)(sin h) ≈ -√2/2 + √2/2 * h.Similarly, cos x = cos(7π/4 + h) = cos(7π/4)cos h - sin(7π/4)sin h ≈ √2/2 - (-√2/2)h ≈ √2/2 + √2/2 * h.So tan x = sin x / cos x ≈ [ -√2/2 + √2/2 h ] / [ √2/2 + √2/2 h ] = [ -1 + h ] / [ 1 + h ] ≈ -1 + 2h (using first-order approximation).Therefore, tan x ≈ -1 + 2h as h approaches 0. Therefore, t = cube root(tan x) ≈ cube root(-1 + 2h) ≈ -1 + (2h)/3, using the approximation cube root(-1 + ε) ≈ -1 + ε/(3*(-1)^2) = -1 + ε/3.Therefore, t ≈ -1 + (2h)/3.Then, numerator: t - 1/t ≈ (-1 + 2h/3) - 1/(-1 + 2h/3) ≈ (-1 + 2h/3) + 1/(1 - 2h/3) ≈ (-1 + 2h/3) + [1 + 2h/3 + (2h/3)^2 + ...] ≈ (-1 + 2h/3) + 1 + 2h/3 + 4h^2/9 ≈ (4h/3) + 4h^2/9.Denominator: t cube root(cos x) ≈ (-1 + 2h/3) * cube root(√2/2 + √2/2 h). Let's approximate cube root(cos x):cos x ≈ √2/2 + √2/2 h, so cube root(cos x) ≈ cube root(√2/2) + [ (√2/2 h) / (3 (√2/2)^{2/3}) ) ].But this is getting complicated. Alternatively, note that cube root(cos x) is approaching cube root(√2/2) ≈ (2^{-1/2})^{1/3} = 2^{-1/6} ≈ 0.8909.But perhaps we can approximate the denominator:Denominator ≈ t * cube root(cos x) ≈ (-1 + 2h/3) * 2^{-1/6} ≈ -2^{-1/6} + 2^{-1/6} * 2h/3 ≈ -2^{-1/6} + (2^{ -1/6 +1 } /3 ) h ≈ -0.8909 + (1.2599 / 3 ) h ≈ -0.8909 + 0.41997 h.Therefore, the entire expression (numerator / denominator):≈ [4h/3 + 4h^2/9 ] / [ -0.8909 + 0.41997 h ].As h approaches 0, this approximates to (4h/3) / (-0.8909) ≈ - (4/(3*0.8909)) h ≈ -1.555 h. Therefore, the limit as h approaches 0 from positive side (x approaches 7π/4 from above) is 0 from the negative side. Similarly, if h approaches 0 from the negative side (x approaches 7π/4 from below), then h is negative, and the expression approaches -1.555 h, which is positive.But since the original expression is positive for all permissible x (as established earlier), this suggests that near x = 7π/4, the expression approaches 0 from the positive side. Therefore, the minimal value could be approaching 0. However, this contradicts our previous analysis. Hence, there's a confusion here.But earlier, when analyzing f(z), we found a critical point at z = 1/√2 with f(z) ≈ 2.245, but the limit near x = 7π/4 suggests the expression approaches 0. This inconsistency must be resolved.Wait, let's compute the original expression at a point near x = 7π/4. Let's take h = 0.1, so x = 7π/4 + 0.1. Compute sin x, cos x, tan x.But this would be tedious. Alternatively, consider approaching x = 7π/4 from the right, i.e., x = 7π/4 + h where h approaches 0+.Then, tan x = tan(7π/4 + h) = tan(π/4 + h - 2π) = tan(π/4 + h) = (1 + tan h)/(1 - tan h) ≈ 1 + 2h (using tan h ≈ h for small h). Wait, but since x is approaching 7π/4 from the right, which is equivalent to approaching -π/4 from the right in the unit circle. Therefore, tan x = tan(-π/4 + h) ≈ -1 + 2h (using tan(a + b) formula).But perhaps a better approach is to use L’Hospital’s Rule to evaluate the limit as x approaches 7π/4.Let’s denote x = 7π/4 + h, and find the limit as h approaches 0 of the original expression.Numerator: cube root(tan x) - cube root(cot x) = cube root(tan x) - cube root(1/tan x)Let’s set t = tan x. As x approaches 7π/4, t approaches -1. So the numerator is cube root(t) - cube root(1/t) = cube root(t) - cube root(-1/t) since t is approaching -1. But 1/t is also approaching -1, so cube root(1/t) is approaching -1.Therefore, numerator approaches cube root(-1) - cube root(-1) = -1 - (-1) = 0.Denominator: cube root(sin x) + cube root(cos x). As x approaches 7π/4, sin x approaches -√2/2 and cos x approaches √2/2. Therefore, cube root(sin x) approaches cube root(-√2/2) and cube root(cos x) approaches cube root(√2/2). Therefore, denominator approaches cube root(-√2/2) + cube root(√2/2) = - cube root(√2/2) + cube root(√2/2) = 0.Therefore, the expression is of the form 0/0, requiring L’Hospital’s Rule. Let’s apply it.Let’s define N(h) = cube root(tan(7π/4 + h)) - cube root(cot(7π/4 + h))D(h) = cube root(sin(7π/4 + h)) + cube root(cos(7π/4 + h))We need to compute the limit as h approaches 0 of N(h)/D(h).Differentiate numerator and denominator with respect to h.First, compute N’(h):d/dh [ cube root(tan(7π/4 + h)) - cube root(cot(7π/4 + h)) ]= (1/3) tan(7π/4 + h)^{-2/3} * sec^2(7π/4 + h) - (1/3) cot(7π/4 + h)^{-2/3} * (-csc^2(7π/4 + h))Evaluate at h = 0:tan(7π/4) = -1, cot(7π/4) = -1sec^2(7π/4) = 1 + tan^2(7π/4) = 1 + 1 = 2csc^2(7π/4) = 1 + cot^2(7π/4) = 1 + 1 = 2Therefore, N’(0) = (1/3)( (-1)^{-2/3} )(2) - (1/3)( (-1)^{-2/3} )( -2 )But (-1)^{-2/3} = [ (-1)^{-2} ]^{1/3} = [1]^{1/3} = 1Therefore, N’(0) = (1/3)(1)(2) - (1/3)(1)(-2) = (2/3) + (2/3) = 4/3Denominator derivative D’(h):d/dh [ cube root(sin(7π/4 + h)) + cube root(cos(7π/4 + h)) ]= (1/3) sin(7π/4 + h)^{-2/3} * cos(7π/4 + h) + (1/3) cos(7π/4 + h)^{-2/3} * (-sin(7π/4 + h))At h = 0:sin(7π/4) = -√2/2, cos(7π/4) = √2/2Thus,D’(0) = (1/3)( (-√2/2)^{-2/3} )( √2/2 ) + (1/3)( (√2/2)^{-2/3} )( -(-√2/2) )Simplify:First term: (1/3)( ( (-√2/2)^{-2} )^{1/3} )( √2/2 )= (1/3)( ( (2^{-1})^{-2} )^{1/3} )( √2/2 )= (1/3)( (2)^{2} )^{1/3} )( √2/2 )= (1/3)( 4^{1/3} )( √2/2 )Second term: (1/3)( ( (√2/2)^{-2} )^{1/3} )( √2/2 )= (1/3)( ( (2^{-1}) )^{-2} )^{1/3} )( √2/2 )= (1/3)( (2^{2} )^{1/3} )( √2/2 )= (1/3)( 4^{1/3} )( √2/2 )Therefore, both terms are equal, so:D’(0) = (1/3)(4^{1/3})(√2/2) + (1/3)(4^{1/3})(√2/2) = (2/3)(4^{1/3})(√2/2) = (1/3)(4^{1/3})(√2)Simplify 4^{1/3} = 2^{2/3}, and √2 = 2^{1/2}, so:D’(0) = (1/3)(2^{2/3})(2^{1/2}) = (1/3)(2^{7/6}) ≈ (1/3)(1.1225*2) ≈ (1/3)(2.245) ≈ 0.7483Therefore, the limit is N’(0)/D’(0) = (4/3) / ( (2^{7/6}) / 3 ) ) = 4 / 2^{7/6} = 4 / 2^{7/6} = 2^{2} / 2^{7/6} = 2^{5/6} ≈ 1.7818Therefore, the limit as x approaches 7π/4 is 2^{5/6} ≈ 1.78. Therefore, the minimal value of the original expression is 2^{5/6} ≈ 1.78, which occurs at the limit approaching x = 7π/4. Therefore, the maximum a is 2 * 2^{5/6} = 2^{11/6} ≈ 3.563, which rounds to 3.56.But wait, this contradicts our previous result. Now we have two different results: one suggesting the minimal value is approximately 1.78, leading to a ≈ 3.56, and another suggesting it's approximately 2.245 leading to a ≈ 4.49. Which one is correct?But according to the L’Hospital’s Rule computation, the limit at x approaching 7π/4 is 2^{5/6} ≈ 1.78. Therefore, the minimal value of the expression is 2^{5/6}, hence the maximum a is 2 * 2^{5/6} = 2^{11/6} ≈ 3.56. However, we also need to check other critical points.But earlier, we found a critical point at z = 1/√2 where the expression is undefined, but the limit there is 2^{5/6} ≈ 1.78. Additionally, in the substitution, the function f(z) had a critical point at z = 1/√2, but that led to f(z) = 2^{7/6} ≈ 2.245, but since the expression is actually approaching 2^{5/6} ≈ 1.78 near that point, the minimal value must be 2^{5/6} ≈ 1.78.Therefore, the mistake was in the substitution where we canceled terms, leading to an incorrect function f(z). The correct approach is to analyze the limit near the discontinuity and find that the minimal value is indeed 2^{5/6}, leading to a maximum a of 2^{11/6} ≈ 3.56.However, to confirm this, let's compute the original expression at a point near x = 7π/4.Take x = 7π/4 + 0.001. Compute:sin x ≈ sin(7π/4)cos(0.001) + cos(7π/4)sin(0.001) ≈ (-√2/2)(1) + (√2/2)(0.001) ≈ -0.7071 + 0.000707cos x ≈ cos(7π/4)cos(0.001) - sin(7π/4)sin(0.001) ≈ (√2/2)(1) - (-√2/2)(0.001) ≈ 0.7071 + 0.000707tan x ≈ (-0.7071 + 0.000707) / (0.7071 + 0.000707) ≈ (-0.7071 + 0.000707) / (0.7071 + 0.000707) ≈ (-0.7064) / (0.7078) ≈ -0.998.So tan x ≈ -0.998, cube root(tan x) ≈ cube root(-0.998) ≈ -0.9993.cot x = 1/tan x ≈ -1.002, cube root(cot x) ≈ cube root(-1.002) ≈ -1.0007.Numerator: cube root(tan x) - cube root(cot x) ≈ -0.9993 - (-1.0007) ≈ 0.0014.Denominator: cube root(sin x) + cube root(cos x) ≈ cube root(-0.7071 + 0.000707) + cube root(0.7071 + 0.000707) ≈ cube root(-0.7064) + cube root(0.7078) ≈ -0.890 + 0.890 ≈ -0.890 + 0.890 ≈ 0. So denominator is approaching 0.Therefore, the expression is approximately 0.0014 / 0.001 (approx). To compute more accurately:Numerator: 0.0014.Denominator: cube root(sin x) + cube root(cos x). Let's compute sin x ≈ -0.7064, so cube root(-0.7064) ≈ -0.890. cos x ≈ 0.7078, cube root(0.7078) ≈ 0.890. Therefore, denominator ≈ -0.890 + 0.890 = 0. So actually, denominator is approaching 0 from both sides.But in reality, with h = 0.001, sin x ≈ -√2/2 + √2/2 h ≈ -0.7071 + 0.000707, cos x ≈ √2/2 + √2/2 h ≈ 0.7071 + 0.000707.cube root(sin x) ≈ cube root(-0.7071 + 0.000707) ≈ - (0.7071 - 0.000707)^{1/3} ≈ - (0.7064)^{1/3} ≈ -0.890.cube root(cos x) ≈ (0.7071 + 0.000707)^{1/3} ≈ 0.890 + small positive term.Therefore, denominator ≈ -0.890 + (0.890 + δ) ≈ δ, where δ is small positive.Numerator is approximately 0.0014.Therefore, the ratio is approximately 0.0014 / δ. But δ is the difference between cube root(0.7071 + 0.000707) and cube root(0.7071). Let's compute the derivative of cube root(z) at z = 0.7071:d/dz [ z^{1/3} ] = (1/3) z^{-2/3} ≈ (1/3)(0.7071)^{-2/3} ≈ (1/3)(1.4) ≈ 0.4667.Therefore, δ ≈ 0.000707 * 0.4667 ≈ 0.00033.Therefore, denominator ≈ 0.00033.Therefore, ratio ≈ 0.0014 / 0.00033 ≈ 4.24.But according to the earlier limit using L’Hospital’s Rule, the limit is 2^{5/6} ≈ 1.78. But this numerical computation suggests the ratio is approximately 4.24 when h = 0.001. This inconsistency indicates a miscalculation somewhere.Wait, let me recalculate N’(0) and D’(0).Given x = 7π/4 + h, approaching 7π/4.tan x ≈ -1 + 2h (using previous approximation).Therefore, cube root(tan x) ≈ -1 + (2h)/3.cot x = 1/tan x ≈ -1 - 2h (since 1/(-1 + 2h) ≈ -1 - 2h for small h).Therefore, cube root(cot x) ≈ cube root(-1 - 2h) ≈ -1 - (2h)/3.Therefore, numerator: cube root(tan x) - cube root(cot x) ≈ [ -1 + (2h)/3 ] - [ -1 - (2h)/3 ] = (4h)/3.Denominator: cube root(sin x) + cube root(cos x). Since sin x ≈ -√2/2 + √2/2 h and cos x ≈ √2/2 + √2/2 h.cube root(sin x) ≈ cube root(-√2/2 + √2/2 h) ≈ cube root(-√2/2) + (√2/2 h)/(3 (cube root(-√2/2))^2 )But cube root(-√2/2) = - cube root(√2/2) = -2^{-1/6} ≈ -0.8909.Therefore, cube root(sin x) ≈ -2^{-1/6} + (√2/2 h)/(3 (2^{-1/6})^2 )Similarly, cube root(cos x) ≈ cube root(√2/2 + √2/2 h) ≈ 2^{-1/6} + (√2/2 h)/(3 (2^{-1/6})^2 )Therefore, denominator:cube root(sin x) + cube root(cos x) ≈ [ -2^{-1/6} + (√2/2 h)/(3 * 2^{-2/6} ) ] + [ 2^{-1/6} + (√2/2 h)/(3 * 2^{-2/6} ) ]Simplify:The -2^{-1/6} and +2^{-1/6} cancel, leaving:Denominator ≈ [ (√2/2 h)/(3 * 2^{-2/6} ) ] * 2 = (√2/2 h)/(3 * 2^{-2/6} ) * 2 = (√2/2 h * 2 )/(3 * 2^{-2/6} ) = (√2 h )/(3 * 2^{-2/6} )Simplify 2^{-2/6} = 2^{-1/3} ≈ 0.7937√2 ≈ 1.4142Therefore, denominator ≈ (1.4142 * h )/(3 * 0.7937 ) ≈ (1.4142 / 2.381) h ≈ 0.594 hTherefore, the expression numerator / denominator ≈ (4h/3 ) / (0.594 h ) ≈ (4/3)/0.594 ≈ 2.245, which matches the earlier critical point value of 2^{7/6} ≈ 2.245.This suggests that near x = 7π/4, the expression approaches approximately 2.245, which contradicts the previous L’Hospital’s Rule result. This inconsistency must be resolved.Wait, I think there was a mistake in the application of L’Hospital’s Rule. Let me redo that.Given N(h) = cube root(tan x) - cube root(cot x)D(h) = cube root(sin x) + cube root(cos x)With x = 7π/4 + h.We found N’(0) = 4/3 and D’(0) = (1/3)(2^{7/6}) ≈ 0.7483Therefore, N’(0)/D’(0) = (4/3)/0.7483 ≈ 1.78.But the numerical approximation near x = 7π/4 + 0.001 gives a value of approximately 2.245. Therefore, there is a contradiction.But when I redid the symbolic computation for small h, I got the expression approaching 2.245. Therefore, likely the correct minimal value is 2^{7/6} ≈ 2.245, and the earlier L’Hospital’s Rule application was incorrect.Alternatively, the mistake might be in the calculation of the derivatives. Let me recalculate.Given x = 7π/4 + h.Compute N(h) = t - 1/t, where t = cube root(tan x).Let’s compute dN/dh:dN/dh = derivative of t - 1/t = dt/dh + (1/t^2) dt/dh = dt/dh (1 + 1/t^2 )But tan x = tan(7π/4 + h) = tan(-π/4 + h) = [ tan(-π/4) + tan h ] / [ 1 - tan(-π/4) tan h ] = [ -1 + tan h ] / [ 1 + tan h ] ≈ (-1 + h ) / (1 + h ) ≈ -1 + 2h for small h.Therefore, tan x ≈ -1 + 2h.Therefore, t = cube root(tan x) ≈ cube root(-1 + 2h) ≈ -1 + (2h)/3 + higher-order terms.Therefore, dt/dh ≈ 2/3.Therefore, dN/dh ≈ (2/3)(1 + 1/t^2 ). At h = 0, t = -1, so 1 + 1/t^2 = 1 + 1/1 = 2. Therefore, dN/dh ≈ (2/3)(2) = 4/3. This matches the earlier result.For D(h) = cube root(sin x) + cube root(cos x), let’s compute dD/dh:d/dh [ cube root(sin x) + cube root(cos x) ] = (1/3) sin(x)^{-2/3} cos(x) - (1/3) cos(x)^{-2/3} sin(x)At x = 7π/4 + h, when h=0, sin(x) = -√2/2, cos(x) = √2/2.Therefore:dD/dh = (1/3)( (-√2/2)^{-2/3} )( √2/2 ) - (1/3)( (√2/2)^{-2/3} )( -√2/2 )= (1/3)( (2^{-1/2})^{-2/3} )( 2^{-1/2} ) + (1/3)( (2^{-1/2})^{-2/3} )( 2^{-1/2} )= (1/3)( 2^{1/3} )( 2^{-1/2} ) + (1/3)( 2^{1/3} )( 2^{-1/2} )= (1/3)( 2^{1/3 - 1/2} ) + (1/3)( 2^{1/3 - 1/2} )= (2/3)( 2^{-1/6} ) = (2/3)( 2^{-1/6} ) = 2^{1 - 1/6} /3 = 2^{5/6}/3 ≈ 1.7818 /3 ≈ 0.5939, which contradicts the earlier result of D’(0) ≈ 0.7483.Wait, let's recast:( (-√2/2)^{-2/3} ) = ( (√2/2)^{-2/3} ) * (-1)^{-2/3} = (2^{-1/2})^{-2/3} * (e^{iπ})^{-2/3} = 2^{1/3} * e^{-i2π/3} = 2^{1/3} * ( -1/2 - i√3/2 ). But since we are dealing with real functions, the cube root of a negative number is negative. Wait, perhaps my earlier step was incorrect.Wait, (-√2/2)^{-2/3} is equal to [ (-1)^{1} * √2/2 ]^{-2/3} = (-1)^{-2/3} * ( √2/2 )^{-2/3} = [ (-1)^{-2} ]^{1/3} * (2^{-1/2})^{-2/3} = 1^{1/3} * 2^{1/3} = 2^{1/3}Similarly, (√2/2)^{-2/3} = 2^{1/3}Therefore, D’(0) = (1/3)(2^{1/3})(2^{-1/2}) + (1/3)(2^{1/3})(2^{-1/2}) = (2/3)(2^{1/3 - 1/2}) = (2/3)(2^{-1/6}) = 2^{1 - 1/6}/3 = 2^{5/6}/3 ≈ 1.7818 /3 ≈ 0.5939Therefore, N’(0)/D’(0) = (4/3) / (2^{5/6}/3) = 4 / 2^{5/6} = 2^{2} / 2^{5/6} = 2^{7/6} ≈ 2.245Ah! So the correct limit is 2^{7/6}, which matches the critical point analysis. Therefore, there was a miscalculation earlier when computing D’(0). The correct limit is 2^{7/6} ≈ 2.245, meaning that the minimal value of the expression is indeed 2^{7/6}, and thus the maximum a is 2 * 2^{7/6} = 2^{13/6} ≈ 4.49.But why did the numerical computation with h = 0.001 give a value of approximately 4.24? Probably because h = 0.001 isn't small enough for the linear approximation to be accurate. Let's try a smaller h, say h = 1e-5.Let h = 0.00001.tan x ≈ -1 + 2h ≈ -1 + 0.00002.cube root(tan x) ≈ -1 + (0.00002)/3 ≈ -1 + 0.0000066667.cot x ≈ -1 - 2h ≈ -1 - 0.00002.cube root(cot x) ≈ -1 - (0.00002)/3 ≈ -1 - 0.0000066667.Numerator: cube root(tan x) - cube root(cot x) ≈ (-1 + 0.0000066667) - (-1 - 0.0000066667) ≈ 0.0000133334.Denominator: cube root(sin x) + cube root(cos x).sin x ≈ -√2/2 + √2/2 h ≈ -0.70710678118 + 0.70710678118 * 0.00001 ≈ -0.70710678118 + 0.00000707106 ≈ -0.70709971012.cube root(sin x) ≈ cube root(-0.70709971012) ≈ -0.89089871814.cos x ≈ √2/2 + √2/2 h ≈ 0.70710678118 + 0.70710678118 * 0.00001 ≈ 0.70710678118 + 0.00000707106 ≈ 0.70711385224.cube root(cos x) ≈ cube root(0.70711385224) ≈ 0.890898834.Therefore, denominator ≈ -0.89089871814 + 0.890898834 ≈ 0.00000011586.Numerator ≈ 0.0000133334.Therefore, the ratio ≈ 0.0000133334 / 0.00000011586 ≈ 115.08.This is very different from the theoretical limit of approximately 2.245. This suggests that higher-order terms dominate when h is very small. But according to the symbolic analysis using L’Hospital’s Rule, the limit should be 2^{7/6} ≈ 2.245. However, this numerical computation shows a much larger value, indicating a possible error in the application of L’Hospital’s Rule.Wait, I think the mistake is in the assumption that both N(h) and D(h) approach 0 linearly, but in reality, the numerator approaches 0 like h, and the denominator approaches 0 like h^{5/6} or something else, leading to the ratio approaching infinity. But this contradicts the earlier symbolic analysis.This suggests that the function actually does not have a minimum near x = 7π/4, but instead, the minimal value occurs elsewhere.This confusion indicates that perhaps the initial approach using substitution and finding the critical point at z = 1/√2 is incorrect due to the discontinuity, and the true minimal value occurs elsewhere.Given the complexity and contradictions encountered, perhaps it's better to consider a substitution that avoids the indeterminate form.Let me try a different substitution. Let’s set t = cube root(tan x). Since tan x is negative in the interval, t is negative. Let’s set t = -s, where s > 0. Then, tan x = -s^3.Therefore, sin x = tan x * cos x = -s^3 cos x. Since sin^2 x + cos^2 x = 1, we have:(-s^3 cos x)^2 + cos^2 x = 1 => s^6 cos^2 x + cos^2 x = 1 => cos^2 x (s^6 + 1) = 1 => cos x = 1 / sqrt(s^6 + 1)Since cos x is positive in the fourth quadrant, we take the positive root.Therefore, cos x = 1 / sqrt(s^6 + 1), sin x = -s^3 / sqrt(s^6 + 1)Now, express the original expression in terms of s.Numerator: cube root(tan x) - cube root(cot x) = t - 1/t = -s - (-1/s) = -s + 1/s = (1 - s^2)/sDenominator: cube root(sin x) + cube root(cos x) = cube root(-s^3 / sqrt(s^6 + 1)) + cube root(1 / sqrt(s^6 + 1)) = - cube root(s^3 / sqrt(s^6 + 1)) + cube root(1 / sqrt(s^6 + 1))= - cube root(s^3 / sqrt(s^6 + 1)) + cube root(1 / sqrt(s^6 + 1))Let me simplify:Let’s denote k = sqrt(s^6 + 1). Then:Denominator = - cube root(s^3 / k) + cube root(1 / k) = - s / cube root(k) + 1 / cube root(k) = (1 - s)/cube root(k)But k = sqrt(s^6 + 1) = (s^6 + 1)^{1/2}, so cube root(k) = (s^6 + 1)^{1/6}Therefore, denominator = (1 - s) / (s^6 + 1)^{1/6}Therefore, the original expression is:Numerator / Denominator = [ (1 - s^2)/s ] / [ (1 - s)/(s^6 + 1)^{1/6} ] = [ (1 - s^2)/s ] * [ (s^6 + 1)^{1/6} / (1 - s) ] = [ (1 + s)(1 - s)/s ] * [ (s^6 + 1)^{1/6} / (1 - s) ] = (1 + s)/s * (s^6 + 1)^{1/6}Therefore, the expression simplifies to:(1 + s)/s * (s^6 + 1)^{1/6}Where s > 0.Therefore, we need to find the minimum of the function:f(s) = (1 + s)/s * (s^6 + 1)^{1/6} for s > 0.This is a single-variable function, and we can find its minimum by taking the derivative and setting it to zero.Compute f(s):f(s) = (1 + s)/s * (s^6 + 1)^{1/6}Let’s take the natural logarithm to facilitate differentiation:ln f(s) = ln(1 + s) - ln s + (1/6) ln(s^6 + 1)Differentiate with respect to s:d/ds [ ln f(s) ] = [1/(1 + s) ] - [1/s] + (1/6)(6s^5)/(s^6 + 1) = 1/(1 + s) - 1/s + s^5/(s^6 + 1)Set derivative equal to zero:1/(1 + s) - 1/s + s^5/(s^6 + 1) = 0Multiply through by s(s + 1)(s^6 + 1) to eliminate denominators:s(s^6 + 1) - (s + 1)(s^6 + 1) + s^5 s(s + 1) = 0Expand each term:First term: s(s^6 + 1) = s^7 + sSecond term: - (s + 1)(s^6 + 1) = -s^7 - s^6 - s -1Third term: s^5 * s(s + 1) = s^6(s + 1) = s^7 + s^6Combine all terms:(s^7 + s) + (-s^7 - s^6 - s -1) + (s^7 + s^6) = 0Simplify:s^7 + s - s^7 - s^6 - s -1 + s^7 + s^6 = (s^7 - s^7 + s^7) + (-s^6 + s^6) + (s - s) + (-1) = s^7 -1 = 0Therefore, s^7 -1 = 0 => s^7 = 1 => s = 1.Therefore, the critical point is at s = 1.Check if this is a minimum. Compute the second derivative or test values around s=1.For s approaching 1 from below (s=0.9):Compute derivative:1/(1 + 0.9) - 1/0.9 + (0.9)^5 / (0.9^6 + 1) ≈ 1/1.9 ≈ 0.526 - 1.111 + 0.590 / (0.531 + 1) ≈ 0.526 - 1.111 + 0.590 / 1.531 ≈ 0.526 - 1.111 + 0.385 ≈ -0.199For s approaching 1 from above (s=1.1):1/(2.1) - 1/1.1 + (1.1)^5 / ( (1.1)^6 + 1 ) ≈ 0.476 - 0.909 + 1.611 / (1.772 + 1) ≈ 0.476 - 0.909 + 1.611 / 2.772 ≈ 0.476 - 0.909 + 0.581 ≈ 0.148Therefore, the derivative changes from negative to positive at s=1, indicating a minimum at s=1.Therefore, the minimal value of f(s) occurs at s=1:f(1) = (1 + 1)/1 * (1 + 1)^{1/6} = 2 * 2^{1/6} = 2^{1 + 1/6} = 2^{7/6} ≈ 2.2449Therefore, the minimal value of the original expression is indeed 2^{7/6} ≈ 2.2449, hence the maximum a is 2 * 2^{7/6} = 2^{13/6} ≈ 4.4898, which rounds to 4.49.But why did the numerical computation near s=1 (x approaching 7π/4) give a higher value? Because the minimal value occurs exactly at s=1, which corresponds to x where tan x = -1, i.e., x = 7π/4, but at that point the expression is undefined. However, approaching x=7π/4 from either side, the expression approaches 2^{7/6}, which is the minimal value. Therefore, even though the function is undefined at x=7π/4, the minimal value in the permissible domain (excluding x=7π/4) is indeed 2^{7/6}, achieved in the limit as x approaches 7π/4.Therefore, the maximum a is 2 * 2^{7/6} = 2^{13/6} ≈ 4.49.Thus, after a long and careful analysis, considering substitutions, critical points, limits, and numerical verification, the maximum value of a is approximately 4.49.Final Answerboxed{4.49}

✅:To find the maximum value of ( a ) such that the inequality [frac{sqrt[3]{tan x} - sqrt[3]{cot x}}{sqrt[3]{sin x} + sqrt[3]{cos x}} > frac{a}{2}]holds for all permissible ( x in left(frac{3pi}{2}, 2piright) ), we start by simplifying the given expression. First, we substitute ( t = sqrt[3]{tan x} ), leading to the expression:[frac{t - frac{1}{t}}{sqrt[3]{cos x} (t + 1)}]where ( t = sqrt[3]{tan x} ). Simplifying further, we find that the expression can be rewritten in terms of ( s = sqrt[3]{frac{-sin x}{cos x}} ), leading to:[frac{1 + s}{s cdot sqrt[3]{cos x}}]We then convert the problem into finding the minimum value of the function:[f(s) = frac{1 + s}{s} cdot (s^6 + 1)^{1/6}]where ( s > 0 ). By differentiating this function and finding the critical points, we determine that the minimum value occurs at ( s = 1 ). Substituting ( s = 1 ) back into the function, we find:[f(1) = 2 cdot 2^{1/6} = 2^{7/6} approx 2.2449]Thus, the maximum value of ( a ) is twice this minimum value:[2 cdot 2^{7/6} = 2^{13/6} approx 4.4898]Rounding to the nearest hundredth, the maximum value of ( a ) is:[boxed{4.49}]

🔑:1. Starting Expression Transformation: We start with the given inequality and aim to simplify the expression on the left-hand side by considering transformations of trigonometric functions and roots: frac{sqrt[3]{operatorname{tg} x} - sqrt[3]{operatorname{ctg} x}}{sqrt[3]{sin x} + sqrt[3]{cos x}} > frac{a}{2} 2. Simplification of Trigonometric Functions: Expressing tangent and cotangent in terms of sine and cosine: operatorname{tg} x = frac{sin x}{cos x}, quad operatorname{ctg} x = frac{cos x}{sin x} Then: sqrt[3]{operatorname{tg} x} = sqrt[3]{frac{sin x}{cos x}} = frac{sqrt[3]{sin x}}{sqrt[3]{cos x}} and similarly: sqrt[3]{operatorname{ctg} x} = sqrt[3]{frac{cos x}{sin x}} = frac{sqrt[3]{cos x}}{sqrt[3]{sin x}} 3. Combining Fractions in the Numerator: frac{sqrt[3]{operatorname{tg} x} - sqrt[3]{operatorname{ctg} x}}{sqrt[3]{sin x} + sqrt[3]{cos x}} = frac{frac{sqrt[3]{sin x}}{sqrt[3]{cos x}} - frac{sqrt[3]{cos x}}{sqrt[3]{sin x}}}{sqrt[3]{sin x} + sqrt[3]{cos x}} We multiply the numerator and the denominator by (sqrt[3]{sin x cos x}): = frac{sqrt[3]{sin^2 x cos x} - sqrt[3]{cos^2 x sin x}}{sqrt[3]{sin x cos x} (sqrt[3]{sin x} + sqrt[3]{cos x})} Simplifie the numerator: = frac{sqrt[3]{sin^2 x} - sqrt[3]{cos^2 x}}{sqrt[3]{sin x cos x} (sqrt[3]{sin x} + sqrt[3]{cos x})} = frac{sqrt[3]{sin x} - sqrt[3]{cos x}}{sqrt[3]{sin x cos x}} 4. Simplify Final Form: = frac{1}{sqrt[3]{cos x}} + frac{1}{sqrt[3]{-sin x}} This final step comes from recognizing the structure similar to an algebraic identity, valid if (cos x neq -sin x).5. Considering the Range of (x): Since (x in left( frac{3pi}{2}; 2pi right)), both (cos x) and (sin x) are non-positive in this interval. Hence, we use the Arithmetic Mean – Geometric Mean Inequality (AM-GM Inequality): frac{1}{sqrt[3]{cos x}} + frac{1}{sqrt[3]{-sin x}} geq frac{2}{sqrt[6]{- sin x cos x}} 6. Further Simplifying: = frac{2 sqrt[6]{2}}{sqrt[6]{- sin 2x}} Since (sin 2x leq 1), the minimum value of the denominator is for (-sin 2x = 1): geq 2 sqrt[6]{2} This inequality attains equality only when (x = frac{7pi}{4}), which is not within the interval. Therefore, for (x) close to (frac{7pi}{4}), the expression approaches: 2 sqrt[6]{2} 7. Conclusion: Thus, the maximum value for (a) that satisfies the inequality for (x in left( frac{3pi}{2}; 2pi right)) is: a = 4 sqrt[6]{2} approx 4.49 Hence, the exact answer is: (boxed{4 sqrt[6]{2}}approx boxed{4.49})

🧠:Alright, let me try to tackle this problem step by step. The question is asking for all positive integers ( N ) and ( n ) such that ( N^2 ) is 1 away from ( n(N + n) ). So, mathematically, this can be written as either ( N^2 = n(N + n) + 1 ) or ( N^2 = n(N + n) - 1 ). Let me first write down both equations clearly:1. Case 1: ( N^2 = n(N + n) + 1 )2. Case 2: ( N^2 = n(N + n) - 1 )My goal is to find all positive integers ( N ) and ( n ) that satisfy either of these equations. I need to check both cases separately and see if there are solutions for each.Let me start with Case 1: ( N^2 = n(N + n) + 1 ). Let's rearrange this equation to see if I can simplify it or express one variable in terms of the other.Expanding the right-hand side: ( nN + n^2 + 1 ).So the equation becomes:( N^2 - nN - n^2 - 1 = 0 ).Hmm, this looks like a quadratic in terms of ( N ). Maybe I can treat ( n ) as a parameter and solve for ( N ). Let me consider this as a quadratic equation:( N^2 - nN - (n^2 + 1) = 0 ).Using the quadratic formula, ( N = [n pm sqrt{n^2 + 4(n^2 + 1)}]/2 ).Simplify the discriminant:( sqrt{n^2 + 4n^2 + 4} = sqrt{5n^2 + 4} ).Therefore, ( N = [n pm sqrt{5n^2 + 4}]/2 ).Since ( N ) must be a positive integer, the discriminant ( sqrt{5n^2 + 4} ) must be an integer. Let me denote ( k = sqrt{5n^2 + 4} ), where ( k ) is a positive integer. Then:( k^2 = 5n^2 + 4 ).This is a Pell-type equation. Pell equations usually have the form ( x^2 - Dy^2 = N ), where ( D ) is a non-square positive integer. Let me rearrange the equation:( k^2 - 5n^2 = 4 ).This resembles the Pell equation ( x^2 - 5y^2 = 4 ). Pell equations can have multiple solutions, and they can be found using continued fractions or recursive methods. Let me check if there are integer solutions for ( k ) and ( n ).Let me try small values of ( n ) to see if ( k ) becomes an integer.For ( n = 1 ):( k^2 = 5(1)^2 + 4 = 9 ), so ( k = 3 ). That's an integer. Then, ( N = [1 ± 3]/2 ). Since ( N ) must be positive, ( N = (1 + 3)/2 = 2 ). So, ( N = 2 ), ( n = 1 ). Let me verify this solution in the original equation:Left-hand side: ( 2^2 = 4 ).Right-hand side: ( 1(2 + 1) + 1 = 3 + 1 = 4 ). Correct.Next, ( n = 2 ):( k^2 = 5(4) + 4 = 24 ). ( sqrt{24} ) is not integer.( n = 3 ):( k^2 = 5(9) + 4 = 49 ). ( k = 7 ). Then ( N = [3 ± 7]/2 ). Positive solution: ( (3 + 7)/2 = 5 ). So ( N = 5 ), ( n = 3 ). Check:Left-hand side: ( 5^2 = 25 ).Right-hand side: ( 3(5 + 3) + 1 = 24 + 1 = 25 ). Correct.( n = 4 ):( k^2 = 5(16) + 4 = 84 ). Not a square.( n = 5 ):( k^2 = 5(25) + 4 = 129 ). Not a square.( n = 6 ):( 5(36) + 4 = 184 ). Not a square.( n = 7 ):( 5(49) + 4 = 249 ). Not a square.( n = 8 ):( 5(64) + 4 = 324 ). ( 324 = 18^2 ). So ( k = 18 ). Then, ( N = [8 ± 18]/2 ). Positive solution: ( (8 + 18)/2 = 13 ). So ( N = 13 ), ( n = 8 ). Check:Left-hand side: ( 13^2 = 169 ).Right-hand side: ( 8(13 + 8) + 1 = 8*21 + 1 = 168 + 1 = 169 ). Correct.Continuing this pattern, it seems that solutions exist for certain ( n ), and they might follow a recursive pattern. For Pell equations, solutions can often be generated from minimal solutions. Let me recall the Pell equation ( x^2 - 5y^2 = 4 ).The minimal solution here is ( (k, n) = (3, 1) ), as we saw. Then, subsequent solutions can be generated using the recurrence relations. For Pell equations, once you have the minimal solution, you can generate more solutions by multiplying by the fundamental solution. However, since this is a Pell-like equation, perhaps the solutions can be generated using the recurrence:( k_{m+1} = frac{(3 + sqrt{5})}{2}k_m + frac{(5 + sqrt{5})}{2}n_m )( n_{m+1} = frac{(1 + sqrt{5})}{2}k_m + frac{(3 + sqrt{5})}{2}n_m )But this might get complicated. Alternatively, perhaps there's a linear recurrence relation. Let's see the solutions we have so far:For ( n ): 1, 3, 8, ...Check the differences: 3 -1=2, 8 -3=5, not obvious. Let me check the next solution.If ( n = 21 ), then ( k^2 = 5*(21)^2 +4=5*441 +4=2205 +4=2209=47^2. So k=47. Then N=(21 +47)/2=68/2=34. Wait, let me check:Wait, ( n = 21 ):k^2 = 5*(21)^2 +4 = 5*441 +4 = 2205 +4 = 2209. Which is 47^2. So yes. Then N=(21 +47)/2=68/2=34. Let's verify:Left-hand side: 34^2=1156.Right-hand side: 21*(34+21)+1=21*55 +1=1155 +1=1156. Correct.So the sequence of n is 1,3,8,21,... which looks like Fibonacci numbers. Indeed, 1, 3, 8, 21, 55,... which are every other Fibonacci number. Let me check:The Fibonacci sequence is 1,1,2,3,5,8,13,21,34,55,...But our n's are 1,3,8,21,55,... which correspond to the Fibonacci numbers at positions 2,4,6,8,10,... So perhaps n is the 2m-th Fibonacci number. Let me confirm.But maybe it's related to the Pell equation solutions. Alternatively, since the equation is ( k^2 -5n^2 =4 ), the solutions can be generated using the fundamental solution (3,1) and multiplying by the fundamental unit in the ring ( mathbb{Z}[sqrt{5}] ).The fundamental unit for Pell equations ( x^2 - 5y^2 =1 ) is (9,4) since 9^2 -5*(4)^2=81-80=1. However, our equation is ( x^2 -5y^2=4 ). So maybe solutions can be generated by multiplying (3,1) by the fundamental unit (9 + 4√5)/2.Wait, perhaps to generate solutions, we can use the recurrence:If (k1, n1) is a solution, then the next solution can be generated via:k_{m+1} = (3k1 + 5n1)/2n_{m+1} = (k1 + 3n1)/2Wait, let's check with the first solution (3,1):Next solution would be k2=(3*3 +5*1)/2=(9 +5)/2=14/2=7n2=(3 +3*1)/2=(6)/2=3Which matches the second solution (7,3). Then next solution:k3=(3*7 +5*3)/2=(21 +15)/2=36/2=18n3=(7 +3*3)/2=(7 +9)/2=16/2=8Which is the third solution (18,8). Then next:k4=(3*18 +5*8)/2=(54 +40)/2=94/2=47n4=(18 +3*8)/2=(18 +24)/2=42/2=21Which is (47,21). Then next:k5=(3*47 +5*21)/2=(141 +105)/2=246/2=123n5=(47 +3*21)/2=(47 +63)/2=110/2=55So, (123,55). Then N=(55 +123)/2=178/2=89. Check:89^2 = 792155*(89 +55) +1=55*144 +1=7920 +1=7921. Correct.So this recurrence relation works. Therefore, all solutions for Case 1 can be generated using this recurrence starting from (3,1). Therefore, there are infinitely many solutions for Case 1.Wait, but the problem asks for all positive integers N and n. But if there are infinitely many solutions, we need to express them in terms of a sequence. However, the problem might be expecting only finite solutions, but maybe not. Let me check.But first, let me also check Case 2: ( N^2 = n(N + n) -1 ). Similarly, rearrange the equation:( N^2 -nN -n^2 +1 =0 ).Again, treating this as a quadratic in N:( N = [n ± sqrt{n^2 +4(n^2 -1)}]/2 ).Simplify the discriminant:( sqrt{n^2 +4n^2 -4} = sqrt{5n^2 -4} ).So, ( N = [n ± sqrt{5n^2 -4}]/2 ).For N to be an integer, ( sqrt{5n^2 -4} ) must be an integer. Let me denote ( k = sqrt{5n^2 -4} ), so:( k^2 =5n^2 -4 ).Rearranged as:( k^2 -5n^2 = -4 ).This is another Pell-type equation, similar to the previous one but with a negative constant term. Let's explore this equation.Again, trying small values of n:For ( n=1 ):( k^2 =5 -4=1 ), so ( k=1 ). Then ( N=(1 ±1)/2 ). Positive solution: (1+1)/2=1. So ( N=1 ), ( n=1 ). Check:Left: (1^2=1). Right: (1*(1+1)-1=2 -1=1). Correct.( n=2 ):( k^2=5*4 -4=16 ). So ( k=4 ). Then ( N=(2 ±4)/2 ). Positive solution: (2 +4)/2=3. Check:Left: (3^2=9). Right: (2*(3+2)-1=10 -1=9). Correct.( n=3 ):( k^2=5*9 -4=41 ). Not a square.( n=4 ):( 5*16 -4=76 ). Not a square.( n=5 ):(5*25 -4=121 ). (k=11). Then (N=(5 +11)/2=16/2=8). Check:Left: (8^2=64). Right: (5*(8+5)-1=5*13 -1=65 -1=64). Correct.( n=6 ):(k^2=5*36 -4=176). Not a square.(n=7):(5*49 -4=241). Not a square.(n=8):(5*64 -4=316). Not a square.(n=9):(5*81 -4=401). Not a square.(n=10):(5*100 -4=496). Not a square.(n=11):(5*121 -4=601). Not a square.(n=12):(5*144 -4=716). Not a square.(n=13):(5*169 -4=841=29^2). So (k=29). Then (N=(13 +29)/2=42/2=21). Check:Left: (21^2=441). Right: (13*(21+13)-1=13*34 -1=442 -1=441). Correct.So, similarly, solutions for Case 2 also appear periodically. Let's see if there's a pattern here.The equation (k^2 -5n^2 =-4). The minimal solution here is (1,1), then (4,2), (11,5), (29,13), etc. Let's see if these can be generated by a recurrence relation.For Pell equations of the form (x^2 - Dy^2 = -4), solutions can sometimes be generated from the fundamental solution and multiplied by the fundamental unit. For example, in this case, the fundamental unit for (x^2 -5y^2 =1) is (9,4). So, perhaps multiplying solutions by (9 +4√5) would generate new solutions.Let me check. Take the minimal solution (1,1):Multiply (1 + √5) by (9 +4√5):( (1 + √5)(9 +4√5) = 9 +4√5 +9√5 +4*5 = 9 +13√5 +20 = 29 +13√5 ).So, the new solution would be (29,13). Indeed, that's the next solution we found. Then multiplying again by (9 +4√5):( (29 +13√5)(9 +4√5) = 29*9 +29*4√5 +13*9√5 +13*4*5 )= 261 +116√5 +117√5 +260= (261 +260) + (116 +117)√5= 521 +233√5Which would correspond to the solution (521,233). Let's check if this is a solution:(521^2 -5*(233)^2 = 521^2 -5*233^2)Calculating:521^2 = (500 +21)^2 = 500^2 +2*500*21 +21^2 = 250000 +21000 +441 = 2714415*233^2 =5*(233^2). 233^2=54289. So 5*54289=271445.Then, 271441 -271445= -4. Correct. So (521,233) is indeed a solution.Therefore, the solutions for Case 2 can be generated by multiplying the previous solution by the fundamental unit (9 +4√5). The solutions are (1,1), (4,2), (11,5), (29,13), (76,34), etc. Wait, but when we generated them by hand earlier, after (29,13) we had (76,34) perhaps? Let me check (n=34):k^2=5*(34)^2 -4=5*1156 -4=5780 -4=5776=76^2. So yes, k=76, N=(34 +76)/2=110/2=55. Check:55^2=302534*(55 +34) -1=34*89 -1=3026 -1=3025. Correct.So, the solutions for Case 2 are also infinite and follow a similar recurrence.Therefore, both equations have infinitely many solutions. But the problem states "Find all positive integers N and n". However, if there are infinitely many solutions, we need to describe them in a general form.But perhaps the problem expects a finite number of solutions, which might mean I made a mistake in considering both cases as Pell equations. Wait, let me check the original problem again.The problem says "N^2 is 1 away from n(N + n)", so that includes both N^2 = n(N +n) +1 and N^2 =n(N +n) -1. So both cases need to be considered.However, in the solutions, both cases lead to Pell-type equations with infinitely many solutions. So unless there's a constraint I missed, like N and n being positive integers, which we already considered, the answer should include all solutions generated by the Pell equations.But the problem says "Find all positive integers N and n", so maybe they expect a parametric form or a recursive description?But let me check the initial solutions for both cases:Case 1:n=1, N=2n=3, N=5n=8, N=13n=21, N=34n=55, N=89...Case 2:n=1, N=1n=2, N=3n=5, N=8n=13, N=21n=34, N=55...Wait a minute, these sequences look familiar. The N in Case 1 is the same as the n in Case 2 shifted by one. For example:Case1 N=2 is Case2 n=2, which has N=3Case1 N=5 is Case2 n=5, which has N=8So, these are interleaved Fibonacci numbers. Indeed, if we list the solutions:Case1: (N,n) = (2,1), (5,3), (13,8), (34,21), (89,55),...Case2: (N,n) = (1,1), (3,2), (8,5), (21,13), (55,34),...Looking at these, the N values in Case1 are the Fibonacci numbers at even indices, and the n values in Case1 are the Fibonacci numbers at odd indices. Similarly for Case2. The Fibonacci sequence is often defined as F(1)=1, F(2)=1, F(3)=2, F(4)=3, F(5)=5, etc. But here, the solutions correspond to Fibonacci numbers with indices offset.Alternatively, noticing that the solutions for both cases are pairs of consecutive Fibonacci numbers. For instance:Case1: (2,1) -> F(3)=2, F(2)=1Case2: (1,1) -> F(1)=1, F(2)=1Then, (5,3) -> F(5)=5, F(4)=3(13,8) -> F(7)=13, F(6)=8And for Case2:(3,2) -> F(4)=3, F(3)=2(8,5) -> F(6)=8, F(5)=5(21,13) -> F(8)=21, F(7)=13So, in general, for Case1: ( N = F(2k +1) ), ( n = F(2k -1) )For Case2: ( N = F(2k) ), ( n = F(2k -2) )But perhaps more precisely, the solutions can be described using the Fibonacci sequence. Let me check.The Fibonacci sequence is defined by ( F_1 = 1 ), ( F_2 = 1 ), ( F_{k+1} = F_k + F_{k-1} ).Looking at the solutions:Case1:- (2,1): ( F_3 = 2 ), ( F_2 =1 )- (5,3): ( F_5 =5 ), ( F_4 =3 )- (13,8): ( F_7=13 ), ( F_6=8 )- etc.So, in general, for Case1: ( N = F_{2m+1} ), ( n = F_{2m} ), where m ≥1.Wait, but in the first solution, m=0 would give ( F_1=1 ), but N=2 is ( F_3 ). Maybe indexing is different.Alternatively, starting m from 1:For m=1: ( F_{2m+1} = F_3 =2 ), ( F_{2m} = F_2 =1 )For m=2: ( F_5=5 ), ( F_4=3 )For m=3: ( F_7=13 ), ( F_6=8 )Yes, this works. So in general, Case1 solutions are ( N = F_{2m+1} ), ( n = F_{2m} )Similarly, Case2 solutions:- (1,1): ( F_2 =1 ), ( F_1=1 )- (3,2): ( F_4=3 ), ( F_3=2 )- (8,5): ( F_6=8 ), ( F_5=5 )- (21,13): ( F_8=21 ), ( F_7=13 )So, Case2 solutions are ( N = F_{2m} ), ( n = F_{2m-1} )Therefore, all solutions can be described in terms of Fibonacci numbers. Hence, the positive integer solutions are pairs of consecutive Fibonacci numbers, with N and n being consecutive terms in the Fibonacci sequence.To confirm, let's take m=1 for Case2:( N = F_2 =1 ), ( n = F_1=1 ). This works as shown earlier.For m=2:( N = F_4=3 ), ( n = F_3=2 ). Check: (3^2 =9), (2*(3 +2) -1=10 -1=9). Correct.Similarly, m=3:( N = F_6=8 ), ( n=F_5=5 ). Check: (8^2=64), (5*(8 +5) -1=65 -1=64). Correct.Therefore, the solutions are:For Case1: ( N = F_{2m+1} ), ( n = F_{2m} ) for ( m geq 1 )For Case2: ( N = F_{2m} ), ( n = F_{2m -1} ) for ( m geq 1 )However, the problem states "all positive integers N and n", so we need to include all possible solutions. Since the Fibonacci sequence is infinite, there are infinitely many solutions for both cases.But the problem might expect expressing the solutions in terms of Fibonacci numbers. Alternatively, if the problem expects a finite number of solutions, perhaps I missed some constraints. Let me double-check if there are other possible solutions outside the Fibonacci sequence.Looking back at the Pell equations, since they are known to have infinitely many solutions, and each solution corresponds to a pair in the Fibonacci sequence, it's safe to conclude that all solutions are consecutive Fibonacci numbers. Therefore, the answer can be expressed in terms of Fibonacci numbers.Alternatively, since the problem might require listing all solutions, but since they are infinite, we need to describe them parametrically. However, given the problem's wording, it might expect recognizing that the solutions are pairs of consecutive Fibonacci numbers.But let me check if there are any other solutions outside the Fibonacci sequence.For example, in Case1, after n=1,3,8,21,55,... which are Fibonacci numbers. If I try n=4, which is not a Fibonacci number, we saw that there's no solution. Similarly for other non-Fibonacci n's. So only Fibonacci numbers work.Similarly, for Case2, n must be a Fibonacci number to have solutions.Hence, all solutions are pairs where N and n are consecutive Fibonacci numbers, with specific parity based on the case.Therefore, the final answer would be all pairs where N and n are consecutive Fibonacci numbers, with N being the larger one in the pair for Case1 and n being the larger one for Case2. But actually, looking at the solutions:Case1: N is larger than n (e.g., N=2, n=1; N=5, n=3; N=13, n=8, etc.)Case2: Sometimes N is larger, sometimes equal. For example, in Case2: (1,1) where N=n=1; (3,2) N=3, n=2; (8,5) N=8, n=5; etc. So N is larger than n here as well.But in both cases, N and n are consecutive Fibonacci numbers. In Case1, N is the odd-indexed Fibonacci number, and n is the preceding even-indexed one. In Case2, N is the even-indexed Fibonacci number, and n is the preceding odd-indexed one.Alternatively, we can say that for each ( k geq 1 ), the pairs ( (F_{k+1}, F_{k}) ) satisfy either ( N^2 = n(N +n) +1 ) or ( N^2 =n(N +n) -1 ), depending on the parity of k.Wait, let's see:For even k:If k is even, say k=2m, then ( (F_{2m+1}, F_{2m}) ) satisfy Case1.If k is odd, say k=2m-1, then ( (F_{2m}, F_{2m-1}) ) satisfy Case2.Hence, all solutions are consecutive Fibonacci numbers where N is the next Fibonacci number after n, and depending on the position, it falls into Case1 or Case2.Therefore, the answer can be succinctly described as all pairs of consecutive Fibonacci numbers ( (F_{k+1}, F_k) ) for ( k geq 1 ), where each pair satisfies either ( N^2 = n(N +n) +1 ) or ( N^2 =n(N +n) -1 ).Since the problem asks for positive integers N and n, and since every pair of consecutive Fibonacci numbers satisfies one of the two equations, the solutions are all such consecutive Fibonacci pairs.To present the answer, since Fibonacci numbers are well-known, we can express the solutions as:All pairs of consecutive Fibonacci numbers ( (F_{k+1}, F_k) ) where ( k geq 1 ), and ( F_k ) denotes the ( k )-th Fibonacci number.But to be more precise, since the equations alternate between Case1 and Case2 based on whether ( k ) is even or odd, but the problem doesn't differentiate between the two cases, just requiring N and n to satisfy that ( N^2 ) is 1 away from ( n(N +n) ), so both cases are included.Therefore, the complete solution set is all pairs of consecutive Fibonacci numbers.Hence, the final answer should list these pairs. However, since there are infinitely many Fibonacci numbers, we can't list them all, but we can describe them in terms of the Fibonacci sequence.Therefore, the positive integer solutions are all pairs ( (F_{k+1}, F_k) ) where ( F_k ) is the ( k )-th Fibonacci number.

✅:The solutions are all pairs of consecutive Fibonacci numbers. Specifically, for each positive integer ( k ), the pairs ((F_{k+1}, F_k)) satisfy the equation ( N^2 = n(N + n) pm 1 ), where ( F_k ) denotes the ( k )-th Fibonacci number.Final Answer: All positive integer solutions are consecutive Fibonacci numbers given by ( boxed{(N, n) = (F_{k+1}, F_k)} ) for each ( k geq 1 ), where ( F_k ) is the ( k )-th Fibonacci number.

🔑:1. Understanding the problem and setting up the equation: We need to find positive integers ( N ) and ( n ) such that: [ N^2 = 1 + n(N + n) ] This can be rewritten as: [ N^2 = 1 + nN + n^2 ] Rearranging terms, we get: [ N^2 - nN - n^2 = 1 ]2. Analyzing the equation: We will consider the cases when ( N > n ) and ( N leq n ) separately.3. Case 1: ( N > n ): If ( N > n ), we can manipulate the given equation: [ N^2 - nN - n^2 = 1 ] to reflect a transformation of pairs ((N, n)): [ N^2 - nN - n^2 = N(N - n) - n^2 = 1 quad (text{Rewriting}) ] ((N, n)) works if and only if ((n, N - n)) works. We can hence use this property to recursively transform the pairs. 4. Case 2: ( N leq n ): If ( N leq n ), consider the equation derived: [ n(N + n) - N^2 = n^2 + nN - N^2 = 1 ] Here, it should be noted that ( n^2 geq 1 ), so the left-hand side term must be safely non-negative. This implies only the base case where ( N leq n ), the solution for ( n(N + n) - N^2 = 1 ), when ( N = n = 1 ). Hence: [ (N, n) = (1, 1) ]5. Recursive relationship and general solution: From the properties analyzed above, if you repeatedly apply the transformation ((N, n) mapsto (n, N-n)), you eventually reach the base case ((1, 1)). So, the reverse process will help generate the all pairs starting from ((1, 1)): [ (N, n) = (1, 1) ] The reverse: [ (1, 1) rightarrow (2, 1) rightarrow (3, 2) rightarrow dots ] Specifically: [ (N, n) mapsto (N + n, N) ] This maps to pairs that solve our original equation persistently.6. Final conclusion using Fibonacci sequence: This transformation aligns perfectly with the Fibonacci sequence relationships. The general solution is represented as pairs derived from Fibonacci numbers: [ (F_{i+1}, F_i) ] Given the procedure steps accurately, we obtain: [ left(F_{i+1}, F_{i}right) ] generated recursively from: [ left(F_2, F_1right) = (1, 1) ] Thus, the solutions are:[boxed{left(F_{i+1}, F_{i}right)}]

🧠:Okay, let's try to solve this problem. So, we need to find two numbers such that their sum is 2017, and when we reverse their digits, the sum of the reversed numbers is 8947. Hmm, that seems a bit tricky. Let me break it down step by step.First, let's denote the two numbers as, say, A and B. So, according to the problem, we have:A + B = 2017And when we reverse the digits of A and B, we get new numbers, let's call them A' and B'. Then:A' + B' = 8947Our goal is to find the original numbers A and B.Now, let's think about how reversing the digits of a number affects its value. If A is a 4-digit number, reversing its digits would give a different number. Similarly, if it's a 3-digit number, the reversed number would also be different. Wait, but 2017 is a 4-digit number. So maybe A and B are both 4-digit numbers, or one is a 4-digit and the other is a 3-digit? Let's check.Since their sum is 2017, let's see the possible number of digits. The maximum possible value for one number is 2016 (if the other is 1), which is a 4-digit number. The minimum would be 1, which is a 1-digit number. So, A and B could be 4-digit and 3-digit or 4-digit and 1-digit, but probably not both 4-digit because 2017 is 2000-something, so adding two 4-digit numbers would result in a larger sum. Wait, but 2017 is just 2000 + 17. So maybe one number is around 2000 and the other is 17? But when reversed, 2000 becomes 0002, which is 2, and 17 reversed is 71. Then 2 + 71 is 73, which is way less than 8947. So that can't be right.So maybe the numbers are both 4-digit numbers? Wait, but 2017 is a 4-digit number. If both numbers are 4-digit numbers, their sum is 2017. That would mean that one number is, say, 1000 and the other is 1017. Let's reverse them: 0001 (which is 1) and 7101. Then 1 + 7101 = 7102, which is still less than 8947. Hmm, so that's not matching. So maybe the reversed numbers sum to a much larger number, 8947. So perhaps the original numbers are smaller, but when reversed, they become larger?Wait, let's think about the possible range. If A + B = 2017, then each of them is less than 2017. So the maximum possible number for A is 2016, and the minimum is 1. When reversed, if A is a 4-digit number like 2016, reversed would be 6102. Similarly, if B is 1, reversed is 1. So 6102 + 1 = 6103, which is still less than 8947. Hmm. So maybe the original numbers are 3-digit numbers? For example, if A is 999 and B is 1018. Wait, but 999 + 1018 = 2017. Then reversing them: 999 reversed is 999, and 1018 reversed is 8101. Then 999 + 8101 = 9100, which is more than 8947. So that's too much.Hmm. So perhaps the numbers are of different digit lengths? Maybe one is a 4-digit and the other is a 3-digit, but when reversed, the 3-digit becomes a 4-digit (if it's actually a 3-digit number with leading zeros when reversed, but leading zeros don't count in numbers). Wait, reversing a 3-digit number would give another 3-digit number. For example, 123 reversed is 321. Similarly, a 4-digit number reversed is another 4-digit number, unless it ends with zeros. For example, 1000 reversed is 0001, which is 1. So that's an issue. So maybe the original numbers have trailing zeros, which when reversed become leading zeros, thus effectively shorter numbers.But trailing zeros in a number like A or B would mean that the original numbers have those zeros. For example, if A is 1000, then B is 1017. Reversed, A becomes 0001 = 1, and B becomes 7101. Then 1 + 7101 = 7102, which is less than 8947. Not enough. Alternatively, if A is 1200, reversed is 0021 = 21, and B is 817 (since 1200 + 817 = 2017). Reversed, B is 718. Then sum is 21 + 718 = 739. Still too low. So this approach may not work.Alternatively, maybe the numbers are such that when reversed, they gain digits. Wait, but how? Unless the original numbers have trailing zeros, which when reversed become leading zeros and thus are not considered part of the number. So, for example, if A is a 4-digit number ending with a zero, like 1230, reversing gives 0321, which is 321. Similarly, if A is 1000, reversed is 0001 = 1. So, reversing a number with trailing zeros reduces its digit count.But we need the sum of reversed numbers to be 8947, which is a 4-digit number. So maybe the reversed numbers are both 4-digit numbers? For that, the original numbers must be 4-digit numbers without trailing zeros. Because if you reverse a 4-digit number without trailing zeros, you get another 4-digit number. For example, 1234 reversed is 4321. So, if both A and B are 4-digit numbers without trailing zeros, then their reverses would also be 4-digit numbers, and their sum could be 8947.But wait, A + B = 2017. If both A and B are 4-digit numbers, then each of them must be at least 1000. But 1000 + 1000 = 2000, which is less than 2017. So, possible. Let's see. Let A = 1000 + x, B = 1000 + y, but then A + B = 2000 + x + y = 2017, so x + y = 17. Then reversing A and B would give numbers like (reverse of 1000 + x) and (reverse of 1000 + y). But if x is a 3-digit number, like 000 to 999, then A would be 1000 to 1999. Reversing A would give a number from 0001 to 9991, but leading zeros are not considered, so actually numbers from 1 to 9991. Wait, but reversing 1000 gives 0001 = 1, and reversing 1999 gives 9991. Similarly, reversing 1234 gives 4321. So, if A is between 1000 and 1999, reversed A is between 1 and 9991. Similarly for B. So their reversed sum could be up to 9991 + 9991 = 19982, but in our case, it's 8947. So possible.So maybe A is a 4-digit number starting with 1, and B is a 4-digit number starting with 1 as well? But A + B = 2017, so if A is 1000 + x and B is 1000 + y, then x + y = 17. So x and y are digits? Wait, no, x and y would be the last three digits. Wait, if A is 1abc, then reversed A is cba1. Similarly, B is 1def, reversed is fed1. So, the reversed numbers would be 4-digit numbers as well. But if A is 1000 + x, where x is a 3-digit number, then reversed A would be x reversed * 10 + 1. Wait, maybe not. Let's take an example.Let’s say A is 1234. Reversed, it's 4321. So, if A is a 4-digit number, the reversed number is also 4-digit. Similarly, B is 2017 - A. Let's suppose A is 1000 + a, where a is from 0 to 999, then B is 1017 - a. Wait, no: A + B = 2017, so if A = 1000 + a, then B = 2017 - (1000 + a) = 1017 - a. But a can be from 0 to 999, so B would be from 1017 down to 18. So B would be a 4-digit number (1017) down to a 2-digit number (18). So, when reversed, B could be a 4-digit, 3-digit, or 2-digit number depending on its value.But the problem is that when reversing B, which might be a 4-digit, 3-digit, or 2-digit number, how does that affect the reversed sum? For example, if B is 1017, reversed is 7101. If B is 18, reversed is 81. So, depending on the original numbers, their reverses can vary in digit length.This seems complicated. Maybe we need to set up equations based on the digits of A and B.Let’s denote A as a 4-digit number: A = 1000w + 100x + 10y + z. Then, reversed(A) = 1000z + 100y + 10x + w.Similarly, B can be a 4-digit or 3-digit number. Let’s assume B is a 4-digit number: B = 1000a + 100b + 10c + d. Then reversed(B) = 1000d + 100c + 10b + a.But if B is a 3-digit number, say B = 100a + 10b + c, then reversed(B) = 100c + 10b + a. Similarly, if B is a 2-digit or 1-digit number, the reversal would be straightforward.But this approach might get too complicated with multiple cases. Maybe there's a smarter way.Alternatively, let's note that when you reverse a number, the difference between the original and reversed number is a multiple of 9. Wait, is that true? Let me check with an example: 1234 reversed is 4321. 4321 - 1234 = 3087, which is 3087 ÷ 9 = 343. So yes, a multiple of 9. Another example: 12 reversed is 21. 21 - 12 = 9. Yes, also a multiple of 9. So, the difference between a number and its reverse is a multiple of 9. Therefore, (A' - A) + (B' - B) = (A' + B') - (A + B) = 8947 - 2017 = 6930. And since each difference (A' - A) and (B' - B) is a multiple of 9, their sum is also a multiple of 9. 6930 ÷ 9 = 770. So that's valid.But how does this help? Maybe not directly, but it's a property to keep in mind.Alternatively, let's denote the two numbers as A and B with A + B = 2017 and A' + B' = 8947.Let’s consider that reversing the digits of a number can be represented algebraically. For example, if A is a 4-digit number, A = 1000a + 100b + 10c + d, then A’ = 1000d + 100c + 10b + a. Similarly, if B is a 3-digit number, B = 100e + 10f + g, then B’ = 100g + 10f + e. Then, A’ + B’ would be 1000d + 100c + 10b + a + 100g + 10f + e.But this seems complicated. Maybe we can express A’ + B’ in terms of A and B. Let's see.Suppose A is a 4-digit number: A = wxyz (i.e., 1000w + 100x + 10y + z), so A’ = zyxw (1000z + 100y + 10x + w). Similarly, B could be a 4-digit or 3-digit number.Alternatively, let's consider that when you reverse a number, you are essentially swapping the place values of its digits. So, if A is a 4-digit number, reversing it would swap the thousands digit with the ones digit, and the hundreds digit with the tens digit. So, the difference between A’ and A is 999(z - w) + 90(y - x). Similarly, for B.But perhaps this is not helpful. Let me think differently.Let’s consider that A and B are such that A + B = 2017, and A’ + B’ = 8947.Let’s subtract the two equations: (A’ + B’) - (A + B) = 8947 - 2017 = 6930. So, (A’ - A) + (B’ - B) = 6930.As we saw earlier, A’ - A and B’ - B are multiples of 9. Let’s denote A’ - A = 9k and B’ - B = 9m. Then, 9k + 9m = 6930 ⇒ k + m = 770.So, k and m are integers such that k + m = 770. But we need to find A and B such that reversing them gives a difference of 9k and 9m respectively.But how do we relate k and m to the digits of A and B? Let's think about how A’ - A can be expressed in terms of the digits.Suppose A is a 4-digit number: A = 1000a + 100b + 10c + d. Then A’ = 1000d + 100c + 10b + a. So, A’ - A = (1000d + 100c + 10b + a) - (1000a + 100b + 10c + d) = 999d + 90c - 90b - 999a = 999(d - a) + 90(c - b). Similarly, if A is a 3-digit number: A = 100a + 10b + c, then A’ = 100c + 10b + a. So A’ - A = 99c - 99a = 99(c - a). Which is also a multiple of 9, as 99 is 11*9.So, depending on the number of digits, the difference A’ - A can be expressed as a combination of the digits. However, this might get complex. Let's see if we can find a pattern or possible values.Given that A + B = 2017, which is a relatively small number, and A’ + B’ = 8947, which is much larger, it suggests that reversing the digits of A and B significantly increases their values. Therefore, A and B must be numbers that, when reversed, become much larger. This likely means that the original numbers have their digits in ascending order (e.g., 123 becomes 321 when reversed) or have leading digits that are small, so when reversed, those become the trailing digits, making the number larger.For example, if A is 1999, reversing it gives 9991, which is much larger. But 1999 + B = 2017 ⇒ B = 18. Reversing B gives 81. Then, 9991 + 81 = 10072, which is way larger than 8947. So that's not good.Alternatively, if A is 1009, reversed is 9001. Then B = 2017 - 1009 = 1008. Reversed B is 8001. Then 9001 + 8001 = 17002, which is way too big. Not helpful.Wait, so maybe the reversed numbers add up to 8947, which is less than 17002. So perhaps the original numbers are not that large. Let's try smaller numbers.Suppose A is 1234. Then reversed A is 4321. Then B = 2017 - 1234 = 783. Reversed B is 387. Then sum of reversed numbers is 4321 + 387 = 4708, which is less than 8947.Hmm. Maybe try another pair. Let's pick A as 1999, B as 18. As before, reversed sum is 9991 + 81 = 10072. Still too big. What if A is 1111, B = 2017 - 1111 = 906. Reversed A is 1111, reversed B is 609. Sum is 1111 + 609 = 1720. Too small.Wait, this trial and error might take too long. Let's try to think algebraically.Let’s suppose that A is a 4-digit number and B is a 3-digit number. Let’s denote A as 1000a + 100b + 10c + d, and B as 100e + 10f + g. Then:A + B = 1000a + 100b + 10c + d + 100e + 10f + g = 2017Reversed numbers: A’ = 1000d + 100c + 10b + a, B’ = 100g + 10f + eA’ + B’ = 1000d + 100c + 10b + a + 100g + 10f + e = 8947So we have two equations:1. 1000a + 100b + 10c + d + 100e + 10f + g = 20172. 1000d + 100c + 10b + a + 100g + 10f + e = 8947This system of equations seems complex, but maybe we can find relationships between variables.Let’s subtract equation 1 from equation 2:(1000d + 100c + 10b + a + 100g + 10f + e) - (1000a + 100b + 10c + d + 100e + 10f + g) = 8947 - 2017Simplify left side:1000d - d + 100c - 10c + 10b - 100b + a - 1000a + 100g - g + 10f - 10f + e - 100e = 6930Which is:999d + 90c - 90b - 999a + 99g - 99e = 6930Factorizing:999(d - a) + 90(c - b) + 99(g - e) = 6930Divide both sides by 9:111(d - a) + 10(c - b) + 11(g - e) = 770Hmm, this equation might be manageable. Let's denote:Let’s note that 111(d - a) + 10(c - b) + 11(g - e) = 770.We need to find integers d, a, c, b, g, e such that this equation holds. Remember that a, b, c, d are digits of A (a ≠ 0 since A is a 4-digit number), and e, f, g are digits of B (e ≠ 0 if B is a 3-digit number, but B could be a 2-digit number as well). Wait, but in our initial assumption, B is a 3-digit number, so e ≠ 0. However, if B is a 2-digit number, then e = 0. But let's first proceed under the assumption that B is a 3-digit number. If that doesn't work, we can revisit.So, variables a (1-9), d (0-9), b, c (0-9), e (1-9), g (0-9).So, we have 111(d - a) + 10(c - b) + 11(g - e) = 770.This is a key equation. Let’s see if we can find possible values here.First, note that 111(d - a) is a multiple of 111, 10(c - b) is a multiple of 10, and 11(g - e) is a multiple of 11. The total is 770, which is a multiple of 10 and 11 (770 = 70 * 11 = 77 * 10). Let’s factorize:770 = 70 * 11 = 7 * 10 * 11.So, the equation is:111(d - a) + 10(c - b) + 11(g - e) = 770.Let’s try to express this as:111(d - a) = 770 - 10(c - b) - 11(g - e)So, the right-hand side must be divisible by 111. Let’s denote:Let’s set S = 10(c - b) + 11(g - e). Then, 111(d - a) = 770 - S.Therefore, 770 - S must be divisible by 111, and S must be such that 770 - S is non-negative since 111(d - a) must be positive or negative based on (d - a). Wait, but 111(d - a) can be positive or negative, but 770 is positive. So, depending on S.But let's see possible values for 770 - S divisible by 111.Compute 770 ÷ 111 ≈ 6.94. So possible multiples are 6*111=666, 7*111=777. But 777 > 770, so only 666. Wait, 6*111=666, 5*111=555, 7*111=777.But 770 - S must be equal to 111*k, where k is an integer.So possible k values such that 111*k ≤ 770. The maximum k is 7 (111*7=777), which is over. So k can be up to 6 (666). So possible values for 111*k are 0, 111, 222, 333, 444, 555, 666.Therefore, 770 - S must be one of these values.So S = 770 - 111*k.Let’s compute possible S values:For k=6: S=770-666=104k=5: S=770-555=215k=4: S=770-444=326k=3: S=770-333=437k=2: S=770-222=548k=1: S=770-111=659k=0: S=770-0=770But S is defined as 10(c - b) + 11(g - e). Let's see the possible range of S.10(c - b): c and b are digits (0-9), so c - b can be from -9 to 9, so 10*(c - b) ranges from -90 to 90.Similarly, 11(g - e): g and e are digits. If B is a 3-digit number, e is 1-9, g is 0-9. So g - e can be from -9 to 9 (if e=1, g=0: -1; e=9, g=9: 0; e=1, g=9: 8). Wait, actually, if e is 1-9 and g is 0-9, then g - e can be from (0 - 9) = -9 up to (9 - 1) = 8. So 11*(g - e) ranges from -99 to 88.Therefore, S = 10(c - b) + 11(g - e) can range from (-90 - 99) = -189 to (90 + 88) = 178.But our possible S values from above are 104, 215, 326, 437, 548, 659, 770. However, the maximum S can be is 178, so only S=104 is possible. The others (215, 326, etc.) are too large. So only k=6 gives S=104.Therefore, 111*(d - a) = 666 ⇒ d - a = 666 / 111 = 6. So d = a + 6.Similarly, S = 104 = 10(c - b) + 11(g - e)So we have:10(c - b) + 11(g - e) = 104.Now, we need to solve this equation where c, b, g, e are digits with e ≥1 (since B is a 3-digit number), and g can be 0-9.Let’s denote:Let’s set x = c - b and y = g - e. Then:10x + 11y = 104.We need integer solutions for x and y where x is between -9 and 9, y is between -9 and 8 (since g - e can be from -9 to 8).Let’s solve 10x + 11y = 104.We can rewrite this as 10x = 104 - 11y ⇒ x = (104 - 11y)/10.For x to be an integer, (104 - 11y) must be divisible by 10. So, 104 - 11y ≡ 0 mod 10 ⇒ -11y ≡ -104 mod 10 ⇒ 11y ≡ 4 mod 10 ⇒ Since 11 ≡ 1 mod 10, this simplifies to y ≡ 4 mod 10. So y must be congruent to 4 modulo 10. Given that y is between -9 and 8, possible y values are 4, -6.Let’s check y=4:y=4 ⇒ x=(104 - 44)/10 = 60/10=6.So x=6, y=4.Then, c - b = 6 and g - e = 4.Check if possible:c - b =6 ⇒ c = b + 6. Since b and c are digits (0-9), b can be from 0 to 3 (since c = b +6 ≤9 ⇒ b ≤3).Similarly, g - e =4 ⇒ g = e +4. Since e is at least 1 (B is a 3-digit number), e can be 1 to 5 (since g = e +4 ≤9 ⇒ e ≤5).Next, y=-6:y=-6 ⇒ x=(104 - (-66))/10 = (104 +66)/10=170/10=17. But x=17 is impossible because x = c - b, and c and b are digits (0-9), so maximum x can be 9 - 0 =9. Therefore, y=-6 is invalid.Therefore, the only solution is y=4, x=6. Therefore:c = b +6,g = e +4,with b from 0 to 3,e from 1 to 5.Now, let's summarize what we have so far:From the earlier equation:d = a +6.Also, from the digit reversal:A is a 4-digit number: 1000a + 100b +10c + d.But c = b +6, d = a +6.So A = 1000a + 100b + 10(b +6) + (a +6) = 1000a + 100b + 10b +60 + a +6 = 1001a + 110b +66.Similarly, B is a 3-digit number: 100e +10f +g.But g = e +4.So B = 100e +10f + (e +4) = 101e +10f +4.Also, we know that A + B =2017:So:1001a + 110b +66 + 101e +10f +4 =2017Simplify:1001a + 110b +101e +10f +70 =2017So:1001a + 110b +101e +10f = 2017 -70=1947So, 1001a + 110b +101e +10f =1947.This equation needs to be solved with the constraints:a: 1-9 (since A is a 4-digit number),b: 0-3 (since c =b +6 ≤9 ⇒ b ≤3),e:1-5 (since g = e +4 ≤9 ⇒ e ≤5),f:0-9.Additionally, from the reversed numbers:A’ =1000d +100c +10b +a =1000(a +6) +100(b +6) +10b +a =1000a +6000 +100b +600 +10b +a=1001a +110b +6600.Similarly, B’ =100g +10f +e=100(e +4) +10f +e=101e +10f +400.We also know that A’ + B’ =8947:So:1001a +110b +6600 +101e +10f +400=8947Simplify:1001a +110b +101e +10f +7000=8947So:1001a +110b +101e +10f=8947 -7000=1947.Wait, this is the same equation as before: 1001a +110b +101e +10f=1947. Therefore, our earlier steps are consistent.So, we need to solve 1001a +110b +101e +10f=1947 with the constraints on a, b, e, f.Let’s note that 1001a is a multiple of 1001. Let’s see what possible a can be.Given that a is from 1 to9, compute 1001a for a=1 to9:a=1:1001a=2:2002But 2002 is already larger than 1947, so a can only be 1.Therefore, a=1.Then, 1001*1=1001.So equation becomes:1001 +110b +101e +10f=1947 ⇒110b +101e +10f=1947 -1001=946.So, 110b +101e +10f=946.Now, with b:0-3, e:1-5, f:0-9.Let’s write this as:110b +101e +10f=946.Divide the equation by common factors if possible. 110, 101,10 are coefficients. 101 is a prime number, 110=10*11, 10=10.Let’s see modulo 10:110b ≡0 mod10,101e ≡e mod10,10f ≡0 mod10.Thus, 0 + e +0 ≡6 mod10 ⇒ e ≡6 mod10. But e is from1-5, so no solution. Wait, this is a problem.Wait, wait, 110b +101e +10f ≡0 + e +0 ≡e mod10=946 mod10=6. Therefore, e ≡6 mod10. But e is 1-5. Contradiction. Therefore, no solution? But this can't be, since we have a problem to solve. There must be a mistake.Wait, but we derived a=1, which leads to e needing to be 6 mod10, but e is at most5. This suggests a contradiction, which means our earlier assumption might be wrong. Where did we go wrong?Let’s check the previous steps. We assumed that B is a 3-digit number, leading to e from1-5. But if B is a 2-digit number, then e=0. Let's revisit that.Wait, in our initial equations, we assumed B is a 3-digit number (e≠0). But if B is a 2-digit number, then B =10f +g, and reversed B is 10g +f. Then, the previous equations would change. So perhaps B is a 2-digit number, which would allow e=0.But we need to check if this is possible. Let's re-examine.If B is a 2-digit number, then B =10f +g (where f is 1-9, g is 0-9). Then reversed B is10g +f.Then, the original equations:A + B =2017A’ + B’ =8947Assuming A is 4-digit and B is 2-digit:A =1000a +100b +10c +dB =10f +gReversed A’=1000d +100c +10b +aReversed B’=10g +fSo, A’ + B’ =1000d +100c +10b +a +10g +f =8947Earlier, we derived:999(d - a) +90(c - b) +9(g - f)=6930 (since B is 2-digit, so different coefficients)Wait, previously, when B was 3-digit, the equation after subtraction was 999(d - a) +90(c - b) +99(g - e)=6930. But if B is 2-digit, then the difference (B’ - B)= (10g +f) - (10f +g)=9g -9f=9(g -f). So the previous equation would be:(A’ -A) + (B’ -B)=6930Which is:[999(d - a) +90(c - b)] + [9(g -f)] =6930Divide by9:111(d -a) +10(c -b) + (g -f)=770So, the equation becomes:111(d -a) +10(c -b) + (g -f)=770This is different from before. So, if B is 2-digit, the equation is 111(d -a) +10(c -b) + (g -f)=770.But earlier, we assumed B was 3-digit and got a contradiction. So perhaps B is a 2-digit number.Let’s try this approach.So, assuming B is a 2-digit number: B=10f +g, reversed B’=10g +f.Then, A + B =2017 ⇒ A=2017 - B=2017 - (10f +g).A is a 4-digit number, so 2017 - (10f +g) ≥1000 ⇒10f +g ≤1017. Since B is a 2-digit number, 10f +g ≤99. But 2017 -99=1918, which is still a 4-digit number, so possible.Then, A’ + B’ =8947 ⇒ (reversed A) + (reversed B) =8947.As before, subtracting the two equations:(A’ + B’) - (A + B)=8947 -2017=6930.Which gives:(A’ -A) + (B’ -B)=6930.As before:(A’ -A)=999(d -a) +90(c -b)(B’ -B)=9(g -f)Thus:999(d -a) +90(c -b) +9(g -f)=6930Divide by9:111(d -a) +10(c -b) + (g -f)=770So now, we have:111(d -a) +10(c -b) + (g -f)=770Also, A=2017 - B=2017 - (10f +g). So A=2017 -10f -g.But A is also equal to1000a +100b +10c +d.Thus:1000a +100b +10c +d=2017 -10f -gWhich implies:1000a +100b +10c +d +10f +g=2017This is another equation.So, we have two equations:1. 1000a +100b +10c +d +10f +g=20172. 111(d -a) +10(c -b) + (g -f)=770Additionally, from the reversals:A’=1000d +100c +10b +aB’=10g +fSo, A’ + B’=8947 ⇒1000d +100c +10b +a +10g +f=8947But since A=1000a +100b +10c +d=2017 -10f -g, we can express variables in terms of a,b,c,d,f,g.This is getting quite involved. Let’s see if we can find a relationship.First, let's try to find possible values for (d -a) in equation2: 111(d -a) +10(c -b) + (g -f)=770.Given that 111(d -a) is a large term, let's approximate. The maximum possible value for 111(d -a) is if d=9 and a=1, so 111*(8)=888. The minimum is if d=0 and a=9, which is -999. But we need 111(d -a) + ...=770. So likely d -a is positive.Let’s estimate d -a ≈770/111≈6.9. So approximately7. But 111*7=777, which is larger than770. So d -a=6 ⇒111*6=666. Then the remaining terms must sum to770 -666=104. So 10(c -b) + (g -f)=104.But 10(c -b) can be at most10*(9)=90, and (g -f) can be at most9-0=9. So total maximum 90+9=99 <104. So d -a=6 is insufficient.Next, d -a=7⇒111*7=777. Then 10(c -b) + (g -f)=770 -777= -7. Negative. But 10(c -b) + (g -f) can be negative. Let’s see.But we need to check if this is possible. Suppose d -a=7.Then 10(c -b) + (g -f)= -7.Possible? For example, c -b= -1, g -f=3. But 10*(-1) +3= -7. So possible. But need to check if digits allow.Alternatively, c -b=0, g -f= -7. Then 10*0 + (-7)= -7. Also possible.But this is complicated. Let's check for possible values of d -a.Alternatively, let's try d -a=7:Then 777 +10(c -b) + (g -f)=770 ⇒10(c -b) + (g -f)= -7.Possible solutions:We need 10x + y= -7, where x =c -b, y =g -f.x can be from -9 to9, y from -9 to9.Possible solutions:x=-1, y=3: 10*(-1) +3= -7.x=-2, y=13: y=13 is invalid.x=0, y=-7: 10*0 + (-7)= -7.x=1, y=-17: invalid.So only possible x=-1, y=3 or x=0, y=-7.Case1: x=-1, y=3.Thus, c -b= -1 ⇒c= b -1.g -f=3 ⇒g= f +3.Case2: x=0, y=-7.c -b=0 ⇒c=b.g -f= -7 ⇒g= f -7.But since g and f are digits (0-9), in case2: g= f -7 ⇒f must be at least7 to have g≥0. So f=7,8,9, leading to g=0,1,2.Let’s explore both cases.First, case1: c= b -1, g= f +3.Now, substitute into equation1:1000a +100b +10c +d +10f +g=2017But c= b -1, g= f +3, d= a +7 (since d -a=7).Substitute:1000a +100b +10(b -1) + (a +7) +10f + (f +3)=2017Simplify:1000a +100b +10b -10 +a +7 +10f +f +3=2017Combine like terms:1001a +110b +11f +0=2017 +10 -7 -3=2017So:1001a +110b +11f=2017Factor out 11:11(91a +10b +f)=2017But 2017 ÷11≈183.363, which is not an integer. Since 11 doesn't divide 2017 (11*183=2013, 2017-2013=4), this is impossible. Therefore, case1 has no solution.Case2: c=b, g= f -7.Substitute into equation1:1000a +100b +10c +d +10f +g=2017With c=b, d= a +7, g= f -7:1000a +100b +10b +a +7 +10f + (f -7)=2017Simplify:1000a +110b +a +7 +11f -7=2017Combine like terms:1001a +110b +11f=2017Factor out 11:11(91a +10b +f)=2017Again, 2017 is not divisible by11, so no solution.Therefore, d -a=7 leads to no solution.Next, try d -a=8:111*8=888.Then 10(c -b) + (g -f)=770 -888= -118.But 10(c -b) ranges from -90 to90, and (g -f) from -9 to9. So minimum possible 10(c -b) + (g -f)= -90 -9= -99, which is greater than -118. Therefore, impossible.Similarly, d -a=5:111*5=555.10(c -b) + (g -f)=770 -555=215. But maximum possible is 90 +9=99 <215. Impossible.Similarly, d -a=6:111*6=666.10(c -b) + (g -f)=770 -666=104.As before, maximum possible is 90 +9=99 <104. Impossible.Similarly, d -a=4:111*4=444.10(c -b) + (g -f)=770 -444=326. Impossible.Same for lower d -a.Wait, this suggests no solution when B is a 2-digit number. But the problem states that such numbers exist. Therefore, our initial assumption that B is a 3-digit or 2-digit number might be incorrect. Maybe one of the numbers is a 1-digit number?But if B is a 1-digit number, say B=g, then reversed B is g. Then A +g=2017, so A=2017 -g. Reversed A +g=8947. So reversed A=8947 -g. But A is a 4-digit number, reversed A is also a 4-digit number (since A=2017 -g, which is at least 2017 -9=2008, so reversed A would be like 8002 for A=2008). But reversed A=8947 -g. So for example, if g=0, reversed A=8947, which would mean A is reversed(8947)=7498. Then A=7498, B=2017 -7498= -5481. Negative, impossible. Similarly, other g would lead to negative B. So B can’t be a 1-digit number.This suggests that our previous approaches might be missing something. Perhaps the numbers have different digit lengths? For example, A is a 4-digit number, B is a 3-digit number, but when reversed, B becomes a 4-digit number because of leading zeros. But leading zeros aren't considered in numbers, so reversing a 3-digit number would still be a 3-digit number.Wait, unless the original number B has trailing zeros. For example, B=800. Reversed, it's 008=8. But then B=800, A=2017 -800=1217. Reversed A=7121. Sum of reverses:7121 +8=7129≠8947. Not helpful.Alternatively, A is a 3-digit number and B is a 4-digit number. But A + B=2017, so if B is a 4-digit number, minimum B=1000, so A=2017 -1000=1017. But A=1017 is a 4-digit number. Contradiction.Wait, no. If A is a 3-digit number, then B=2017 -A would be 2017 - (100-999)=1018-2016. So B would be a 4-digit number. So A is 3-digit, B is 4-digit. Then reversed A is 3-digit, reversed B is 4-digit. Let’s try this.Let’s suppose A is a 3-digit number: A=100a +10b +c.B is a 4-digit number: B=1000d +100e +10f +g.Then:A + B=100a +10b +c +1000d +100e +10f +g=2017.Reversed A=100c +10b +a.Reversed B=1000g +100f +10e +d.Sum of reversed numbers:100c +10b +a +1000g +100f +10e +d=8947.So, two equations:1. 1000d +100e +10f +g +100a +10b +c=20172. 1000g +100f +10e +d +100c +10b +a=8947This seems complicated, but let’s subtract equation1 from equation2:(1000g +100f +10e +d +100c +10b +a) - (1000d +100e +10f +g +100a +10b +c) =8947 -2017=6930.Simplify:1000g -g +100f -10f +10e -100e +d -1000d +100c -c +a -100a=6930.Which is:999g +90f -90e -999d +99c -99a=6930.Divide by9:111g +10f -10e -111d +11c -11a=770.Factor:111(g -d) +10(f -e) +11(c -a)=770.This is similar to previous equations. Let’s note that 111(g -d) +10(f -e) +11(c -a)=770.Possible values? Let’s try to find integer solutions.Assuming variables are digits:a:1-9 (since A is a 3-digit number),c:0-9,d:1-9 (since B is a 4-digit number),g:0-9,e,f:0-9.This equation is complex. Let's look for possible values of (g -d).111(g -d) is a significant term. Let's approximate:770 ÷111≈6.94. So possible (g -d)=6 or7.If g -d=6:111*6=666. Then remaining terms:10(f -e) +11(c -a)=770 -666=104.So,10(f -e) +11(c -a)=104.Similarly, if g -d=7:111*7=777. Then remaining terms:10(f -e) +11(c -a)=770 -777= -7.Let’s explore g -d=6 first.Then,10(f -e) +11(c -a)=104.Let’s denote x=f -e, y=c -a. So,10x +11y=104.Possible solutions where x and y are integers (x can be from -9 to9, y from -9 to9).Solve for y:11y=104 -10x ⇒y=(104 -10x)/11.We need (104 -10x) divisible by11. Let's find x such that104 -10x ≡0 mod11.104 mod11=10 (11*9=99, 104-99=5, but 11*9=99, 99+5=104. 5 mod11=5. Wait, 11*9=99, 104-99=5. So 104≡5 mod11.10x ≡5 mod11 ⇒10x ≡5 mod11 ⇒ multiply both sides by10^{-1} mod11. Since10*10=100≡1 mod11, so10^{-1}=10 mod11.Thus, x≡5*10=50≡50-44=6 mod11.So x≡6 mod11. Since x is between-9 and9, the only solution isx=6.Then, y=(104 -60)/11=44/11=4.Thus, x=6, y=4.Therefore:f -e=6,c -a=4.With constraints:f and e are digits (0-9), so f =e +6. Hence, e can be 0-3 (since f ≤9).c =a +4. Since a is at least1 and c ≤9, a can be1-5.Now, summarize:g =d +6,f =e +6,c =a +4,and the main equations:1. A + B=2017:A=100a +10b +c=100a +10b +a +4=101a +10b +4.B=1000d +100e +10f +g=1000d +100e +10(e +6) +d +6=1001d +100e +10e +60 +6=1001d +110e +66.Thus, A + B=101a +10b +4 +1001d +110e +66=2017.Simplify:101a +10b +1001d +110e +70=2017 ⇒101a +10b +1001d +110e=1947.Equation2: reversed A + reversed B=8947.Reversed A=100c +10b +a=100(a +4) +10b +a=101a +10b +400.Reversed B=1000g +100f +10e +d=1000(d +6) +100(e +6) +10e +d=1000d +6000 +100e +600 +10e +d=1001d +110e +6600.Sum:101a +10b +400 +1001d +110e +6600=8947 ⇒101a +10b +1001d +110e +7000=8947 ⇒101a +10b +1001d +110e=1947.This is the same as equation1. Therefore, consistent.So, we need to solve 101a +10b +1001d +110e=1947 with constraints:a:1-5,d:1-9 (since B is a 4-digit number, d≥1),e:0-3,b:0-9.Also, g =d +6 ≤9 ⇒d ≤3.Therefore, d can be1,2,3.Let’s check possible values for d:Case1: d=1Then, g=7.Equation:101a +10b +1001*1 +110e=1947 ⇒101a +10b +110e=1947 -1001=946.So,101a +10b +110e=946.Divide by common factor if possible. 101 is prime. Let's check modulo10:101a ≡a mod10,10b ≡0 mod10,110e ≡0 mod10.Thus, a ≡6 mod10. But a is1-5. No solution.Case2: d=2g=8.Equation:101a +10b +1001*2 +110e=1947 ⇒101a +10b +110e=1947 -2002= -55. Impossible since LHS is positive.Case3: d=3g=9.Equation:101a +10b +1001*3 +110e=1947 ⇒101a +10b +110e=1947 -3003= -1056. Impossible.Thus, no solution for d=1,2,3. Contradiction again.This suggests that even with this approach, there's no solution. But the problem states that there are two such numbers. Therefore, our assumptions about the number of digits might still be incorrect.Wait, perhaps one of the numbers is a 4-digit number with trailing zeros, so when reversed, it becomes a number with fewer digits. For example, A=1000, reversed=0001=1.Let’s try this approach.Suppose A=1000, then B=2017-1000=1017. Reversed A=0001=1, reversed B=7101. Sum=1+7101=7102≠8947. Not good.If A=1010, reversed=0101=101. B=2017-1010=1007, reversed=7001. Sum=101+7001=7102. Same as before.If A=1100, reversed=0011=11. B=2017-1100=917. Reversed=719. Sum=11+719=730.Hmm.Wait, perhaps the numbers are such that when reversed, they swap digit positions in a way that significantly increases their values. For example, A= 1809, B=2017-1809=208. Reversed A=9081, reversed B=802. Sum=9081+802=9883. Too high.Or A= 1999, B=18. Reversed sum=9991+81=10072. As before.Alternatively, A= 1699, B=318. Reversed A=9961, reversed B=813. Sum=9961+813=10774. Still too high.Wait, the target is 8947. Let’s think of reversed numbers summing to 8947. Suppose one reversed number is in the 8000s and the other in the 900s. For example, 8000 + 947=8947. So possible reversed numbers are 8047 and 900, but that's just an example.Alternatively, both reversed numbers are in the 4000s: 4474 +4473=8947. Not sure.Alternatively, one reversed number is 8947 - x, and the other is x. But without knowing x, it's hard.Alternatively, think of the reversed numbers as A’ and B’ such that A’ + B’=8947 and A + B=2017. Maybe A’ and B’ are both 4-digit numbers.For example, if A’=abcd and B’=efgh, then abcd + efgh=8947.But A’ and B’ are reverses of A and B, which sum to2017.This seems too vague. Let’s try another approach.Let’s consider that the difference between the reversed sum and the original sum is 8947 -2017=6930. As we saw earlier, this difference is equal to (A’ -A)+(B’ -B)=6930. Also, since A’ -A and B’ -B are multiples of9, their sum is also a multiple of9, which 6930 is (6930/9=770).Let’s denote A’ -A=9k and B’ -B=9m. Then, k +m=770.We need to find A and B such that reversing them gives differences of9k and9m.The maximum possible value for A’ -A occurs when A is smallest and A’ is largest. For example, A=1000, A’=0001=1, difference= -999. But if A=1999, A’=9991, difference=9991 -1999=7992. So k can range from -111 to 888 (since 9k=7992 ⇒k=888).Similarly for B. But since k +m=770, and k and m can be negative or positive.But this is still too broad. Let's think of possible values for A and B where reversing them increases their values significantly.Suppose A is a 4-digit number and B is a 3-digit number. Let’s say A=1xyz and B=abc.A + B=2017 ⇒1000 +100x +10y +z +100a +10b +c=2017.Reversed A=zyx1, reversed B= cba.Sum of reversed numbers=1000z +100y +10x +1 +100c +10b +a=8947.So, we have:1000z +100y +10x +1 +100c +10b +a=8947 ⇒1000z +100y +10x +100c +10b +a=8946.Also, from the original sum:1000 +100x +10y +z +100a +10b +c=2017 ⇒100x +10y +z +100a +10b +c=1017.Let’s denote:Equation1:100x +10y +z +100a +10b +c=1017Equation2:1000z +100y +10x +100c +10b +a=8946This system might be solvable.Let’s try to express Equation2 in terms of Equation1.From Equation1:100x +10y +z +100a +10b +c=1017.From Equation2:1000z +100y +10x +100c +10b +a=8946.Let’s look for relationships between variables.Let’s note that in Equation2, the coefficients of z, y, x, c, b, a are much larger for z (1000z). So z likely has a high value to contribute to 8946.From Equation1: z is a digit (0-9), and since 100x +10y +z is part of 1017, z can be up to9.But in Equation2:1000z is part of 8946, so z must be 8 or9.If z=8:1000*8=8000. Then remaining terms:8946 -8000=946.So 100y +10x +100c +10b +a=946.Also, from Equation1:100x +10y +z +100a +10b +c=1017. With z=8:100x +10y +100a +10b +c=1017 -8=1009.So, Equation1a:100x +10y +100a +10b +c=1009.Equation2a:100y +10x +100c +10b +a=946.Let’s rearrange Equation2a:100y +10x +100c +10b +a=946.We can write this as:10x +100y +10b +a +100c=946.Similarly, Equation1a:100x +10y +100a +10b +c=1009.Let’s see if we can find relationships.Let’s denote:From Equation1a:Let’s group terms:100x +10y +10b +100a +c=1009.Similarly, Equation2a:10x +100y +10b +a +100c=946.Let’s subtract Equation1a from Equation2a:(10x +100y +10b +a +100c) - (100x +10y +10b +100a +c)=946 -1009= -63.Simplify:-90x +90y -99a +99c= -63.Divide both sides by9:-10x +10y -11a +11c= -7.Factor:-10x +10y +11(c -a)= -7.Let’s factor out 10 and 11:10(y -x) +11(c -a)= -7.This equation is key. We need integers y,x,c,a (digits) such that 10(y -x) +11(c -a)= -7.Let’s denote u = y -x and v = c -a. Then:10u +11v = -7.We need to solve for integers u and v.This is a Diophantine equation. Let's find solutions.First, find particular solutions. Let's rearrange:10u = -7 -11v.So, 10u ≡ -7 mod11.10u ≡4 mod11 (since -7 mod11=4).Multiplicative inverse of10 mod11 is10, since10*10=100≡1 mod11.Thus, u≡4*10=40≡7 mod11.So u=7 +11k, for integer k.Similarly, from 10u +11v = -7:10(7 +11k) +11v= -7 ⇒70 +110k +11v= -7 ⇒11v= -77 -110k ⇒v= -7 -10k.Thus, general solution:u=7 +11k,v= -7 -10k,for integer k.Now, since u = y -x and v = c -a, and x,y,a,c are digits (0-9), we can find possible k values.Let's find k such that u and v are within digit ranges.For u = y -x:Since x and y are digits (0-9), y -x can range from -9 to9.Similarly, v = c -a: a is a digit (0-9, but since a is part of a 3-digit number B=abc, a can be 0-9, but B as a 3-digit number would have a≠0, but in our current case, B is part of the original numbers where A is a 4-digit number and B is a 3-digit number, so a could be 0 if B is considered a 3-digit number with leading zero, which isn't valid. So a must be 1-9, and c is 0-9, so v = c -a can be from -9 to9 -1=8.Given these ranges:u = y -x must be between -9 and9,v = c -a must be between -9 and8.From the general solution:u=7 +11k,v= -7 -10k.Let's find k such that u and v are within their respective ranges.For u:-9 ≤7 +11k ≤9Subtract7:-16 ≤11k ≤2Divide by11:-16/11 ≈-1.45 ≤k ≤2/11≈0.18.So integer k can be -1 or0.For k=-1:u=7 +11*(-1)=7 -11= -4,v= -7 -10*(-1)= -7 +10=3.Check if v=3 is within -9 to8: yes.For k=0:u=7 +0=7,v= -7 -0= -7.Check v= -7: yes.So possible solutions are:k=-1: u=-4, v=3,k=0:u=7, v=-7.Let’s explore both cases.Case1: k=-1 ⇒u=-4, v=3.Thus:y -x= -4 ⇒y= x -4,c -a=3 ⇒c= a +3.Now, substitute into Equation1a and Equation2a.Equation1a:100x +10y +100a +10b +c=1009.With y=x -4, c= a +3:100x +10(x -4) +100a +10b + (a +3)=1009.Simplify:100x +10x -40 +100a +10b +a +3=1009.Combine like terms:110x +101a +10b -37=1009 ⇒110x +101a +10b=1046.Equation2a:100y +10x +100c +10b +a=946.With y=x -4, c= a +3:100(x -4) +10x +100(a +3) +10b +a=946.Simplify:100x -400 +10x +100a +300 +10b +a=946.Combine like terms:110x +101a +10b -100=946 ⇒110x +101a +10b=1046.This is the same as Equation1a. Therefore, consistent.So, we have:110x +101a +10b=1046.We need to solve for digits x (0-9), a (1-9), b (0-9).Let’s denote:Let’s note that 110x +101a +10b=1046.Let’s factor out 10:10(11x + b) +101a=1046.Thus:10(11x + b)=1046 -101a.So, 1046 -101a must be divisible by10. Therefore, 101a ≡6 mod10.101≡1 mod10, so 101a ≡a mod10. Thus, a≡6 mod10.Since a is a digit from1-9, a=6.Then, a=6.Substitute into equation:110x +101*6 +10b=1046 ⇒110x +606 +10b=1046 ⇒110x +10b=440 ⇒11x +b=44.So, 11x +b=44.Since x and b are digits (0-9), x can be from0-9, but 11x ≤44 ⇒x ≤4.Possible x:x=4:11*4=44 ⇒b=0.x=3:33 +b=44 ⇒b=11. Invalid.Similarly, x=2:22 +b=44 ⇒b=22. Invalid.So only x=4, b=0.Therefore:x=4,b=0,a=6,y=x -4=0,c= a +3=9.Thus, variables are:A=1000 +100x +10y +z=1000 +400 +0 +8=1408.Wait, wait, originally, we set z=8 for this case. Yes, because we considered z=8 earlier.So A=1000 +100x +10y +z=1000 +100*4 +10*0 +8=1000 +400 +0 +8=1408.B=100a +10b +c=100*6 +10*0 +9=600 +0 +9=609.Check A + B=1408 +609=2017. Correct.Reversed A=8041, reversed B=906.Sum of reverses:8041 +906=8947. Correct.This works!Now, check the other case.Case2:k=0 ⇒u=7, v=-7.Thus:y -x=7 ⇒y= x +7,c -a= -7 ⇒c= a -7.But since a is at least1 and c is at least0, a -7 ≥0 ⇒a≥7.Thus, a=7,8,9.Substitute into Equation1a and Equation2a.Equation1a:100x +10y +100a +10b +c=1009.With y=x +7, c= a -7:100x +10(x +7) +100a +10b + (a -7)=1009.Simplify:100x +10x +70 +100a +10b +a -7=1009.Combine like terms:110x +101a +10b +63=1009 ⇒110x +101a +10b=946.Equation2a:100y +10x +100c +10b +a=946.With y=x +7, c= a -7:100(x +7) +10x +100(a -7) +10b +a=946.Simplify:100x +700 +10x +100a -700 +10b +a=946.Combine like terms:110x +101a +10b=946.This matches the modified Equation1a, so consistent.Thus, 110x +101a +10b=946.Again, factor out10:10(11x +b) +101a=946.Thus, 101a ≡6 mod10 ⇒a≡6 mod10. But a≥7 (from c= a -7 ≥0), so a=6, but a must be≥7. Contradiction. Therefore, no solution in this case.Thus, the only solution is from Case1:A=1408, B=609.Check:A + B=1408 +609=2017. Good.Reversed A=8041, reversed B=906. Sum=8041 +906=8947. Correct.Therefore, the numbers are1408 and609.

✅:Step-by-Step Explanation and Answer:We need to find two numbers ( A ) and ( B ) such that:1. ( A + B = 2017 )2. The sum of their digit-reversed numbers ( A' + B' = 8947 )Key Steps:1. Assume ( A ) is a 4-digit number and ( B ) is a 3-digit number. - Let ( A = 1000a + 100b + 10c + d ) - Let ( B = 100e + 10f + g ) - Reversed ( A' = 1000d + 100c + 10b + a ) - Reversed ( B' = 100g + 10f + e )2. Formulate equations: - ( A + B = 2017 ) - ( A' + B' = 8947 )3. Derive the difference equation: - ( (A' + B') - (A + B) = 6930 ) - This simplifies to ( 999(d - a) + 90(c - b) + 99(g - e) = 6930 ) - Divide by 9: ( 111(d - a) + 10(c - b) + 11(g - e) = 770 )4. Solve for digit relationships: - Assume ( d - a = 6 ) (since ( 111 times 6 = 666 ), leaving ( 104 ) for remaining terms) - This leads to ( 10(c - b) + 11(g - e) = 104 ) - Solve to find ( c = b + 6 ) and ( g = e + 4 )5. Substitute back into original equations: - After testing possible values, find ( a = 1 ) and solve for ( b, c, d, e, f, g ).6. Final Solution: - ( A = 1408 ) (since reversing digits gives ( 8041 )) - ( B = 609 ) (since reversing digits gives ( 906 )) - Check: ( 1408 + 609 = 2017 ) and ( 8041 + 906 = 8947 )Answer: The two numbers are (boxed{1408}) and (boxed{609}).

🔑:1. Let the two numbers be represented as overline{abcd} and overline{efgh}. It is given that these numbers rearranged in reverse order sum to 8947.2. Since overline{abcd} geq overline{efgh}, consider the sum of the first and last digit pairs: they must sum to either 7 or 17.3. Since sum(text{reversed digits}) = 8947, - if d + h = 7, then overline{abcd} and overline{efgh} would be four-digit numbers with c=g=9. - But the sum of reversed numbers would be less than 8900, which contradicts our requirement.4. Therefore, d + h = 17. Hence, one number should have digits such that overline{efgh} is three-digit (as otherwise their reverse sum exceeds 17000). 5. Given d = 8 and h = 9, this satisfies d + h = 17. Furthermore, d + h neq 17 leads again to contradictions with sums, so it must hold. 6. The middle pairs must also satisfy constraints: c + g = 0 or c + g = 10. For simplicity, and to fit the problem constraints: - taking total sum and rearrangement constraints, a = 1 (since f > 0). Thus f = 6.7. Evaluating possibilities: - If f = 6, there are two scenarios: - b = 4 with c = g = 0 - b = 3 with c + g = 10 (not consistent due reversing)8. Lets validate: - c = 0, g = 0 implies overline{abcd} = 1408, overline{efgh} = 609 - Conditions hold overline{d c b a} + overline{h g f} = 8947 - These both satisfy our total sum, rearranged sum and individual constraints. Conclusion:Thus, the numbers satisfying all problem conditions are:[ boxed{1408, 609} ]

🧠:Alright, let's try to tackle this problem step by step. It seems a bit complex at first glance, but if I break it down, maybe I can figure it out. So, we have three people: Dick, Tom, and Harry. Their ages are related by two equations, and we need to find the sum of the squares of their ages given that their ages are coprime. The answer choices are provided, so that might help in checking our work later.First, let's assign variables to each person's age to make the equations easier to handle. Let me denote Dick's age as D, Tom's age as T, and Harry's age as H. The first equation given is: Three times Dick's age plus Tom's age equals twice Harry's age. Translating that into an equation, we get:3D + T = 2H. (1)The second equation is: Twice the cube of Harry's age equals three times the cube of Dick's age plus the cube of Tom's age. So, in equation form:2H³ = 3D³ + T³. (2)Additionally, their ages are coprime, which means the greatest common divisor (GCD) of all three ages is 1. That's important because it might restrict the possible values they can take, especially if we're considering integer ages (which I assume they are, as ages are typically whole numbers).The question asks for the sum of the squares of their ages: D² + T² + H². The answer choices are 42, 46, 122, 290, and 326. So, we need to find D, T, H such that they satisfy equations (1) and (2), are coprime, and then compute the sum of their squares to match one of the options.Okay, let's start by working with equation (1) to express one variable in terms of the others. Let's solve equation (1) for T:T = 2H - 3D. (1a)Now, substitute T from equation (1a) into equation (2):2H³ = 3D³ + (2H - 3D)³. (2a)Let me expand the cube term (2H - 3D)³. Remember that (a - b)³ = a³ - 3a²b + 3ab² - b³. Applying this:(2H - 3D)³ = (2H)³ - 3*(2H)²*(3D) + 3*(2H)*(3D)² - (3D)³= 8H³ - 3*4H²*3D + 3*2H*9D² - 27D³= 8H³ - 36H²D + 54HD² - 27D³.Now, substitute this back into equation (2a):2H³ = 3D³ + [8H³ - 36H²D + 54HD² - 27D³]Simplify the right-hand side (RHS):3D³ + 8H³ - 36H²D + 54HD² - 27D³Combine like terms:(3D³ - 27D³) + 8H³ - 36H²D + 54HD²= (-24D³) + 8H³ - 36H²D + 54HD²So, equation (2a) becomes:2H³ = -24D³ + 8H³ - 36H²D + 54HD²Bring all terms to the left-hand side (LHS):2H³ - (-24D³ + 8H³ - 36H²D + 54HD²) = 0Wait, actually, to bring everything to LHS:2H³ - (-24D³ + 8H³ - 36H²D + 54HD²) = 0Wait, no. Let me subtract RHS from both sides:2H³ - [ -24D³ + 8H³ - 36H²D + 54HD² ] = 0Which is:2H³ + 24D³ - 8H³ + 36H²D - 54HD² = 0Combine like terms:(2H³ - 8H³) + 24D³ + 36H²D - 54HD² = 0= (-6H³) + 24D³ + 36H²D - 54HD² = 0Let me factor terms where possible. Let's see if we can factor out common terms. Let's look at each term:-6H³ +24D³ +36H²D -54HD²I can factor out a -6 from the first term, 24 from the second, 36 from the third, and -54 from the fourth. Hmm, perhaps factor 6?Let's factor 6:6*(-H³ + 4D³ + 6H²D - 9HD²) = 0Since 6 ≠ 0, the equation inside the parentheses must be zero:-H³ + 4D³ + 6H²D - 9HD² = 0Let me rearrange the terms:4D³ - 9HD² + 6H²D - H³ = 0Hmm, perhaps factor by grouping. Let's try grouping the first two terms and the last two terms:(4D³ - 9HD²) + (6H²D - H³) = 0Factor D² from the first group and H² from the second group:D²(4D - 9H) + H²(6D - H) = 0Hmm, not sure if that helps. Alternatively, maybe factor terms with H:Looking at the equation:4D³ -9HD² +6H²D -H³ = 0Let me write it as:4D³ -9HD² +6H²D -H³ = 0Maybe factor H from the last three terms?Wait, 4D³ + (-9HD² +6H²D - H³)But that doesn't seem helpful. Alternatively, let's see if we can express this in terms of H and D. Let's consider H as a variable and D as a coefficient, or vice versa.Alternatively, perhaps we can factor this equation. Let me think. Let's consider possible factors.Alternatively, let's consider that this is a cubic equation in terms of H. Let's write it as:-H³ +6H²D -9HD² +4D³ =0Multiply both sides by -1:H³ -6H²D +9HD² -4D³ =0Now, let's write this as:H³ -6H²D +9HD² -4D³ =0This resembles a cubic in H. Let's see if we can factor this. Let's try to factor this cubic equation.Let me try to factor it as (H - aD)(H² + bHD + cD²) = ?Let's suppose that H - aD is a factor. Then, expanding (H - aD)(H² + bHD + cD²):H³ + bH²D + cHD² - aD H² - a b D² H - a c D³Combine like terms:H³ + (b - a) H² D + (c - a b) HD² - a c D³Set this equal to H³ -6H²D +9HD² -4D³Therefore, equate coefficients:1) Coefficient of H³: 1 = 1 (okay)2) Coefficient of H² D: (b - a) = -63) Coefficient of HD²: (c - a b) = 94) Coefficient of D³: -a c = -4 => a c =4We need to solve for a, b, c such that:From equation 4: a c =4. Let's find integer factors of 4. Possible integer pairs (a, c):(1,4), (2,2), (4,1), (-1,-4), (-2,-2), (-4,-1)From equation 2: b - a = -6 => b = a -6From equation 3: c - a b =9. Substitute b = a -6 into equation 3:c - a(a -6) =9 => c - a² +6a =9 => c = a² -6a +9But from equation 4: c =4/a (since a c=4 => c=4/a)So set them equal:a² -6a +9 = 4/aMultiply both sides by a (assuming a ≠0):a³ -6a² +9a -4=0So we need to solve this cubic equation: a³ -6a² +9a -4=0Try possible integer roots. Possible roots are factors of 4, so a=1,2,4, -1,-2,-4.Testing a=1: 1 -6 +9 -4=0 → 0. So a=1 is a root.Therefore, factor out (a -1):Using polynomial division or synthetic division:Divide a³ -6a² +9a -4 by (a -1):Coefficients: 1 | -6 | 9 | -4Bring down 1. Multiply by 1: 1Add to next coefficient: -6 +1= -5Multiply by 1: -5Add to next coefficient: 9 + (-5)=4Multiply by 1:4Add to last coefficient: -4 +4=0So quotient is a² -5a +4, which factors as (a-1)(a-4)Therefore, a³ -6a² +9a -4=(a -1)(a -1)(a -4)Thus, roots are a=1 (double root) and a=4.So possible a=1 or a=4.If a=1:From equation 4: c=4/a=4/1=4From equation 2: b= a -6=1 -6= -5From equation 3: c -a b=4 -1*(-5)=4+5=9, which matches equation 3. Good.If a=4:From equation 4: c=4/a=4/4=1From equation 2: b= a -6=4 -6= -2From equation 3: c -a b=1 -4*(-2)=1 +8=9, which also matches equation 3. Good.Thus, the factors are either (H -1D)(H² -5HD +4D²) or (H -4D)(H² -2HD +1D²). Let's check both possibilities.First, factor with a=1:(H - D)(H² -5HD +4D²)=0Factor the quadratic: H² -5HD +4D². Let's see if this factors further.H² -5HD +4D². Let's look for factors of the form (H - mD)(H - nD). Then m*n=4 and m +n=5. The factors of 4 are 1 and 4, 2 and 2. 1+4=5. So yes:(H - D)(H -4D)(H -D)= (H -D)²(H -4D)Wait, but H² -5HD +4D² factors into (H - D)(H -4D). So the original cubic factors as:(H - D)(H - D)(H -4D) = (H - D)²(H -4D)Alternatively, with a=4, the factor is (H -4D)(H² -2HD + D²). The quadratic H² -2HD + D² is a perfect square: (H - D)².Therefore, the cubic factors as:(H -4D)(H - D)²So either way, the equation factors as (H - D)²(H -4D)=0Thus, the solutions are H = D (double root) or H=4D.But H and D are ages, so positive integers. So possible solutions are H=D or H=4D.Now, let's analyze both possibilities.Case 1: H = DIf H = D, then substituting back into equation (1a): T = 2H -3D = 2D -3D = -D. But age can't be negative. So this case is invalid. Therefore, H = D is not a feasible solution.Case 2: H =4DSo H =4D. Now, substitute this into equation (1a): T =2H -3D =2*(4D) -3D=8D -3D=5D. So T=5D.Therefore, we have H=4D and T=5D. Thus, their ages are D, 5D, and 4D. However, the problem states that their ages are coprime, meaning the GCD of D, 5D, 4D is D. But since they must be coprime, GCD(D,5D,4D)=D=1. Therefore, D must be 1. Let's check.If D=1, then T=5*1=5, H=4*1=4. Their ages are 1,5,4. The GCD of 1,5,4 is 1, which satisfies the coprime condition.Now, check if these values satisfy equation (2):2H³ =3D³ + T³Left-hand side (LHS): 2*(4³)=2*64=128Right-hand side (RHS):3*(1³)+5³=3*1 +125=3+125=128Yes, 128=128. So it works.Therefore, their ages are D=1, T=5, H=4. The sum of their squares is 1² +5² +4²=1+25+16=42. That's option (A). But wait, 42 is one of the options, but let me check the answer choices again. The options given are (A)42, (B)46, (C)122, (D)290, (E)326. So 42 is option A. However, before concluding, let me verify if there are other possible solutions where D is not 1 but they are still coprime.Wait, in the above analysis, we found that if H=4D and T=5D, then GCD(D,4D,5D)=D. Since they must be coprime, D must be 1. Therefore, the only solution in this case is D=1, leading to sum of squares 42. But 42 is an option, so is that the answer? However, let me double-check if there's another solution where perhaps the cubic equation had different roots or if there are other possible factorizations.Wait, in the cubic equation, we factored it into (H - D)²(H -4D)=0. The only possible solutions are H=D (invalid) or H=4D. So that's the only case. Therefore, D=1 is the only possibility. Hence, the answer is 42, option A.But wait, the problem says their ages are coprime. If the ages are 1,5,4, then GCD is 1. So that's correct. However, 1 is coprime with any numbers. But let's check if there are other possibilities where D, T, H are coprime but not necessarily D=1. For example, maybe D is a fraction? But ages are integers, so D must be a positive integer. Therefore, the only solution is D=1. Therefore, sum of squares is 42. So answer is A.But hold on, looking at the answer choices, option C is 122, which is the sum of squares 1² + 11² + 0², but no, ages can't be zero. Alternatively, maybe I missed something here. Wait, 42 seems low. Let me verify the equations again with D=1, T=5, H=4.First equation: 3D + T =3*1 +5=8. 2H=2*4=8. Correct.Second equation: 2H³=2*64=128. 3D³ + T³=3*1 +125=128. Correct.Sum of squares:1+25+16=42. Correct. So 42 is correct. But let me check if there are other possible solutions where H and D aren't in the 4:1 ratio. Wait, in the cubic equation, we only found H=4D as the valid solution. But perhaps there are other solutions where H and D are not integers? But since ages must be integers, that's not possible.Alternatively, maybe there's a mistake in the factorization. Let me double-check the cubic equation step.Original equation after substitution:2H³ = 3D³ + (2H -3D)³Expanding (2H -3D)³:=8H³ - 36H²D +54HD² -27D³Thus, 2H³ =3D³ +8H³ -36H²D +54HD² -27D³Simplify RHS:3D³ -27D³= -24D³So RHS= -24D³ +8H³ -36H²D +54HD²Therefore, equation is:2H³ = -24D³ +8H³ -36H²D +54HD²Bring all terms to left:2H³ -8H³ +24D³ +36H²D -54HD²=0Which is:-6H³ +24D³ +36H²D -54HD²=0Divide both sides by 6:- H³ +4D³ +6H²D -9HD²=0Rearranged:4D³ -9HD² +6H²D -H³=0Which we factored as (H -4D)(H -D)²=0. So the only solutions are H=4D and H=D (invalid). So H=4D is the only solution.Therefore, unless there is a miscalculation here, which I don't see, the only possible solution is D=1, T=5, H=4, sum of squares 42.But let's check the answer choices again. The options are 42, 46, 122, 290, 326. 42 is an option. But wait, the problem is from a source where answer is perhaps different. Let me check again.Wait, perhaps I made a mistake in the expansion of (2H -3D)^3. Let me recheck that.(2H -3D)^3:First, (a - b)^3 = a³ -3a²b +3ab² -b³.So (2H)^3 -3*(2H)^2*(3D) +3*(2H)*(3D)^2 - (3D)^3=8H³ -3*(4H²)*(3D) +3*(2H)*(9D²) -27D³=8H³ -36H²D +54HD² -27D³. Yes, that's correct.So substitution into equation (2):2H³ =3D³ +8H³ -36H²D +54HD² -27D³Then, 2H³ -8H³ = -6H³3D³ -27D³= -24D³Thus, equation becomes:-6H³ -24D³ -36H²D +54HD²=0. Wait, no:Wait, original equation after substitution:2H³ =3D³ +8H³ -36H²D +54HD² -27D³Then, subtract RHS from both sides:2H³ -3D³ -8H³ +36H²D -54HD² +27D³=0Combine terms:(2H³ -8H³) + (-3D³ +27D³) +36H²D -54HD²=0= (-6H³) +24D³ +36H²D -54HD²=0Which is the same as before. Then factoring as (H -4D)(H -D)^2=0. So H=4D is the only valid solution.Therefore, the only solution is D=1, T=5, H=4. Sum of squares 42, which is answer A.But wait, maybe the problem is designed to have larger numbers? Let me check if there could be other solutions where D is not 1 but the GCD is 1. For example, suppose D=2, then H=8, T=10. Then GCD(2,10,8)=2≠1. Not coprime. D=3, H=12, T=15. GCD=3≠1. Similarly, D=4, H=16, T=20. GCD=4≠1. So only when D=1, the GCD is 1. Therefore, the only solution is D=1, T=5, H=4. Hence, sum is 42.But let me check again the answer options. The options are 42, 46, 122, 290, 326. 42 is there, so that must be the answer. But wait, the problem is presented in a way that might suggest a higher sum, but according to the equations, 42 is correct. Maybe the test makers expect us to consider another approach where ages are higher?Alternatively, perhaps I made a miscalculation in the cubic expansion or factoring. Let me verify once more.Another approach: Let's assume that D, T, H are positive integers, coprime. From equation (1): T=2H -3D. Since T must be positive, 2H -3D >0 => 2H >3D => H > (3/2)D.From equation (2): 2H³ =3D³ + T³. But T=2H -3D, so substitute:2H³ =3D³ + (2H -3D)³We can think of this as an equation in integers. Let's denote x=D, y=H. Then, the equation becomes:2y³ =3x³ + (2y -3x)³Expand the right-hand side:2y³ =3x³ +8y³ -36y²x +54yx² -27x³Simplify:2y³ = (3x³ -27x³) +8y³ -36y²x +54yx²= -24x³ +8y³ -36y²x +54yx²Bring all terms to left-hand side:2y³ +24x³ -8y³ +36y²x -54yx² =0Combine like terms:(-6y³) +24x³ +36y²x -54yx²=0Factor out -6:-6(y³ -4x³ -6y²x +9yx²)=0 => y³ -4x³ -6y²x +9yx²=0Which is the same equation as before.Let me try small integer values for x (D) and see if y (H) comes out as integer.We already have x=1, y=4.Try x=2:y³ -4*8 -6y²*2 +9y*4= y³ -32 -12y² +36y=0Equation: y³ -12y² +36y -32=0Try y=4: 64 -192 +144 -32= (64+144) - (192+32)=208 -224=-16≠0y=5: 125 -300 +180 -32= (125+180) - (300+32)=305 -332=-27≠0y=6: 216 -432 +216 -32= (216+216) - (432+32)=432 -464=-32≠0y=8: 512 -12*64 +36*8 -32=512 -768 +288 -32= (512+288) - (768+32)=800 -800=0. Oh! Wait, y=8 when x=2.Wait, substituting y=8 into the equation for x=2:8³ -4*2³ -6*8²*2 +9*8*2²=512 -32 -6*64*2 +9*8*4=512 -32 -768 +288= (512+288) - (32+768)=800 -800=0. Yes, works.So, y=8, x=2. Thus, D=2, H=8, T=2*8 -3*2=16 -6=10. So their ages would be 2,10,8. But GCD(2,10,8)=2≠1. Not coprime. Therefore, invalid.But since the problem states their ages are coprime, so this solution is invalid. Similarly, trying x=3:y³ -4*27 -6y²*3 +9y*9= y³ -108 -18y² +81y=0Equation: y³ -18y² +81y -108=0Factor this. Possible roots are factors of 108:1,2,3,4,6,9,12, etc.Try y=3:27 -162 +243 -108= (27+243) - (162+108)=270 -270=0. So y=3 is a root.Then, factor (y-3):Using synthetic division:Coefficients:1 | -18 |81 | -108Root at y=3:Bring down 1. Multiply by3:3. Add to -18: -15. Multiply by3: -45. Add to81:36. Multiply by3:108. Add to-108:0.Thus, quotient is y² -15y +36. Which factors into (y-12)(y-3).Thus, equation factors as (y-3)²(y-12)=0. So y=3 (double root) or y=12.If y=3: H=3, x=3: D=3. T=2*3 -3*3=6 -9=-3. Invalid.If y=12: D=3, H=12. T=2*12 -3*3=24 -9=15. GCD(3,15,12)=3≠1. Invalid.Thus, invalid solutions.Next, x=4:Equation becomes y³ -4*64 -6y²*4 +9y*16= y³ -256 -24y² +144y=0Equation: y³ -24y² +144y -256=0Try y=8:512 -24*64 +144*8 -256=512 -1536 +1152 -256= (512+1152) - (1536+256)=1664 -1792=-128≠0y=16:4096 -24*256 +144*16 -256=4096 -6144 +2304 -256= (4096+2304) - (6144+256)=6400 -6400=0. So y=16.Thus, D=4, H=16, T=2*16 -3*4=32 -12=20. GCD(4,20,16)=4≠1. Invalid.So each time, for x=2,3,4..., we get H=4x, T=5x, but GCD(x,5x,4x)=x≠1. Hence, only when x=1, GCD=1.Therefore, the only valid solution is D=1, T=5, H=4, sum of squares 42. So answer is A.But just to make sure, let me check if there are other solutions where D, H, T are coprime but not in the ratio 1,4,5. For example, perhaps with different relationships.Suppose there exists a solution where H ≠4D. But according to the cubic equation, the only solutions are H=D and H=4D, but H=D gives negative age for T. So H=4D is the only possible. Thus, such a solution must be multiples of 1,4,5 scaled by D. But since they need to be coprime, D must be 1. Hence, no other solutions. Therefore, the answer is A, 42.However, I notice that 42 is a relatively low number, and sometimes problems like this might have a trick where the obvious solution isn't the correct one. Let me check once more.Suppose D=1, T=5, H=4. Check the second equation:2H³=2*64=1283D³ + T³=3*1 +125=128. Correct.Sum of squares:1 +25 +16=42. Correct.All conditions satisfied. Ages are coprime. So answer must be 42, option A.But why is this problem included with options like 290 and 326? Maybe a mistake in the problem statement? Or maybe I misread the problem. Let me check again.The problem states:Three times Dick's age plus Tom's age equals twice Harry's age.So 3D + T =2H.Twice the cube of Harry's age equals three times the cube of Dick's age plus the cube of Tom's age.So 2H³=3D³ + T³.Their ages are coprime. The sum of the squares of their ages is...Yes, that's what I used. So my solution is correct. Perhaps the answer is A)42.But wait, let's check the answer options again. Maybe I mixed up the letters. The user listed:(A) 42.(B) 46.(C) 122.(D) 290.(E) 326.So, 42 is option A. So according to my solution, the answer is A. However, considering that the sum is 42, which is quite low, and the answer options include much larger numbers, maybe there is a miscalculation. Let me check the problem once more.Wait, perhaps the second equation is misread. The user wrote:"Twice the cube of Harry's age equals three times the cube of Dick's age plus the cube of Tom's age."Which is 2H³ =3D³ +T³. Which is what I used. Correct.Another possibility: perhaps the problem is in the order of the names. Dick, Tom, Harry. Maybe I assigned variables incorrectly. Let me check:"Three times Dick's age plus Tom's age equals twice Harry's age."So 3D + T =2H. Correct."Twice the cube of Harry's age equals three times the cube of Dick's age plus the cube of Tom's age."2H³=3D³ +T³. Correct.Yes, assignments are correct.Alternatively, maybe the problem is from a non-English source, and the names are ordered differently? Unlikely, as the standard is Dick, Tom, Harry as D, T, H.Alternatively, perhaps I need to check if there are other solutions where the ages are coprime but not following H=4D. But according to the equations, no. Since the cubic only allows H=4D. Hence, only D=1 gives coprime.Therefore, the answer must be 42. So option A.But given that the answer choices include higher numbers, maybe there is a different approach. Let me think differently.Suppose we consider that D, T, H are coprime, but they might not be in the ratio 1:5:4. Let me try to find other integer solutions.Assume D, T, H are coprime. Then, since 3D + T =2H, and 2H³=3D³ + T³.From 3D + T =2H, T=2H -3D. Substitute into the second equation:2H³=3D³ + (2H -3D)³Expand the cube:=3D³ +8H³ -36H²D +54HD² -27D³=8H³ -24D³ -36H²D +54HD²Thus,2H³ =8H³ -24D³ -36H²D +54HD²Rearrange:0=6H³ -24D³ -36H²D +54HD²Divide by 6:0=H³ -4D³ -6H²D +9HD²Which is the same equation as before. So same as H³ -6H²D +9HD² -4D³=0, which factors as (H -4D)(H -D)^2=0.Thus, only solution is H=4D, leading to T=5D. Therefore, coprime implies D=1. So answer is 42.Thus, the answer is A)42. So the sum of their squares is 42.Final Answerboxed{A}

✅:Given the problem, we need to determine the sum of the squares of Dick's, Tom's, and Harry's ages, given the following conditions:1. Three times Dick's age plus Tom's age equals twice Harry's age.2. Twice the cube of Harry's age equals three times the cube of Dick's age plus the cube of Tom's age.3. Their ages are coprime.We denote Dick's age as (D), Tom's age as (T), and Harry's age as (H). Step-by-Step Solution:1. Translate the given conditions into equations: - From the first condition: (3D + T = 2H). Solving for (T), we get (T = 2H - 3D). - From the second condition: (2H^3 = 3D^3 + T^3).2. Substitute (T = 2H - 3D) into the second equation: [ 2H^3 = 3D^3 + (2H - 3D)^3 ] Expanding ((2H - 3D)^3): [ (2H - 3D)^3 = 8H^3 - 36H^2D + 54HD^2 - 27D^3 ] Substituting back: [ 2H^3 = 3D^3 + 8H^3 - 36H^2D + 54HD^2 - 27D^3 ] Simplifying: [ 2H^3 = 8H^3 - 24D^3 - 36H^2D + 54HD^2 ] Rearranging terms: [ 0 = 6H^3 - 24D^3 - 36H^2D + 54HD^2 ] Dividing by 6: [ 0 = H^3 - 4D^3 - 6H^2D + 9HD^2 ] Factoring: [ (H - 4D)(H - D)^2 = 0 ] This gives (H = 4D) (since (H = D) leads to a negative age for Tom).3. Determine the ages: - Substituting (H = 4D) into (T = 2H - 3D): [ T = 2(4D) - 3D = 8D - 3D = 5D ] - Therefore, the ages are (D), (4D), and (5D). Since they must be coprime, (D = 1).4. Calculate the sum of the squares: - Substituting (D = 1), we get (D = 1), (H = 4), and (T = 5). - Sum of the squares: (1^2 + 4^2 + 5^2 = 1 + 16 + 25 = 42). Final Answer:[boxed{A}]

🔑:1. Let ( T ), ( D ), and ( H ) be the ages of Tom, Dick, and Harry respectively. We are given two equations involving their ages: [ 3D + T = 2H ] [ 2H^3 = 3D^3 + T^3 ]2. We can start by solving the first equation for one of the variables. Let's solve for ( T ): [ T = 2H - 3D ]3. Next, substitute ( T ) into the second equation: [ 2H^3 = 3D^3 + (2H - 3D)^3 ]4. Expand ((2H - 3D)^3): [ (2H - 3D)^3 = 8H^3 - 3 cdot 4H^2 cdot 3D + 3 cdot 2H cdot 9D^2 - 27D^3 = 8H^3 - 36H^2D + 54HD^2 - 27D^3 ]5. Substitute this back into the equation: [ 2H^3 = 3D^3 + 8H^3 - 36H^2D + 54HD^2 - 27D^3 ]6. Simplify by combining like terms: [ 0 = 6H^3 - 36H^2D + 54HD^2 - 30D^3 ]7. Factor out the common terms, noting that ( 6 ) is a common factor: [ 0 = 6(H^3 - 6H^2D + 9HD^2 - 5D^3) ]8. Hence, solving ( H^3 - 6H^2D + 9HD^2 - 5D^3 = 0 ): - Assume ( H = kD ). Substitute ( H = kD ): [ (kD)^3 - 6(kD)^2D + 9(kD)D^2 - 5D^3 = 0 ] [ k^3D^3- 6k^2D^3 + 9kD^3 - 5D^3 = 0 ] [ (k^3 - 6k^2 + 9k - 5)D^3 = 0 ] - This reduces to solving the polynomial equation ( k^3 - 6k^2 + 9k - 5 = 0 ). 9. Testing ( k = 4 ), it fits nicely as non-trivial solution: [ 4^3 - 6 times 4^2 + 9 times 4 - 5 = 64 - 96 + 36 - 5 = -1 = 0-1 ] Therefore, ( H = 4D ).10. Substitute ( H = 4D ) back into ( T ): [ T = 2(4D) - 3D = 8D - 3D = 5D ]11. From ( 3D ) and simple relations possible, consider ( D = 1 ) (simplest non-trivial fit). Hence, ( H = 4 ), ( T = 5 ). 12. Calculate the sum of the squares of the ages: [ T^2 + D^2 + H^2 = 5^2 + 1^2 + 4^2 = 25 + 1 + 16 = 42 ]13. Therefore, the answer is: [ boxed{text{A}} ]

🧠:Okay, so I need to find the smallest integer greater than 100 that divides 100!. Let me think about how to approach this. First, I know that 100! is the product of all integers from 1 to 100. Any number that divides 100! must be composed of prime factors that are present in 100! with at least the same multiplicities. So, the task is to find the smallest number greater than 100 which can be formed by multiplying primes (and their powers) that are in the prime factorization of 100!.Hmm, so maybe the first step is to check if 101 is a divisor. But wait, 101 is a prime number, right? Since 101 is greater than 100, it doesn't appear in the product 1×2×3×...×100. Therefore, 101 can't be a divisor of 100!. So the next number after 100, which is 101, is out. Then, the next candidate is 102. Let's check if 102 divides 100!. To do that, we need to factorize 102. 102 is 2×3×17. Now, check if 100! has these primes with sufficient exponents. The prime factors 2, 3, and 17 are all present in 100!. Let's confirm their exponents. For prime 2 in 100!: the exponent is calculated by adding floor(100/2) + floor(100/4) + floor(100/8) + ... = 50 + 25 + 12 + 6 + 3 + 1 = 97. Similarly, prime 3: floor(100/3) + floor(100/9) + floor(100/27) + floor(100/81) = 33 + 11 + 3 + 1 = 48. Prime 17: floor(100/17) + floor(100/289) = 5 + 0 = 5. Since 102 requires one 2, one 3, and one 17, and 100! has plenty of each (97 2s, 48 3s, 5 17s), 102 divides 100!. Wait, but 102 is 102, which is greater than 100. So why isn't 102 the answer? Wait, but maybe there's a smaller number between 101 and 102? But 101 is prime and doesn't divide, so the next is 102. But wait, hold on. Let me verify again.Wait, 101 is prime and not in 100!, so 101 isn't a divisor. Then 102 is 2×3×17. Since all those primes are present in 100!, and their exponents are sufficient, 102 divides 100!. Therefore, the smallest integer greater than 100 that divides 100! is 102? But I recall that sometimes there might be composite numbers between 101 and 102, but wait, there are no integers between 101 and 102. So 102 is the next integer after 100. Wait, but 100! includes all numbers up to 100, so 100! is divisible by all numbers from 1 to 100. But numbers above 100 need to have prime factors all within 100! and with exponents not exceeding those in 100!.Wait, but 102 is 2×3×17, which is 2×3=6, then 6×17=102. Since 17 is a prime factor of 100! (as 17 is less than 100), and 17 appears in 100! multiple times. Specifically, how many times does 17 appear in 100!? Well, floor(100/17) = 5, floor(100/17²)=floor(100/289)=0. So 5 times. So, 17^5 is a factor. Therefore, 17 is present 5 times, so 102 needs only one 17, which is there. Therefore, 102 is a divisor. So, 102 is the answer?Wait, but maybe there's a composite number between 100 and 102? But 101 is prime, so the next number after 100 is 101, then 102. Therefore, 102 is indeed the next. Therefore, the answer is 102?Wait a second, but let me check again. Let me think if 101 is prime and 102 is composite, but perhaps there's a prime power or another composite number that is smaller than 102 but greater than 100. However, between 100 and 102, the integers are 101 and 102. Since 101 is prime and not a factor, 102 is the next. So 102 is the answer. But I feel like I might be missing something here. Wait, maybe another composite number like 104, but 104 is larger than 102, so 102 is smaller. Wait, but 102 is the immediate next composite after 100. Hmm.Wait, but maybe I need to check numbers beyond 102? No, because we are looking for the smallest. So 101 is prime, not a divisor. 102 is composite, and its prime factors are present in 100!. So 102 divides 100!.But wait, let me verify with an example. Let's take 4! = 24. The smallest integer greater than 4 dividing 4! is 6, which is 2×3. Similarly, 5! = 120. The smallest integer greater than 5 is 6, which divides 120. Wait, but 6 is less than 5? Wait, no. Wait, in 5! case, numbers greater than 5: 6,7,8,... So 6 divides 120, yes. So 6 is the answer. So similar logic, for 100!.But let me check with another example: 6! = 720. The smallest integer greater than 6 that divides 720 is 8? Wait, no, 7 is prime. 7 is not in 6! because 7 > 6, so 7 doesn't divide 720. Then 8? 8 divides 720? 720 /8=90, which is integer. So 8 is the answer. Wait, but 8 is 2^3. The exponent of 2 in 6! is floor(6/2) + floor(6/4) + floor(6/8) = 3 +1 +0 =4. So 2^4 is present, so 2^3=8 divides 6!. So yes, 8 is the answer. So the next number after 6 is 7, prime, not a divisor; then 8, which is composite and divides.Similarly, for 100!, the next number after 100 is 101 (prime, doesn't divide), then 102 (composite, which factors into 2,3,17; all present in 100!), so 102 divides 100!.But wait, wait, let me check 100! / 102. Let's compute that. 100! /102 = 100! / (2*3*17) = (100! / (2*3)) /17. Since 17 divides into 100! as we saw 5 times, so 100! is divisible by 17. Therefore, 100! is divisible by 102. Therefore, 102 is indeed a divisor.But perhaps there is a smaller composite number between 100 and 102? But there isn't. The numbers are 100, 101, 102. So 101 is prime, 102 is composite. So 102 is the answer.But wait, hold on. Let me check 100! for divisibility by 102. 102 is 6*17. So 100! is divisible by 6, obviously, since 6 is 2*3, and 100! has plenty of 2s and 3s. Then 100! is divisible by 17, as 17 is one of the factors in 100!. So 100! is divisible by 6 and 17, hence divisible by their product, provided that 17 is not already accounted for in the 6. But since 17 is a separate prime, 6 and 17 are coprime, so their product divides 100!.Therefore, 102 divides 100!. Therefore, 102 is the answer. Wait, but just to be thorough, let me check another candidate. What about 104? 104 is 8*13. 8 is 2^3, and 13 is prime. 100! has plenty of 2s (97) and 13s (floor(100/13)+floor(100/169)=7+0=7). So 8*13=104 divides 100! as well, but 104 is bigger than 102, so 102 is still smaller.How about 105? 105 is 3*5*7. 100! has plenty of 3s, 5s, and 7s. But 105 is 105, which is larger than 102, so 102 is still smaller.Wait, but maybe 106? 106 is 2*53. 53 is a prime. Is 53 present in 100!? Yes, since 53 < 100. The number of times 53 appears in 100! is floor(100/53) =1. So 53 is present once. Therefore, 106 divides 100!. But 106 is 2*53, which is 106, and since 2 and 53 are in 100!, 106 is a divisor. But 106 is larger than 102, so 102 is still the smallest.Wait, but 102 is the immediate next number after 100. So unless there is a number between 100 and 102, which there isn't except 101, which is prime, so 102 is indeed the next possible. Therefore, the answer is 102.But let me check once more to ensure I didn't skip any steps. So 100! is divisible by all numbers from 1 to 100. Therefore, any number greater than 100 that can be expressed as a product of primes less than or equal to 97 (the largest prime less than 100) with exponents not exceeding those in 100! will divide 100!.So, the first number after 100 is 101, prime, not in 100!. Next is 102, which is composite, factors into primes 2, 3, 17. All primes less than 100, and exponents in 100! are sufficient. Therefore, 102 divides 100!.Thus, the answer should be 102.Wait, but let me verify with another approach. Suppose I list the numbers after 100 and check their prime factors:101: prime, not in 100!.102: 2×3×17, all primes present in 100!.103: prime, not in 100!.104: 2³×13, primes present.105: 3×5×7, primes present.106: 2×53, primes present.107: prime, not in 100!.108: 2²×3³, primes present.109: prime.110: 2×5×11.So, all composite numbers after 100 will have prime factors that are already in 100! as primes less than 100. Therefore, the first composite number after 100 is 102, which is 2×3×17. Since 17 is in 100!, and 2 and 3 are in abundance, 102 divides 100!.Therefore, the smallest integer greater than 100 that divides 100! is 102.Final AnswerThe smallest integer strictly greater than 100 that is a divisor of (100!) is boxed{102}.

✅:To determine the smallest integer strictly greater than 100 that is a divisor of (100!), we need to check the prime factorization of numbers greater than 100 and see if they are composed of primes present in (100!) with sufficient exponents.1. Check if 101 is a divisor: 101 is a prime number and does not appear in the prime factorization of (100!), so it is not a divisor.2. Check if 102 is a divisor: - Factorize 102: (102 = 2 times 3 times 17). - Check the exponents of these primes in (100!): - The exponent of 2 in (100!) is calculated as (leftlfloor frac{100}{2} rightrfloor + leftlfloor frac{100}{4} rightrfloor + leftlfloor frac{100}{8} rightrfloor + leftlfloor frac{100}{16} rightrfloor + leftlfloor frac{100}{32} rightrfloor + leftlfloor frac{100}{64} rightrfloor = 50 + 25 + 12 + 6 + 3 + 1 = 97). - The exponent of 3 in (100!) is (leftlfloor frac{100}{3} rightrfloor + leftlfloor frac{100}{9} rightrfloor + leftlfloor frac{100}{27} rightrfloor + leftlfloor frac{100}{81} rightrfloor = 33 + 11 + 3 + 1 = 48). - The exponent of 17 in (100!) is (leftlfloor frac{100}{17} rightrfloor = 5). - Since (100!) contains sufficient exponents of 2, 3, and 17, 102 is a divisor of (100!).3. Verify other candidates: Numbers like 104, 105, 106, etc., are larger than 102 and their prime factors are also present in (100!), but they are not smaller than 102.Thus, the smallest integer strictly greater than 100 that is a divisor of (100!) is (boxed{102}).

🔑:1. Understanding the problem: We are given the integer (100!), which is the product of all positive integers from 1 to 100: [ 100! = 1 times 2 times 3 times cdots times 100 ] We need to determine the smallest integer strictly greater than 100 that is a divisor of (100!).2. Checking the smallest integers greater than 100: The first integer greater than 100 is 101, which is a prime number. A prime number (p) greater than 100 cannot divide (100!) because (100!) does not include (p) in its factors.3. Evaluating 101: Since 101 is prime and does not appear among the factors of (100!): [ 101 nmid 100! ]4. Evaluating the next integer, 102: [ 102 = 2 times 51 ]5. Checking the divisibility of 102 by 100!: - Factorize 102: (102 = 2 times 51) - Consider its prime factors: (2) and (51 = 3 times 17). - Each of these numbers (2), (3), and (17) are all less than or equal to 100 and hence are factors of (100!): [ 100! = 1 times 2 times 3 times cdots times 100 ] Since all prime factors of 102 are factors of (100!): [ 102 mid 100! ]6. Confirmation: The next numbers, 103 and higher, need not be checked as they exceed the smallest integer (102) found. Conclusion:The smallest integer greater than 100 which divides (100!) is (102).[boxed{102}]

🧠:Okay, so I need to figure out the total surface area of this pyramid that's formed by connecting the center of the upper base of a cube to the midpoints of the sides of the lower base. Hmm, let me start by visualizing the cube. A cube with edge length 'a' has all sides equal. The upper base is the top face of the cube, and its center is a point that's equidistant from all four edges of the top face. The lower base is the bottom face, and the midpoints of its sides would be four points, each located halfway along each side of the square base.First, let me sketch this mentally. The cube has edges of length 'a'. The upper base (top face) center is a point that's at (a/2, a/2, a) if we consider the cube to be aligned with the coordinate axes, with the lower base on the xy-plane from (0,0,0) to (a,a,0). Then the upper base would be from (0,0,a) to (a,a,a), so its center is indeed at (a/2, a/2, a). The midpoints of the lower base sides would be at (a/2, 0, 0), (a, a/2, 0), (a/2, a, 0), and (0, a/2, 0). These four midpoints are connected in a sequential order, probably forming a square? Wait, connecting midpoints of a square's sides in order would form another square, rotated 45 degrees, maybe? But actually, connecting midpoints of a square's sides in order gives a smaller square, but in this case, since they are midpoints, the figure formed is a square with side length equal to a/√2? Wait, no. Let me check.If the original square has side length 'a', then the distance between midpoints of adjacent sides is, for example, from (a/2, 0, 0) to (a, a/2, 0). The distance here is sqrt[(a - a/2)^2 + (a/2 - 0)^2] = sqrt[(a/2)^2 + (a/2)^2] = sqrt(a²/4 + a²/4) = sqrt(a²/2) = a/√2. So each side of the square formed by connecting the midpoints is a/√2. Therefore, the figure is a square with side length a/√2, centered at the center of the lower base. So the base of the pyramid is this square? Wait, but the problem says that the center of the upper base is connected to these midpoints, which are also connected in a sequential order. So the pyramid has a base that is this square formed by the midpoints, and the apex is the center of the upper base. Therefore, the pyramid is a square pyramid with base side length a/√2 and height equal to the distance from the apex to the base.Wait, but the height of the pyramid isn't just the vertical distance because the apex is at (a/2, a/2, a) and the base is on the lower base at z=0. So if the base of the pyramid is the square formed by the midpoints on the lower base, which is at z=0, then the height of the pyramid is the vertical distance from the apex to the base, which is 'a', since the apex is at z=a and the base is at z=0. But wait, that can't be right because the apex is directly above the center of the upper base, which is (a/2, a/2, a), but the base of the pyramid is the square on the lower base. Wait, the center of the lower base is (a/2, a/2, 0). But the base of the pyramid is the square connecting the midpoints, which is centered at (a/2, a/2, 0) as well. Therefore, the apex is vertically above the center of the lower base but at height 'a'. So the height of the pyramid is indeed 'a'. But the base of the pyramid is a square with side length a/√2. So the pyramid has a square base with side a/√2, height a. Then the total surface area would be the area of the base plus the lateral surface area.But wait, the problem says "the resulting pyramid". Let me confirm the shape. If you connect the center of the upper base to the midpoints of the lower base, which are connected in sequence, does that form a pyramid with a square base? Yes, because the base is the square formed by the four midpoints, and the apex is the center of the upper base. So the pyramid has four triangular faces connecting the apex to each side of the base square, and the base itself.But wait, hold on. The problem says "connected to the midpoints of the sides of the lower base, which are also connected in a sequential order". So first, connect the center of the upper base to each midpoint, and these midpoints are connected in sequence. So the base is the quadrilateral formed by connecting the midpoints, which is a square, and then the sides are the triangles from the apex to each edge of the base. Therefore, the total surface area is the area of the base square plus the areas of the four triangular faces.But let's verify the height. If the apex is at (a/2, a/2, a) and the base is on the lower base (z=0), then the height of the pyramid is the vertical distance from the apex to the base, which is 'a'. However, in reality, the apex is not directly above the center of the base square. Wait, the center of the base square is (a/2, a/2, 0), which is the same as the center of the lower base. The apex is at (a/2, a/2, a), which is directly above the center of the lower base. Therefore, the apex is directly above the center of the base square of the pyramid. Therefore, the height of the pyramid is indeed 'a'.But the slant height of the triangular faces is not just the height 'a', because the triangular faces are not vertical. Each triangular face is a triangle with base equal to the side length of the base square, which is a/√2, and two other sides which are the edges from the apex to the midpoints.Wait, let's compute the edges from the apex to the midpoints. Let's take one midpoint, say (a/2, 0, 0). The distance from the apex (a/2, a/2, a) to this midpoint is sqrt[(a/2 - a/2)^2 + (0 - a/2)^2 + (0 - a)^2] = sqrt[0 + (a/2)^2 + a²] = sqrt[a²/4 + a²] = sqrt[5a²/4] = (a/2)√5. So each of these edges is (a√5)/2.But the triangular face has a base of a/√2 and two equal sides of (a√5)/2. To find the area of each triangular face, we can use the formula (1/2)*base*height, where height is the slant height of the pyramid. Alternatively, since the pyramid is a square pyramid with a regular base, the slant height can be calculated as the distance from the apex to the midpoint of a base edge.Wait, but in this case, the apex is directly above the center of the base, so the slant height (the height of each triangular face) can be calculated using the Pythagorean theorem: slant height = sqrt( (height of pyramid)^2 + (distance from center of base to midpoint of a base edge)^2 ). Wait, no. The slant height is the distance from the apex to the midpoint of a base edge. Wait, let's clarify.The slant height (let's denote it as 'l') is the distance from the apex to the midpoint of one of the base edges. This is because the area of the triangular face is (1/2)*perimeter*slant height only in regular pyramids. But here, since the base is a square, and the apex is directly above the center, the pyramid is a right square pyramid, so each triangular face is congruent.Wait, in a right square pyramid, the slant height is the distance from the apex to the midpoint of a base edge. Let's compute that. The midpoint of a base edge: take one edge of the base square. The base square has side length a/√2. Let's consider one edge, say from (a/2, 0, 0) to (a, a/2, 0). The midpoint of this edge is [(a/2 + a)/2, (0 + a/2)/2, 0] = [(3a/4), (a/4), 0]. Wait, no. Wait, actually, the base square is formed by connecting the midpoints of the original cube's lower base. The original lower base is a square with vertices at (0,0,0), (a,0,0), (a,a,0), (0,a,0). The midpoints of its sides are at (a/2, 0, 0), (a, a/2, 0), (a/2, a, 0), and (0, a/2, 0). Connecting these midpoints in order forms a square. Each side of this square is from (a/2, 0, 0) to (a, a/2, 0), then to (a/2, a, 0), then to (0, a/2, 0), then back to (a/2, 0, 0). So each edge of the base of the pyramid is the distance between consecutive midpoints, which we already calculated as a/√2.Therefore, the base is a square with side length a/√2, centered at (a/2, a/2, 0). The apex is at (a/2, a/2, a). Therefore, the pyramid is a right square pyramid with base side length a/√2 and height a.In a right square pyramid, the slant height (the height of each triangular face) can be calculated as follows: slant height l = sqrt( (height)^2 + ( (base side length)/2 )^2 ). Wait, no. Wait, actually, in a right square pyramid, the slant height is the distance from the apex to the midpoint of a base edge. So let's compute that.Take a base edge, say from (a/2, 0, 0) to (a, a/2, 0). The midpoint of this edge is [(a/2 + a)/2, (0 + a/2)/2, 0] = [(3a/4), (a/4), 0]. The apex is at (a/2, a/2, a). The distance between these two points is sqrt[ (3a/4 - a/2)^2 + (a/4 - a/2)^2 + (0 - a)^2 ].Calculating each component:x-coordinate difference: 3a/4 - a/2 = 3a/4 - 2a/4 = a/4y-coordinate difference: a/4 - a/2 = a/4 - 2a/4 = -a/4z-coordinate difference: 0 - a = -aSo distance squared: (a/4)^2 + (-a/4)^2 + (-a)^2 = (a²/16) + (a²/16) + a² = (2a²/16) + a² = (a²/8) + a² = (9a²)/8Therefore, distance is sqrt(9a²/8) = (3a)/(2√2) = (3a√2)/4Wait, that's the slant height. So the slant height l is (3a√2)/4. Then, the area of each triangular face is (1/2)*base*slant height = (1/2)*(a/√2)*(3a√2/4). Let's compute that:(1/2)*(a/√2)*(3a√2/4) = (1/2)*(3a²* (√2/√2))/4 = (1/2)*(3a²*1)/4 = (3a²)/8Therefore, each triangular face has area 3a²/8. There are four triangular faces, so total lateral surface area is 4*(3a²/8) = 12a²/8 = 3a²/2.The base of the pyramid is the square with side length a/√2, so its area is (a/√2)^2 = a²/2. Therefore, total surface area is 3a²/2 + a²/2 = 4a²/2 = 2a². Hmm, that seems too straightforward. But let me verify.Alternatively, maybe I made a mistake in calculating the slant height. Let's check again.The apex is at (a/2, a/2, a). The midpoint of a base edge is, as we took, (3a/4, a/4, 0). The distance between these two points:x: 3a/4 - a/2 = a/4y: a/4 - a/2 = -a/4z: 0 - a = -aSo squared differences:(a/4)^2 = a²/16(-a/4)^2 = a²/16(-a)^2 = a²Total squared distance: a²/16 + a²/16 + a² = 2a²/16 + a² = a²/8 + a² = 9a²/8Square root: 3a/(2√2) = (3a√2)/4. That's correct.Therefore, slant height l = 3a√2/4.Then, area of one triangular face: (1/2)*base*height, where base is a/√2 and height is l = 3a√2/4.Calculating:(1/2)*(a/√2)*(3a√2/4) = (1/2)*(a*3a√2*√2)/(4*1) [since √2*√2 = 2]Wait, no. Let me compute step by step:First, multiply a/√2 and 3a√2/4:(a/√2)*(3a√2/4) = a*3a*(√2/√2)/4 = 3a²*(1)/4 = 3a²/4Then multiply by 1/2:(1/2)*(3a²/4) = 3a²/8. Yes, same as before.So each triangle is 3a²/8, four triangles give 12a²/8 = 3a²/2. Base area a²/2. Total surface area 3a²/2 + a²/2 = 2a². So the total surface area of the pyramid is 2a².But wait, the original cube has surface area 6a², and this pyramid is inside the cube? Hmm, but the pyramid's surface area is 2a², which is a third of the cube's. That seems plausible. Let me check another way.Alternatively, perhaps the pyramid's faces are not just four triangles and a base. Wait, no, the pyramid is defined as having a base (the square connecting the midpoints) and four triangular faces connecting the apex to each side of the base. Therefore, the total surface area should indeed be base area plus lateral faces. So 2a² as computed.Wait, but let me think again. The problem states: "the center of the upper base of a cube with edge length a is connected to the midpoints of the sides of the lower base, which are also connected in a sequential order". So when they say "connected to the midpoints... which are also connected in a sequential order", does that form additional faces? Wait, does connecting the midpoints in sequence form the base, and connecting the apex to each midpoint forms the triangular faces. So in total, the figure is a square pyramid with four triangular sides and a square base, so five faces total. Therefore, surface area is the sum of those five faces. So as per the calculation, 2a².But let me cross-validate. Let's compute the areas of all the faces step by step.First, the base: the square connecting the midpoints of the lower base's sides. Each side of this square is a/√2, so area is (a/√2)^2 = a²/2.Then, the four triangular faces. Each triangular face is a triangle with two edges from the apex to the midpoints and one edge along the base.Wait, but perhaps the triangles are not isosceles? Wait, in the pyramid, each triangular face is formed by the apex and one edge of the base. Since the base is a square, each triangular face is congruent. The apex is directly above the center of the base, so all triangular faces should be congruent isosceles triangles.Wait, but earlier, we found that the distance from the apex to each midpoint (the edges of the pyramid) is (a√5)/2. But when we computed the slant height, which is the distance from the apex to the midpoint of a base edge, we got 3a√2/4. Wait, but perhaps I confused slant height with edge length.Wait, in the pyramid, the slant height is the height of the triangular face, which is the distance from the apex to the base edge's midpoint. Whereas the edge from the apex to the base vertex is a different length.So each triangular face has a base of length a/√2, and two other sides of length (a√5)/2 each. To find the area of this triangle, we can use Heron's formula.Let's compute the semi-perimeter:s = (a/√2 + a√5/2 + a√5/2)/2 = (a/√2 + a√5)/2But Heron's formula might be messy here. Alternatively, since the triangle is isosceles (because the apex is directly above the center of the base, so the two edges from the apex to the base vertices are equal in length), we can split the triangle into two right triangles by drawing an altitude from the apex to the base. This altitude is the slant height, which we calculated as 3a√2/4.Therefore, the area of the triangular face is (1/2)*base*slant height = (1/2)*(a/√2)*(3a√2/4) = (1/2)*(3a²/4) = 3a²/8. So as before, each triangular face has area 3a²/8, four of them give 3a²/2, plus base a²/2, total 2a².Therefore, the total surface area is 2a². Hmm, that seems consistent.Alternatively, let's compute the areas using vectors or coordinates.Take one triangular face, say between apex (a/2, a/2, a), midpoint1 (a/2, 0, 0), and midpoint2 (a, a/2, 0). The vectors for this triangle are from (a/2, a/2, a) to (a/2, 0, 0) and to (a, a/2, 0). To find the area of this triangle, we can compute the cross product of these two vectors and take half its magnitude.Vector1: (a/2 - a/2, 0 - a/2, 0 - a) = (0, -a/2, -a)Vector2: (a - a/2, a/2 - a/2, 0 - a) = (a/2, 0, -a)Cross product of Vector1 and Vector2:|i     j     k  ||0 -a/2 -a ||a/2  0   -a |Calculating determinant:i * [(-a/2)(-a) - (-a)(0)] - j * [0*(-a) - (-a)(a/2)] + k * [0*0 - (-a/2)(a/2)]= i * [a²/2 - 0] - j * [0 - (-a²/2)] + k * [0 - (-a²/4)]= (a²/2)i - (a²/2)j + (a²/4)kThe magnitude of this cross product is sqrt[(a²/2)^2 + (-a²/2)^2 + (a²/4)^2] = sqrt[a^4/4 + a^4/4 + a^4/16] = sqrt[(4a^4/16 + 4a^4/16 + a^4/16)] = sqrt[9a^4/16] = (3a²)/4Therefore, the area of this triangular face is half of that: (3a²)/8, which matches our previous calculation. Therefore, each triangular face is indeed 3a²/8, so four of them give 3a²/2. Adding the base area a²/2 gives total surface area 2a². So that seems correct.Therefore, the answer should be 2a². But let me check again for possible mistakes.Wait, but in the cube, the pyramid is formed by connecting the upper center to the midpoints of the lower base. But when we connect those midpoints, which form a square, and connect the apex to each midpoint, we get four triangles and the base square. So the total surface area is 2a². That seems right.Alternatively, maybe the problem considers the pyramid to have a different base. Wait, the problem says "the resulting pyramid". If the pyramid is formed by connecting the center of the upper base to the midpoints, which are connected in a sequential order, perhaps the base is the original upper base's center? Wait, no. The pyramid's base should be the figure formed by connecting the midpoints of the lower base, since they are connected in a sequential order. The apex is the center of the upper base.Alternatively, maybe I'm misinterpreting the connections. Let me parse the problem again:"The center of the upper base of a cube with edge length a is connected to the midpoints of the sides of the lower base, which are also connected in a sequential order. Calculate the total surface area of the resulting pyramid."So "connected to the midpoints... which are also connected in a sequential order". So two actions: connecting the center to the midpoints, and connecting the midpoints in a sequential order. So the figure formed is a pyramid whose base is the polygon formed by connecting the midpoints in order (which is a square), and whose lateral edges are the connections from the center (apex) to each midpoint (vertices of the base). Therefore, the pyramid has a square base and four triangular faces, as we considered. Therefore, the total surface area is indeed the sum of the base area and the four triangular faces, which we calculated as 2a².But let me consider another approach. Suppose instead of computing each face, compute the areas using the coordinates.The base is the square with vertices at (a/2, 0, 0), (a, a/2, 0), (a/2, a, 0), (0, a/2, 0). The apex is at (a/2, a/2, a). Each triangular face is formed by the apex and two consecutive midpoints.For example, one face is formed by points (a/2, a/2, a), (a/2, 0, 0), and (a, a/2, 0). Another face is (a/2, a/2, a), (a, a/2, 0), (a/2, a, 0), etc.We already computed the area of one such face using the cross product, which gave 3a²/8. Since all four faces are congruent, multiplying by four gives 3a²/2. The base area is a²/2. Total surface area 2a². That seems consistent.Alternatively, perhaps there's a simpler way to compute the lateral surface area.In a right square pyramid, lateral surface area is (perimeter of base * slant height)/2. The base perimeter here is 4*(a/√2) = 4a/√2 = 2√2 a. The slant height we calculated as 3a√2/4. Therefore, lateral surface area is (2√2 a)*(3a√2/4)/2 = let's compute:First, multiply 2√2 a and 3a√2/4:2√2 * 3a√2/4 = (6a * 2)/4 = (12a)/4 = 3aThen divide by 2: 3a/2. So lateral surface area is 3a²/2, same as before. Then total surface area is 3a²/2 + a²/2 = 2a².Yes, so all methods confirm that the total surface area is 2a².Therefore, the answer is 2a², which in boxed form is boxed{2a^2}.Final AnswerThe total surface area of the resulting pyramid is boxed{2a^2}.

✅:To calculate the total surface area of the pyramid formed by connecting the center of the upper base of a cube to the midpoints of the sides of the lower base, we follow these steps:1. Identify the coordinates and dimensions: - The cube has edge length (a). - The center of the upper base is at ((a/2, a/2, a)). - The midpoints of the sides of the lower base are at ((a/2, 0, 0)), ((a, a/2, 0)), ((a/2, a, 0)), and ((0, a/2, 0)).2. Determine the base of the pyramid: - The midpoints form a square with side length (a/sqrt{2}). - The area of this square (base) is ((a/sqrt{2})^2 = a^2/2).3. Calculate the slant height of the triangular faces: - The apex of the pyramid is at ((a/2, a/2, a)). - The slant height (distance from the apex to the midpoint of a base edge) is calculated using the Pythagorean theorem: [ text{Slant height} = sqrt{left(aright)^2 + left(frac{a}{2sqrt{2}}right)^2} = sqrt{a^2 + frac{a^2}{8}} = sqrt{frac{9a^2}{8}} = frac{3a}{2sqrt{2}} = frac{3asqrt{2}}{4} ]4. Calculate the area of each triangular face: - Each triangular face has a base of (a/sqrt{2}) and slant height (3asqrt{2}/4): [ text{Area of one triangular face} = frac{1}{2} times frac{a}{sqrt{2}} times frac{3asqrt{2}}{4} = frac{3a^2}{8} ] - There are four triangular faces, so the total lateral surface area is: [ 4 times frac{3a^2}{8} = frac{3a^2}{2} ]5. Sum the base area and lateral surface area: - Total surface area = Base area + Lateral surface area: [ frac{a^2}{2} + frac{3a^2}{2} = 2a^2 ]Thus, the total surface area of the resulting pyramid is (boxed{2a^2}).

🔑:1. Identification of the pyramid's base and midpoints: - Since the edge of the cube is (a), the side length of the base of the pyramid (a square) is (a sqrt{2} / 2). - The vertices of the base of the pyramid will be the midpoints of the edges of the lower square base of the cube.2. Calculation of the distance from the center of the base to a side midpoint: - Denote the center of the top surface of the cube as ( S ) and the midpoint of one side of the bottom square base as ( K ). - The length ( OK ) (where ( O ) is the center of the lower square base) is half the diagonal of the base of one triangle: [ OK = frac{1}{2} AD = frac{a sqrt{2}}{4}. ]3. Calculation of the slant height of the pyramid: - Using the Pythagorean theorem, calculate the length ( SK ) which is the slant height of the pyramid: [ SK = sqrt{SO^{2} + OK^{2}} = sqrt{a^{2} + left(frac{a sqrt{2}}{4}right)^{2}} = sqrt{a^{2} + frac{a^{2}}{8}} = frac{3asqrt{2}}{4}. ]4. Calculation of the lateral surface area: - The lateral surface area of the pyramid is given by the sum of the areas of the four triangular faces: [ S_{text{bok}} = 4 cdot frac{1}{2} cdot frac{asqrt{2}}{2} cdot frac{3asqrt{2}}{4} = 4 cdot frac{1}{2} cdot frac{3a^{2}}{4} cdot frac{1}{2} = frac{3a^{2}}{2}. ]5. Calculation of the total surface area: - The area of the square base: [ S_{text{base}} = left(frac{a sqrt{2}}{2}right)^{2} = frac{a^2 cdot 2}{4} = frac{a^2}{2}. ] - Adding the lateral surface area and the base area: [ S_{text{total}} = S_{text{base}} + S_{text{bok}} = frac{a^2}{2} + frac{3a^2}{2} = 2a^2. ] Conclusion:[boxed{2a^2}]