🧠:Alright, let me try to figure this out. The problem says that if the sum of the squares of two numbers is divisible by 7, then each of those numbers must be divisible by 7. Hmm, I need to determine if this statement is true or not. First, let me restate it in my own words. Suppose we have two integers, say a and b. If a² + b² is divisible by 7, then both a and b must individually be divisible by 7. The question is whether this implication holds. So, if 7 divides a² + b², does 7 necessarily divide both a and b?I remember something about modular arithmetic, maybe that can help. If a number is divisible by 7, then modulo 7 it would be 0. So, if a² + b² ≡ 0 mod 7, then does that imply a ≡ 0 mod 7 and b ≡ 0 mod 7?Let me think. To check this, maybe I should look at the possible squares modulo 7. That is, compute all possible residues of squares when divided by 7. Let's list them out.The possible residues mod 7 are 0, 1, 2, 3, 4, 5, 6. Let's square each and see what they are mod 7:0² = 0 ≡ 0 mod71² = 1 ≡ 1 mod72² = 4 ≡ 4 mod73² = 9 ≡ 2 mod7 (since 9-7=2)4² = 16 ≡ 2 mod7 (16-14=2)5² = 25 ≡ 4 mod7 (25-21=4)6² = 36 ≡ 1 mod7 (36-35=1)So the possible quadratic residues modulo 7 are 0, 1, 2, and 4. That is, when you square a number mod7, the result can only be 0, 1, 2, or 4.Now, the problem states that a² + b² ≡ 0 mod7. Let's see what combinations of these residues add up to 0 mod7. So possible pairs (x, y) where x and y are in {0,1,2,4} and x + y ≡ 0 mod7.Let me list all possible pairs:0 + 0 = 0 → 0 mod71 + 6 = 7, but 6 isn't a quadratic residue. Wait, but we can only use the residues 0,1,2,4. So possible sums:0 + 0 = 01 + 6 → but 6 isn't a quadratic residue. Next:1 + 1 = 21 + 2 = 31 + 4 = 52 + 2 = 42 + 4 = 64 + 4 = 8 ≡ 1 mod7So the only way to get a sum that's 0 mod7 is if both squares are 0 mod7. Because the other combinations don't add up to 0. For example, 1 + 6 isn't possible since 6 isn't a quadratic residue. Similarly, 2 + 5 isn't possible. So the only possible pair is (0,0). Therefore, if a² + b² ≡ 0 mod7, then both a² ≡ 0 mod7 and b² ≡ 0 mod7. But if a² ≡ 0 mod7, then a ≡ 0 mod7 because 7 is prime. Similarly for b. Therefore, both a and b must be divisible by 7.Wait, but let me check if that's the case. Suppose, for example, a² ≡ 0 mod7. Then 7 divides a², which implies 7 divides a, since 7 is prime. Yes, that's correct. If a prime divides a square, it must divide the base. So indeed, a and b must each be divisible by 7.Therefore, the original statement is true: if the sum of the squares of two numbers is divisible by 7, then each number is divisible by 7.But let me test this with some examples to be sure. Take a = 7, b = 0. Then a² + b² = 49 + 0 = 49, which is divisible by 7. Both numbers are divisible by 7. Okay, that works.Another example: a = 7, b = 7. Then 49 + 49 = 98, which is divisible by 7. Both are divisible by 7. Still holds.What if I take a number not divisible by 7. Let's say a = 1, b = some number. Then a² = 1. So to get the sum divisible by 7, b² must be ≡ -1 mod7, which is 6 mod7. But from earlier, the quadratic residues are 0,1,2,4. 6 isn't among them. So there's no such b. Therefore, if a is not divisible by 7, then a² ≡ 1,2,4 mod7. Then b² would need to be ≡ -1, -2, -4 mod7, which aren't quadratic residues. Therefore, impossible. Hence, such b doesn't exist.Therefore, there's no pair of numbers not divisible by 7 whose squares sum to a multiple of 7. Therefore, the statement holds.But wait, let me check another example. Suppose a = 3, which is not divisible by 7. Then a² = 9 ≡ 2 mod7. So we need b² ≡ -2 ≡ 5 mod7. But 5 isn't a quadratic residue mod7. So no solution.Similarly, a = 2, a² = 4 mod7. Then b² ≡ -4 ≡ 3 mod7. But 3 isn't a quadratic residue. So again, no solution.So in all cases, unless both a and b are ≡0 mod7, the sum of squares can't be ≡0 mod7. Therefore, the original statement is true.I think that's thorough. So the conclusion is that the statement is true.

✅:The statement is true. Here's the step-by-step reasoning:1. Quadratic Residues Modulo 7: Compute all possible residues of squares modulo 7: - (0^2 equiv 0 mod 7) - (1^2 equiv 1 mod 7) - (2^2 equiv 4 mod 7) - (3^2 equiv 9 equiv 2 mod 7) - (4^2 equiv 16 equiv 2 mod 7) - (5^2 equiv 25 equiv 4 mod 7) - (6^2 equiv 36 equiv 1 mod 7) The quadratic residues are (0, 1, 2, 4).2. Sum of Residues: For (a^2 + b^2 equiv 0 mod 7), the pairs of residues ((x, y)) must satisfy (x + y equiv 0 mod 7). - The only valid pair is ((0, 0)), as other combinations (e.g., (1 + 6), (2 + 5)) involve non-residues. - Thus, (a^2 equiv 0 mod 7) and (b^2 equiv 0 mod 7).3. Prime Divisibility: Since (7) is prime, if (7 mid a^2), then (7 mid a). Similarly, (7 mid b^2 implies 7 mid b). 4. Conclusion: If (7 mid a^2 + b^2), then (7 mid a) and (7 mid b). No counterexamples exist because non-zero residues cannot sum to (0 mod 7).Final Answer: The statement is true. If (a^2 + b^2) is divisible by 7, both (a) and (b) must individually be divisible by 7. (boxed{text{True}})

🔑:Let's denote the two numbers as ( a ) and ( b ).1. Assumption: - Assume that both ( a ) and ( b ) are not divisible by 7. 2. Application of Fermat's Little Theorem: - By Fermat's Little Theorem, for any integer ( x ) that is not divisible by a prime ( p ), ( x^{p-1} equiv 1 pmod{p} ). - In our case, ( p = 7 ). Therefore, for any ( a ) and ( b ) not divisible by 7: [ a^6 equiv 1 pmod{7} quad text{and} quad b^6 equiv 1 pmod{7} ]3. Summation and Equation Formation: - Thus, [ a^6 - 1 equiv 0 pmod{7} quad text{and} quad b^6 - 1 equiv 0 pmod{7} ] - We can then write: [ a^6 equiv 1 pmod{7} quad text{and} quad b^6 equiv 1 pmod{7} ] - Summing these congruences: [ a^6 + b^6 - 2 equiv 0 pmod{7} ] - This can be simplified to: [ a^6 + b^6 equiv 2 pmod{7} ]4. Transforming the Problem: - Notice that: [ a^6 + b^6 - 2 = (a^2 + b^2)(a^4 - a^2 b^2 + b^4) - 2 ] - Since ( a^2 + b^2 ) is given as divisible by 7, let ( a^2 + b^2 = 7k ) for some integer ( k ).5. Contradiction: - Substituting into the equation gives: [ 7k (a^4 - a^2 b^2 + b^4) - 2 ] - For ( a^6 + b^6 - 2 equiv 0 pmod{7} ): [ 7k (a^4 - a^2 b^2 + b^4) equiv 2 pmod{7} ] - This implies that: [ 0 equiv 2 pmod{7} ] - This is a contradiction since 2 is not congruent to 0 modulo 7.6. Conclusion: - Therefore, the assumption that ( a ) and ( b ) are not divisible by 7 cannot be true. - Thus, both ( a ) and ( b ) must be divisible by 7. [ boxed{text{Both } a text{ and } b text{ must be divisible by 7.}} ]

🧠:Alright, so I need to solve this inequality: log base 2 of (5-12x)/(12x-8) plus log base 1/2 of x is less than or equal to 0. Hmm, okay. Let me try to break this down step by step. First, I remember that logarithms have certain domains. So, I need to make sure that the arguments inside the logs are positive. Let me write that down. For log base 2 of (5-12x)/(12x-8) to be defined, the argument (5-12x)/(12x-8) must be positive. Similarly, for log base 1/2 of x, the argument x must be positive as well. So, the domain restrictions are:1. (5 - 12x)/(12x - 8) > 02. x > 0Let me solve these inequalities first to find the domain where the original inequality is defined.Starting with the first inequality: (5 - 12x)/(12x - 8) > 0.To solve this, I need to find where the numerator and denominator are both positive or both negative. Let's find the critical points by setting numerator and denominator equal to zero.Numerator: 5 - 12x = 0 → x = 5/12 ≈ 0.4167Denominator: 12x - 8 = 0 → x = 8/12 = 2/3 ≈ 0.6667So, the critical points are at x = 5/12 and x = 2/3. These divide the real number line into three intervals:1. x < 5/122. 5/12 < x < 2/33. x > 2/3Now, let's test each interval.First interval: x < 5/12. Let's pick x = 0. Then numerator is 5 - 0 = 5 (positive), denominator is 0 - 8 = -8 (negative). So overall, 5/-8 = negative. So this interval doesn't satisfy the inequality.Second interval: 5/12 < x < 2/3. Let's pick x = 0.5 (which is 1/2). Then numerator: 5 - 12*(0.5) = 5 - 6 = -1 (negative). Denominator: 12*(0.5) - 8 = 6 - 8 = -2 (negative). So, (-1)/(-2) = 0.5 (positive). So this interval satisfies the inequality.Third interval: x > 2/3. Let's pick x = 1. Numerator: 5 - 12*1 = -7 (negative). Denominator: 12*1 -8 = 4 (positive). So, -7/4 = negative. Doesn't satisfy.So, the first domain restriction is 5/12 < x < 2/3.Second domain restriction is x > 0. But since 5/12 is approximately 0.4167, which is greater than 0, the combined domain is 5/12 < x < 2/3.So, x must be between 5/12 and 2/3. That's important because any solution must lie within this interval.Now, moving on to solving the inequality: log₂[(5-12x)/(12x-8)] + log_{1/2}(x) ≤ 0.Hmm, logarithms with different bases. Maybe I can convert the second logarithm to base 2 to make it easier. Since log_{1/2}(x) is the same as log base 2 of x divided by log base 2 of 1/2. But log base 2 of 1/2 is -1, right? Because 2^(-1) = 1/2. So log_{1/2}(x) = log₂x / log₂(1/2) = log₂x / (-1) = -log₂x.So, the inequality becomes:log₂[(5-12x)/(12x-8)] - log₂x ≤ 0.That's better because now both terms are base 2. Let's write that as:log₂[(5-12x)/(12x-8)] - log₂x ≤ 0.Combine the logs using logarithm rules. Since log A - log B = log(A/B), so:log₂[ ( (5-12x)/(12x-8) ) / x ] ≤ 0.Simplify the argument inside the log:(5-12x)/(12x -8) divided by x is equal to (5 -12x)/(x*(12x -8)).So, log₂[ (5 -12x)/(x*(12x -8)) ] ≤ 0.Now, since log₂(something) ≤ 0, that means that "something" is ≤ 2^0 = 1, because log₂(y) ≤ 0 implies y ≤ 1. But also, since log₂ is only defined for positive numbers, the argument must still be positive. But we already considered the domain earlier, so we can proceed.Therefore, the inequality becomes:(5 -12x)/(x*(12x -8)) ≤ 1.But we have to be careful here. Because the argument of the log is (5 -12x)/(x*(12x -8)), which must be positive. Wait, let's check: since in our domain, 5/12 < x < 2/3, let's verify the sign of the denominator and numerator.First, 5 -12x: when x >5/12, 12x >5, so 5 -12x <0.12x -8: when x < 2/3, 12x <8, so 12x -8 <0.So, numerator is negative, denominator is x*(negative). Since x is positive in the domain (from the second restriction), x*(negative) is negative. Therefore, the entire argument (5 -12x)/(x*(12x -8)) is (negative)/(negative) = positive. So that's good, the log is defined here.So, the inequality simplifies to:(5 -12x)/(x*(12x -8)) ≤ 1.But let's write it as:(5 -12x)/(x*(12x -8)) ≤ 1.Let me rearrange this inequality. Let me first note that 12x -8 is negative in the domain, as we saw. So, x*(12x -8) is negative. So, the denominator is negative, numerator is also negative, so the entire expression is positive.But let's proceed step by step. Let me write the inequality:(5 -12x)/(x*(12x -8)) ≤ 1.Let me subtract 1 from both sides to get:(5 -12x)/(x*(12x -8)) -1 ≤ 0.Combine the terms into a single fraction:[(5 -12x) - x*(12x -8)] / [x*(12x -8)] ≤ 0.Compute the numerator:5 -12x -12x² +8x = 5 -4x -12x².So, the inequality is:(-12x² -4x +5)/[x*(12x -8)] ≤ 0.Hmm, let me write the numerator as -12x² -4x +5. Alternatively, factor out a negative sign: - (12x² +4x -5). Let me check if 12x² +4x -5 can be factored.Looking for factors of 12*(-5) = -60 that add up to 4. Let's see: 10 and -6. 10*(-6) = -60 and 10 -6 =4. So, split the middle term:12x² +10x -6x -5.Group:(12x² +10x) + (-6x -5) = 2x(6x +5) -1(6x +5) = (2x -1)(6x +5).Wait, let me verify:12x² +10x -6x -5 = 12x² +4x -5. Yes, correct. Then grouping:First two terms: 2x(6x +5)Last two terms: -1(6x +5)So, (2x -1)(6x +5). Therefore, 12x² +4x -5 = (2x -1)(6x +5). Therefore, the numerator is - (2x -1)(6x +5).So, the inequality becomes:- (2x -1)(6x +5) / [x*(12x -8)] ≤ 0.Let me simplify this expression. Let's also note that the denominator is x*(12x -8) = x*4*(3x -2) = 4x(3x -2). But maybe that's not necessary. Let's write all factors:Numerator: - (2x -1)(6x +5)Denominator: x*(12x -8) = x*4*(3x -2) = 4x(3x -2). But the denominator is 4x(3x -2). But 4 is a positive constant, so it doesn't affect the inequality's sign. So, we can write the inequality as:- (2x -1)(6x +5) / [4x(3x -2)] ≤ 0.Since 4 is positive, we can multiply both sides by 4 without changing the inequality, but actually, maybe it's better to just keep it as it is. Alternatively, since 4 is positive, the sign is determined by the rest:- (2x -1)(6x +5) / [x(3x -2)] ≤ 0.So, let's focus on:- (2x -1)(6x +5) / [x(3x -2)] ≤ 0.Multiply numerator and denominator by -1 to make it easier, but remember that multiplying both sides of an inequality by a negative number reverses the inequality sign. However, here, it's part of the expression. Alternatively, distribute the negative sign:(-1)*(2x -1)(6x +5) / [x(3x -2)] ≤ 0.Which is equivalent to:(1 - 2x)(6x +5) / [x(3x -2)] ≤ 0.So, the inequality is now:(1 - 2x)(6x +5) / [x(3x -2)] ≤ 0.Let me note the critical points of this expression. The numerator is (1 -2x)(6x +5), and the denominator is x(3x -2). So, setting each factor equal to zero:Numerator:1 -2x =0 → x=1/26x +5=0 → x= -5/6Denominator:x=03x -2=0 → x=2/3So, critical points at x= -5/6, 0, 1/2, 2/3.But our domain is 5/12 <x <2/3. So, in this interval, x is between approx 0.4167 and 0.6667. So, within this interval, the critical points inside are x=1/2 (0.5) and x=2/3 (approx 0.6667). But x=2/3 is the upper bound of the domain, which is excluded. Similarly, x=1/2 is within the domain.But wait, in the numerator, we have (1 -2x)(6x +5). Let's see the values in the domain:For x in (5/12, 2/3):1 -2x: when x >1/2, 1 -2x becomes negative. When x <1/2, positive. Since 5/12 ≈0.4167 <1/2=0.5, so for x in (5/12,1/2), 1 -2x is positive. For x in (1/2, 2/3), 1 -2x is negative.6x +5: for x > -5/6, which is certainly true here since x is positive. So 6x +5 is positive in the entire domain.Denominator:x(3x -2). x is positive. 3x -2: when x <2/3, 3x -2 is negative. So denominator is positive * negative = negative.Therefore, let's analyze the sign of the expression (1 -2x)(6x +5)/[x(3x -2)] in the domain interval (5/12, 2/3).First, split the domain into two intervals based on the critical point x=1/2:1. 5/12 <x <1/22. 1/2 <x <2/3In the first interval (5/12, 1/2):(1 -2x) is positive (since x <1/2)(6x +5) is positiveDenominator x(3x -2) is negative (since x positive and 3x -2 negative)So, overall, the expression is (positive * positive)/negative = positive/negative = negative.Therefore, in 5/12 <x <1/2, the expression is negative.In the second interval (1/2, 2/3):(1 -2x) is negative(6x +5) is positiveDenominator is still negativeSo, (negative * positive)/negative = (negative)/negative = positive.Therefore, in 1/2 <x <2/3, the expression is positive.Now, the inequality is:(1 -2x)(6x +5)/[x(3x -2)] ≤ 0.So, we want where the expression is ≤0. From the analysis:In (5/12,1/2), the expression is negative, so satisfies ≤0.At x=1/2, the expression is:(1 -2*(1/2))(6*(1/2)+5)/[ (1/2)(3*(1/2) -2) ] = (0)(3 +5)/[ (1/2)(1.5 -2) ] = 0 / [ (1/2)(-0.5) ] = 0. So, x=1/2 is included as it gives 0, which satisfies ≤0.In (1/2, 2/3), the expression is positive, which doesn't satisfy ≤0.At x=5/12, the original domain starts, but x=5/12 is excluded since the domain is 5/12 <x <2/3. Similarly, x=2/3 is excluded.Therefore, the solution to the inequality (1 -2x)(6x +5)/[x(3x -2)] ≤0 within the domain 5/12 <x <2/3 is [5/12 <x ≤1/2].But wait, we have to check if x=1/2 is in the domain. Since 1/2=0.5 is between 5/12≈0.4167 and 2/3≈0.6667, yes, it's included. So, the solution to the transformed inequality is 5/12 <x ≤1/2.But we need to check if this is the final solution, considering all the steps.Wait, let's recap. The original inequality after simplifying became:log₂[(5-12x)/(12x-8)] + log_{1/2}x ≤0.Which transformed into:(1 -2x)(6x +5)/[x(3x -2)] ≤0.With the domain 5/12 <x <2/3.And solving that gives 5/12 <x ≤1/2.Therefore, the solution set is (5/12, 1/2].But let's verify this with a test point in the interval (5/12,1/2) and another in (1/2,2/3).First, take x=0.45 (which is 9/20, 0.45 is between 5/12≈0.4167 and 1/2=0.5).Compute the original inequality:log₂[(5 -12*0.45)/(12*0.45 -8)] + log_{1/2}(0.45) ≤0.Compute numerator: 5 -5.4= -0.4Denominator: 5.4 -8= -2.6So, ( -0.4 ) / ( -2.6 ) ≈0.1538. log₂(0.1538) ≈-2.7log_{1/2}(0.45)= ln(0.45)/ln(1/2)= (ln0.45)/(-ln2)≈ (-0.7985)/(-0.6931)≈1.152So total: -2.7 +1.152≈-1.548≤0. True.Now, take x=0.6 (which is in (1/2,2/3)).log₂[(5 -12*0.6)/(12*0.6 -8)] + log_{1/2}(0.6) ≤0.Numerator:5 -7.2= -2.2Denominator:7.2 -8= -0.8So, (-2.2)/(-0.8)=2.75. log₂(2.75)≈1.45log_{1/2}(0.6)= ln(0.6)/ln(0.5)= (-0.5108)/(-0.6931)≈0.737Total:1.45 +0.737≈2.187>0. So, doesn't satisfy the inequality. Correct.At x=0.5:log₂[(5 -6)/(6 -8)] + log_{1/2}(0.5) = log₂(-1/-2) + log_{1/2}(0.5)= log₂(0.5) +1= (-1) +1=0. So equality holds. Correct.Therefore, the solution is 5/12 <x ≤1/2.But just to make sure there are no mistakes in the algebraic manipulations.Original inequality:log₂[(5-12x)/(12x-8)] + log_{1/2}x ≤0.Convert log_{1/2}x to -log₂x.So, log₂[(5-12x)/(12x-8)] - log₂x ≤0.Combine logs: log₂[ (5-12x)/( (12x -8)x ) ] ≤0.Which implies (5-12x)/(x(12x -8)) ≤1, since log₂(y) ≤0 ⇨ y ≤1.So, solving (5-12x)/(x(12x -8)) ≤1.Which led us to (5 -12x)/(x(12x -8)) -1 ≤0 → (-12x² -4x +5)/[x(12x -8)] ≤0.Which was simplified to (1 -2x)(6x +5)/[x(3x -2)] ≤0.All steps checked out. Then considering the domain 5/12 <x <2/3, the solution is 5/12 <x ≤1/2.Thus, the final solution is x in (5/12, 1/2].But let me just check if 5/12 is approximately 0.4167 and 1/2 is 0.5. So, writing in fractions:5/12 is 5/12, 1/2 is 6/12, so the interval is (5/12, 6/12].But 6/12 is 1/2. So, yes.But let me confirm if there's a strict inequality at x=5/12 and x=2/3 because the original domain is 5/12 <x <2/3. So, the endpoints are excluded. But in the transformed inequality, after simplifying, we found that the solution is up to x=1/2, including 1/2. So, x=1/2 is included because at that point the expression equals zero.Therefore, the final solution is (5/12, 1/2].But just to make sure, let's check x=5/12. But x=5/12 is excluded from the domain. At x approaching 5/12 from the right, let's see the behavior.As x approaches 5/12+, (5 -12x) approaches 0 from the negative side (since x >5/12, 12x >5), denominator (12x -8) approaches 12*(5/12) -8 =5 -8=-3. So, (5 -12x)/(12x -8) approaches 0 from the positive side (since numerator approaches 0 from negative, denominator approaches -3, so overall 0+/-3=0 from positive side). So, log₂(approaching 0+) tends to -infinity. Then log_{1/2}x is log_{1/2}(5/12+) which is finite. So, the left-hand side tends to -infinity, which is certainly ≤0. Wait, but x=5/12 is excluded from the domain, so even though the limit is -infinity, we can't include x=5/12.Wait, but the domain is 5/12 <x <2/3, so x=5/12 is not included. Therefore, even if the expression approaches -infinity near x=5/12, the solution is up to but not including 5/12.But in our transformed inequality, when we solved (1 -2x)(6x +5)/[x(3x -2)] ≤0, and found that in (5/12,1/2] the inequality holds. So, x=1/2 is included, but x=5/12 is excluded.Therefore, the solution is 5/12 <x ≤1/2.Yes, that seems correct. So, converting 5/12 and 1/2 to fractions, 5/12 is already in simplest form, and 1/2 is 1/2. So, the interval is (5/12, 1/2].Thus, the final answer is x ∈ (5/12, 1/2].Final AnswerThe solution to the inequality is boxed{left( dfrac{5}{12}, dfrac{1}{2} right]}.

✅:To solve the inequality log _{2} frac{5-12 x}{12 x-8}+log _{1 / 2} x leqslant 0,we first determine the domain where the logarithms are defined. 1. The argument of log _{2} frac{5-12 x}{12 x-8} must be positive: frac{5 - 12x}{12x - 8} > 0. Solving this, we find the critical points x = frac{5}{12} and x = frac{2}{3}. The solution is the interval (frac{5}{12}, frac{2}{3}).2. The argument of log _{1 / 2} x must be positive: x > 0. Combining these, the domain is (frac{5}{12}, frac{2}{3}).Next, we convert the inequality using logarithmic properties. Noting that log_{1/2} x = -log_2 x, we rewrite the inequality as:log _{2} frac{5-12 x}{12 x-8} - log _{2} x leqslant 0.Combining the logarithms, we get:log _{2} left( frac{5 - 12x}{x(12x - 8)} right) leqslant 0.This implies:frac{5 - 12x}{x(12x - 8)} leqslant 1.Subtracting 1 from both sides and simplifying, we get:frac{-12x^2 - 4x + 5}{x(12x - 8)} leqslant 0.Factoring the numerator and denominator, we analyze the critical points and test intervals within the domain (frac{5}{12}, frac{2}{3}). We find the solution interval where the expression is non-positive, which is (frac{5}{12}, frac{1}{2}].Verifying with test points confirms the solution. Thus, the final solution to the inequality is:boxed{left( dfrac{5}{12}, dfrac{1}{2} right]}.```

🔑:Given the inequality:log_{2}left(frac{5 - 12x}{12x - 8}right) + log_{frac{1}{2}}(x) leq 0Step 1: Simplify the log expression.We recall that the logarithm with base frac{1}{2} can be rewritten using the logarithm with base 2, because log_{frac{1}{2}}(x) = -log_{2}(x):log_{2}left(frac{5 - 12x}{12x - 8}right) - log_{2}(x) leq 0Step 2: Combine the logarithms.Using the property of logarithms log_{a}(b) - log_{a}(c) = log_{a}left(frac{b}{c}right):log_{2}left(frac{frac{5 - 12x}{12x - 8}}{x}right) leq 0This simplifies to:log_{2}left(frac{5 - 12x}{x(12x - 8)}right) leq 0Step 3: Exponentiate both sides to isolate the argument of the logarithm.Since the exponent base 2 is positive, the inequality remains the same:frac{5 - 12x}{x(12x - 8)} leq 1Step 4: Transform the inequality.Rearranging and finding a common denominator, we have:frac{5 - 12x}{12x^2 - 8x} leq 1Multiply through by the denominator to clear the fraction:5 - 12x leq 12x^2 - 8xRearrange all terms to one side of the inequality:12x^2 - 8x + 12x + 5 geq 012x^2 + 4x + 5 geq 0Step 5: Solve the quadratic inequality.For this inequality to hold, we examine the conditions under which the nominator and denominator change signs. Let's start with the nominator:5 - 12x > 0 implies x < frac{5}{12}For the denominator:12x - 8 > 0 implies x > frac{2}{3}Combining these gives:frac{5}{12} < x < frac{2}{3}Step 6: Check intersection with 5 - 12x geq x(12x - 8) for valid x.Using the conditions derived:5 - 12x geq x(12x - 8)5 - 12x geq 12x^2 - 8x5 - 4x geq 12x^25 geq 4x + 12x^212x^2 + 4x - 5 leq 0Using the quadratic formula:x = frac{-4 pm sqrt{16 + 240}}{24}x = frac{-4 pm sqrt{256}}{24}x = frac{-4 pm 16}{24}x_1 = frac{frac{-4 + 16}{24} = frac{12}{24} = frac{1}{2}x_2 = frac{frac{-4 - 16}{24} = - frac{20}{24} = - frac{5}{6}Thus, the values of x must be within:x in left( frac{5}{12}, frac{2}{3} right) Conclusion:The solution to the inequality is:boxed{frac{5}{12} < x leq frac{1}{2}}

🧠:Okay, let's see. The problem is about finding how many elements in the set T, which consists of powers of 9 from 9^0 up to 9^4000, have 9 as their most significant digit. We are given that 9^4000 has 3817 digits and its most significant digit is 9. Hmm, interesting.First, I need to recall that the most significant digit (MSD) of a number is the first digit of the number when written in decimal form. For example, the MSD of 987 is 9. So the question is asking, out of all the numbers 9^0, 9^1, 9^2, ..., 9^4000, how many of these start with the digit 9?I remember that the number of digits of a number N can be found using the formula floor(log10(N)) + 1. Since 9^4000 has 3817 digits, this formula must hold: floor(log10(9^4000)) + 1 = 3817. Let me verify that. log10(9^4000) is 4000*log10(9). Let me compute log10(9). Since 9 is 3^2, log10(9) = 2*log10(3). Log10(3) is approximately 0.4771, so log10(9) is about 0.9542. Therefore, 4000*0.9542 ≈ 3816.8. Taking the floor gives 3816, then adding 1 gives 3817 digits. Yep, that checks out. So that's consistent with the given information.Now, the problem states that 9^4000 has 3817 digits and its MSD is 9. So, that tells us that when we write 9^4000 in decimal, it starts with a 9. The question is how many other exponents k (from 0 to 4000) result in 9^k starting with 9.I remember that the leading digit of a number relates to the fractional part of its logarithm. Let me recall the concept. If we take a number N = 9^k, then log10(N) = k*log10(9). Let me write this as log10(N) = m + f, where m is an integer and 0 ≤ f < 1 is the fractional part. Then, the leading digit of N is determined by 10^f. Specifically, if 10^f starts with a 9, then the leading digit of N is 9. So, 10^f ∈ [9, 10), which implies that f ∈ [log10(9), 1). Therefore, the fractional part of k*log10(9) must lie in [log10(9), 1) for the leading digit to be 9.So, for each k from 0 to 4000, we compute k*log10(9) and look at the fractional part. If the fractional part is in [log10(9), 1), then 9^k has leading digit 9. Since log10(9) ≈ 0.9542, this interval is [0.9542, 1). The length of this interval is 1 - 0.9542 ≈ 0.0458. So, if the fractional parts of k*log10(9) are uniformly distributed, we would expect approximately 0.0458 of all k's to satisfy this condition. However, since log10(9) is irrational (because 9 is not a power of 10), the fractional parts of k*log10(9) are uniformly distributed modulo 1 due to equidistribution theorem. Therefore, the number of k's with fractional part in [0.9542, 1) should be approximately 0.0458*4001 ≈ 183. However, since the problem gives us exact information about 9^4000 starting with 9, maybe we need to adjust this estimate. Also, the equidistribution might not be exact, but for large k, it should be close.But wait, the problem is a competition problem, so it's likely expecting an exact answer rather than an approximation. Therefore, there must be a way to compute the exact number using properties of logarithms and possibly some periodicity or recursion.Alternatively, since 9 = 3^2, maybe there's a relationship between powers of 9 and powers of 3, but I'm not sure if that helps here.Wait, another thought: If 9^k starts with a 9, then 9^k is between 9*10^m and 10^{m+1} for some integer m. Taking logarithms, this inequality becomes:log10(9*10^m) ≤ log10(9^k) < log10(10^{m+1})Which simplifies to:log10(9) + m ≤ k*log10(9) < m + 1Subtract m from all parts:log10(9) ≤ (k*log10(9) - m) < 1But (k*log10(9) - m) is the fractional part of k*log10(9), since m is the integer part. Therefore, fractional part of k*log10(9) ∈ [log10(9), 1). Which is the same as before.Therefore, the problem reduces to finding the number of integers k ∈ {0, 1, ..., 4000} such that {k*log10(9)} ∈ [log10(9), 1), where {x} denotes the fractional part of x.Now, log10(9) ≈ 0.9542425094. Let's denote α = log10(9). So, we need to count the number of k such that {kα} ∈ [α, 1). The length of the interval is 1 - α ≈ 0.0457574906.As α is irrational (since 9 is not a power of 10), the sequence {kα} is uniformly distributed modulo 1. Therefore, the asymptotic proportion of k's satisfying the condition is (1 - α). So, the approximate number of such k's is (1 - α)*4001 ≈ 0.0457574906*4001 ≈ 0.0457574906*4000 + 0.0457574906 ≈ 183.03 + 0.045 ≈ 183.075. So approximately 183. But the problem states that 9^4000 has 3817 digits and starts with 9, so k=4000 is included. Therefore, the exact answer might be 184? But we need to check.Wait, but the problem is from 1990, so maybe the answer is 184? But how do we find the exact number?Alternatively, perhaps there is a pattern or cyclic behavior in the leading digits of 9^k. However, since 9 and 10 are coprime, the leading digits cycle with some period, but due to the irrationality of log10(9), the fractional parts {kα} never exactly repeat, so the cycle length is infinite. Therefore, equidistribution is the key here. However, equidistribution is a limiting behavior as k approaches infinity. For finite k, especially up to 4000, the discrepancy could be small, but the exact count may differ by 1 or 2 from the approximate value.But the problem gives us that 9^4000 has MSD 9, so at least one element (k=4000) is in the set. So if our approximate answer is 183, but since k=4000 is included, maybe it's 184? Alternatively, maybe the equidistribution gives us exactly 184, considering the fractional parts.Wait, let's compute 4001*(1 - log10(9)). Let's calculate 1 - log10(9) ≈ 1 - 0.9542425094 ≈ 0.0457574906. Multiply by 4001: 0.0457574906 * 4001.First, compute 0.0457574906 * 4000 = 0.0457574906 * 4 * 1000 = 0.1830299624 * 1000 = 183.0299624.Then, 0.0457574906 * 1 = 0.0457574906.Add them together: 183.0299624 + 0.0457574906 ≈ 183.07571989. So approximately 183.0757. Since we can't have a fraction of an element, the exact number is either 183 or 184. However, because the problem states that 9^4000 is included and it has MSD 9, that would correspond to the fractional part {4000*log10(9)}. Let's compute that.Since log10(9^4000) = 4000*log10(9) ≈ 4000*0.9542425094 ≈ 3816.9700376. Therefore, the fractional part is 0.9700376. Wait, but 0.9700376 is greater than log10(9) ≈ 0.9542425, so indeed, the fractional part is in [alpha, 1). Therefore, k=4000 is included. So the approximate value is 183.075, so the actual number is 184? But wait, 0.075 is the decimal part, but since each k contributes either 0 or 1, the exact number is floor(4001*(1 - alpha)) + 1? Or maybe not. Let's think.Alternatively, since the fractional parts are equidistributed, the number of k's with fractional part in [alpha, 1) is approximately N*(1 - alpha), where N is the number of terms (4001). Since the problem is from a competition, and given that 9^4000 is included, maybe we can take the floor of N*(1 - alpha) + 0.5) or something. But maybe the answer is 184.But perhaps there is a more exact approach.Alternatively, since we have log10(9^k) = k*log10(9). Let's denote m_k = floor(k*log10(9)), so that log10(9^k) = m_k + f_k, where f_k is the fractional part. Then, 9^k = 10^{m_k + f_k} = 10^{m_k} * 10^{f_k}. Therefore, the leading digit is determined by 10^{f_k}. If the leading digit is 9, then 10^{f_k} ∈ [9, 10), so f_k ∈ [log10(9), 1).So the question is equivalent to finding the number of k where f_k ∈ [log10(9), 1). Let me note that since log10(9) is irrational, the sequence f_k is uniformly distributed in [0,1). Therefore, the proportion of k's with f_k in [log10(9),1) is (1 - log10(9)), so approximately 0.045757. Multiply by 4001 gives approximately 183.075. Since we can't have a fraction, but since the endpoint k=4000 is included and it's one of the numbers that does satisfy the condition, then we need to round up? So 184.But is there a way to confirm this?Alternatively, perhaps the exact answer is floor(4001*(1 - log10(9))) + 1. Let's compute 4001*(1 - log10(9)) ≈ 4001*0.045757 ≈ 183.075, so floor(183.075) = 183, then add 1 gives 184. Hmm, that might be the case. But why would we add 1? Because the fractional parts {k*alpha} for k=0,...,4000. The total number is 4001 terms. If the total is approximately 183.075, then the exact number could be 184. However, this is heuristic.Alternatively, note that the map k → {k*alpha} is a rotation on the unit circle. The number of times the rotation lands in the interval [alpha,1) is equal to floor((n + 1)*beta) - floor(beta), where beta = 1 - alpha. Wait, maybe applying some formula from equidistribution or Beatty sequences.Alternatively, using the fact that the number of solutions to {k*alpha} ∈ [c, d) for k = 0, 1, ..., N-1 is approximately (d - c)*N, and the error term is bounded by something. But in competitive math problems like this, especially from the 1990 AIME, the answer is likely to be 184. But let's check with the given data.Wait, the problem states that 9^4000 has 3817 digits and its most significant digit is 9. So, 9^4000 is included, so that's one element. Then, if the approximate number is 183.075, which is approximately 183, but since we know that the last term is included, maybe the exact number is 184. Alternatively, maybe it's 184.But let's think of another approach. Let me recall that when dealing with leading digits, the fractional parts of k*log10(9) determine the leading digit. So, if we consider the fractional parts {k*log10(9)}, each time k increases by 1, we add log10(9) to the previous fractional part, modulo 1. So, this is like a circular walk where each step is of length log10(9) ≈ 0.9542 on a circle of circumference 1. Each time the walk passes through the interval [log10(9),1), which is an arc of length 1 - log10(9) ≈ 0.045757, we count a hit. The number of hits is approximately the number of times the walk falls into that arc.But since the step length is irrational, the walk is dense and uniformly distributed. Therefore, the number of times it lands in the arc is approximately proportional to the length of the arc. But again, how do we get the exact count?Wait, perhaps we can use the fact that the total number of elements is 4001 (from k=0 to 4000 inclusive). Given that 9^4000 is included, which is one of them. Then, the fractional parts {k*log10(9)} for k=0 to 4000. Let me compute the exact fractional parts.But that would be tedious. Alternatively, think in terms of the sequence {k*alpha} mod 1, with alpha = log10(9). The number of solutions to {k*alpha} ∈ [alpha,1) is equal to the number of k where k*alpha mod 1 ∈ [alpha,1). Which is equivalent to k*alpha - m ∈ [alpha,1) for some integer m. Rearranging, m < k*alpha < m + 1 - alpha. Wait, not sure.Alternatively, think of the fractional parts {k*alpha} = k*alpha - floor(k*alpha). We want {k*alpha} >= alpha. So, k*alpha - floor(k*alpha) >= alpha. Rearranged, k*alpha - alpha >= floor(k*alpha). Let me see: k*alpha - floor(k*alpha) >= alpha => k*alpha - alpha >= floor(k*alpha) => (k - 1)alpha >= floor(k*alpha). Hmm, not sure if that helps.Alternatively, perhaps using the concept of Beatty sequences or some reciprocal relation.Alternatively, since log10(9) + log10(9^k) = log10(9^{k+1}), so if we can relate the fractional parts of log10(9^k) to each other. Wait, but log10(9^k) = k*log10(9), which is the same as we already have.Alternatively, note that if we let f(k) = fractional part of k*log10(9), then f(k+1) = f(k) + log10(9) mod 1. So, each step, we add approximately 0.9542 to the fractional part, modulo 1. Therefore, each step, the fractional part increases by 0.9542, wraps around if it exceeds 1.Therefore, starting from f(0) = 0, f(1) ≈ 0.9542, f(2) ≈ 0.9085, f(3) ≈ 0.8627, f(4) ≈ 0.817, f(5) ≈ 0.7712, f(6) ≈ 0.7255, f(7) ≈ 0.6797, f(8) ≈ 0.634, f(9) ≈ 0.5882, f(10) ≈ 0.5425, f(11) ≈ 0.4967, f(12) ≈ 0.451, f(13) ≈ 0.4052, f(14) ≈ 0.3595, f(15) ≈ 0.3137, f(16) ≈ 0.268, f(17) ≈ 0.2222, f(18) ≈ 0.1765, f(19) ≈ 0.1307, f(20) ≈ 0.085, f(21) ≈ 0.0392, f(22) ≈ 0.9934, f(23) ≈ 0.9477, etc. Wait, this is approximate, but each step, subtracting roughly 0.045757 from the fractional part (since adding 0.9542 mod 1 is the same as subtracting 0.045757).Wait, actually, since log10(9) ≈ 0.9542425094, so 1 - log10(9) ≈ 0.0457574906. Therefore, adding log10(9) mod 1 is equivalent to subtracting (1 - log10(9)) from the current fractional part. Therefore, each step, the fractional part decreases by approximately 0.045757. So, starting from 0 (k=0), the fractional parts go 0, 0.9542, 0.9542 - 0.045757 ≈ 0.9084, then 0.9084 - 0.045757 ≈ 0.8626, and so on, subtracting each time approximately 0.045757.Therefore, the sequence of fractional parts is decreasing by approximately 0.045757 each step. So, every time the fractional part crosses below alpha (0.9542), we wrap around by adding 1 (mod 1). Wait, no. Wait, when we subtract 0.045757 from the previous fractional part, if it goes below 0, we add 1. But since we start at 0, then f(1) = 0 + 0.9542 ≈ 0.9542, then f(2) = 0.9542 - 0.045757 ≈ 0.9084, f(3) ≈ 0.8626, ..., f(21) ≈ 0.0392, f(22) ≈ 0.0392 - 0.045757 ≈ -0.006557, which mod 1 is 0.9934, then f(23) ≈ 0.9934 - 0.045757 ≈ 0.9476, and so on.So, the fractional parts decrease by approximately 0.045757 each time, but when they cross below 0, they wrap around to 1 - 0.045757 ≈ 0.9542. Wait, actually, each time the subtraction causes a wrap-around, the fractional part jumps up by 1 - 0.045757 ≈ 0.9542. Therefore, the sequence of fractional parts is a linearly decreasing sequence with step -0.045757, wrapping around when necessary.But this seems similar to a sawtooth wave decreasing, with each tooth of length 1 / 0.045757 ≈ 21.85 steps. So approximately every 22 steps, the fractional part wraps around from below 0 to near 1.Given that, we can model the number of times the fractional part falls into [alpha, 1). Each time the fractional part is in that interval, we have a hit.Since the fractional parts decrease by approximately 0.045757 each step, the interval [alpha, 1) is of length 0.045757. Therefore, each time the fractional part passes through this interval, it stays there for one step. Wait, but if the step is exactly 0.045757, then starting from a point just above alpha, subtracting 0.045757 would take it to just above alpha - 0.045757. But since the interval is of the same length as the step, perhaps each time the fractional part is in [alpha, 1), it only occurs once per cycle.Wait, this is getting a bit convoluted. Maybe there's a better way.Alternatively, consider that each time the fractional part wraps around (i.e., when k increases by 1 / (1 - alpha) ≈ 21.85), we get one hit in the interval [alpha, 1). Therefore, over 4001 terms, the number of hits would be approximately 4001 / (1 / (1 - alpha)) ) = 4001*(1 - alpha). Which is exactly the same as before. So, this again gives us approximately 183.075, so 183 or 184.But the problem is from 1990, so the answer is likely 184, given that the fractional calculation is about 183.075, and we have to include k=4000. But how do we confirm?Alternatively, maybe the answer is 184, which is the nearest integer.But let's think differently. Let's compute the number of k's such that the fractional part of k*log10(9) is in [log10(9), 1). Let's denote alpha = log10(9). Then, the condition is {k*alpha} >= alpha.But since alpha is irrational, the sequence {k*alpha} is dense in [0,1). So, the number of solutions is roughly (1 - alpha)*N, where N=4001.Alternatively, we can use the concept of circular overlap. Since each step, we subtract approximately beta = 1 - alpha ≈ 0.045757 from the fractional part. Therefore, each time we subtract beta, and when we cross below 0, we add 1. Therefore, each "cycle" consists of floor(1 / beta) = floor(1 / 0.045757) ≈ 21.85 steps. Therefore, each cycle of approximately 22 steps, there is one hit where the fractional part is in [alpha, 1). Therefore, the number of cycles in 4001 steps is approximately 4001 / 21.85 ≈ 183.1 cycles. Therefore, approximately 183 hits. But since we start at k=0, which is f=0, and end at k=4000, which is a hit, maybe we have 184 hits. Wait, 183 cycles give 183 hits, but since the last hit is at k=4000, which is beyond 183 cycles, maybe it's 184.Alternatively, compute the number of times the fractional part wraps around. Each wrap-around occurs after 1 / beta ≈ 21.85 steps. So, the number of wrap-arounds is floor(4001 * beta). Wait, not sure.Alternatively, the number of times the fractional part crosses from above alpha to below alpha is equal to the number of wrap-arounds, which is floor(k_max * beta + 1). Wait, this is getting too vague.Alternatively, use the formula for the number of solutions to {k*alpha} ∈ [c, d) for k = 0, 1, ..., N-1, which is approximately N*(d - c) + error, where the error is bounded by the discrepancy of the sequence. For irrational rotations, the discrepancy is known to be O((log N)/N)), so the error is O(log N). Therefore, for N=4001, the error is O(log 4001) ≈ 8. So, the exact number of solutions is approximately 183.075 ±8. Therefore, the answer could be between 175 and 191. But the problem states that 9^4000 is included, so we can check if the error term could account for the 0.075, making the answer 183 or 184. However, given that the problem is from a competition, and that 9^4000 is included, which is one of the numbers we need to count, the answer is likely 184.But how to confirm? Wait, let's note that 9^k cycles its leading digit in a way that depends on the fractional part of k*log10(9). Since log10(9) is irrational, the fractional parts never repeat, so the leading digits don't cycle periodically. However, over time, the leading digits follow a logarithmic distribution, where the probability of a leading digit d is log10(1 + 1/d). For d=9, this probability is log10(1 + 1/9) ≈ log10(10/9) ≈ 0.045757, which matches (1 - log10(9)). Therefore, the number of elements with leading digit 9 is approximately 4001 * 0.045757 ≈ 183.075, so 183 or 184. But since 9^4000 is included, and 0.075 is almost 0.1, it's 184.Alternatively, note that the fractional parts {k*alpha} for k=0,1,...,n-1 have a certain number of times they fall into [alpha,1). According to the theory of equidistribution, the discrepancy D(n) for this sequence is bounded by something like O(1/n) or O(log n /n). For large n, the error term is small. However, since the problem gives us that 9^4000 has 3817 digits and starts with 9, which is the last term, that term is included. So, if our approximate count is 183.075, we can infer that the exact count is 184. Since 0.075 is more than 0.05, we round up.Alternatively, compute the exact fractional part for k=4000. Wait, let's compute log10(9^4000) = 4000*log10(9). We are told that this has 3817 digits, so the characteristic (integer part) is 3816. Therefore, log10(9^4000) = 3816 + f, where f ≈ 0.9700376. Which is in [alpha,1) as given. So, that's one count. Now, the approximate total is 183.075, so perhaps 184.But how can we check this more precisely?Alternatively, note that the number of solutions is equal to the number of integers k in [0,4000] such that there exists an integer m with 9*10^m ≤ 9^k <10^{m+1}. Dividing by 9, we get 10^m ≤ 9^{k-1} < (10/9)*10^m. Taking log10:m ≤ (k -1)*log10(9) < log10(10/9) + mTherefore,m - (k -1)*log10(9) ≤0 < log10(10/9) + m - (k -1)*log10(9)Wait, not sure.Alternatively, rewrite the inequality 9^k >= 9*10^m and 9^k <10^{m+1}.Take log10:k*log10(9) >= log10(9) + mandk*log10(9) < m +1So,m <=k*log10(9) - log10(9)andk*log10(9) <m +1So,k*log10(9) - log10(9) >=mandm >k*log10(9) -1So combining,k*log10(9) -1 <m <=k*log10(9) - log10(9)But m must be integer. Therefore, there is an integer m in the interval (k*log10(9) -1, k*log10(9) - log10(9)]Therefore, the number of solutions is equal to the number of integers k in [0,4000] for which there exists an integer m in that interval. The length of the interval is log10(9) -1 ≈ 0.9542 -1= -0.0458. Wait, that doesn't make sense. Wait, perhaps I made a mistake in the inequality.Wait, starting again:We need 9*10^m <=9^k <10^{m+1}Divide by 9: 10^m <=9^{k-1} <(10/9)*10^mTake log10:m <= (k -1)*log10(9) < m + log10(10/9)Subtract m:0 <= (k -1)*log10(9) -m <log10(10/9)Let n = (k -1)*log10(9) -m.Then, 0 <=n <log10(10/9)≈0.045757.Therefore, the fractional part of (k -1)*log10(9) must be less than 0.045757.But (k -1)*log10(9) =k*log10(9) -log10(9). So, fractional part of (k*log10(9) -log10(9)) = fractional part of k*log10(9) -log10(9) + floor(k*log10(9) -log10(9)).But since fractional part is {x} = x - floor(x), then:{(k -1)*log10(9)} = {k*log10(9) -log10(9)} = { {k*log10(9)} - log10(9) + floor(k*log10(9)) - log10(9) }Wait, this seems complicated. Maybe an alternative approach.Alternatively, note that if we let l =k -1, then the condition is {l*log10(9)} <log10(10/9). So, the number of k where {l*log10(9)} <0.045757, where l =k -1, so l ranges from -1 to 3999. But k >=1 (since l >=-1), but k=0 corresponds to l=-1, which would be { -log10(9) } = {1 -log10(9)} ≈0.045757, which is not less than 0.045757. Therefore, for k=0, there is no m. For k >=1, we need to check if { (k -1)*log10(9) } <0.045757.Therefore, the number of solutions is equal to the number of integers l=0 to 3999 where {l*log10(9)} <0.045757, plus possibly k=0. But since k=0: 9^0=1, which does not start with 9, so it's excluded. Therefore, the total number of elements in T with MSD 9 is equal to the number of l in 0 to 3999 where {l*log10(9)} <0.045757.But 0.045757 is equal to 1 - log10(9). Therefore, the number of l where {l*log10(9)} <1 - log10(9). This is the same as the number of l where {l*alpha} <1 - alpha, where alpha=log10(9). But since alpha is irrational, this is again approximately N*(1 - alpha). Therefore, for l=0 to 3999, N=4000, so approximately 4000*0.045757≈183.03. So, approximately 183. Then, adding k=4000, which is another count, gives 183 +1=184.Ah! This is a better approach. Since the original problem is counting k=0 to 4000 where 9^k starts with 9. By rephrasing the inequality, we find that this is equivalent to counting l=0 to 3999 where {l*log10(9)} <1 - log10(9)≈0.045757, and then adding 1 because k=4000 is included. Therefore, the total is approximately 4000*0.045757≈183.03 +1≈184.03. Therefore, the exact answer is 184.Therefore, the number of elements in T with MSD 9 is 184.Final Answerboxed{184}

✅:To determine how many elements in the set ( T = {9^k mid 0 leqslant k leqslant 4000} ) have 9 as their most significant digit, we analyze the fractional parts of ( k log_{10}(9) ).1. Understanding the Most Significant Digit: - The most significant digit (MSD) of ( 9^k ) is 9 if and only if the fractional part of ( k log_{10}(9) ) lies in the interval ([ log_{10}(9), 1 )). - This is because ( 9^k ) starting with 9 implies ( 9 times 10^m leq 9^k < 10^{m+1} ) for some integer ( m ). Taking the logarithm base 10, this translates to the fractional part condition.2. Key Values and Equidistribution: - ( log_{10}(9) approx 0.9542 ). - The interval ([ log_{10}(9), 1 )) has length ( 1 - log_{10}(9) approx 0.045757 ). - Since ( log_{10}(9) ) is irrational, the fractional parts ( { k log_{10}(9) } ) are uniformly distributed modulo 1.3. Approximate Count: - The number of ( k ) such that ( { k log_{10}(9) } in [ log_{10}(9), 1 ) ) is approximately ( (1 - log_{10}(9)) times 4001 approx 0.045757 times 4001 approx 183.075 ).4. Exact Count Consideration: - Given the problem states ( 9^{4000} ) has 3817 digits and its MSD is 9, ( k = 4000 ) is included. - The approximate value 183.075 suggests rounding to 184, considering the inclusion of ( k = 4000 ).5. Final Calculation: - The number of elements in ( T ) with MSD 9 is determined by the equidistribution theorem and rounding considerations, resulting in 184.Thus, the number of elements in ( T ) that have 9 as their most significant digit is (boxed{184}).

🔑:We need to determine how many elements in the set (T = {9^k mid k text{ is an integer}, 0 leq k leq 4000}) have 9 as their leftmost digit.First, we note the following:1. If (9^t) has 9 as its leftmost digit, then (9^{t-1}) must have 1 as its leftmost digit.To see why, assume that the leftmost digit of (9^{t-1}) is not 1. Specifically, if the leftmost digit of (9^{t-1}) were any of (2, 3, 4, 5, 6, 7, 8), then the leftmost digit of (9^{t-1} cdot 9 = 9^t) will still be one of these digits and not 9. For instance:- If (9^{t-1} = 2 cdot 10^n, 3 cdot 10^n, ldots, 8 cdot 10^n,) multiplying by 9 results in values like (1.8 cdot 10^{n+1}, 2.7 cdot 10^{n+1}, ldots, 7.2 cdot 10^{n+1}) respectively, none of these would start with the digit 9.2. If (9^{t-1}) starts with the digit 9, then consider:[ 9^{t-1} leq 9.999 cdot 10^{d-1} Rightarrow 9 cdot 9^{t-1} leq 89.991 cdot 10^{d} = 8.9911 cdot 10^{d+1}.]In this case, the leftmost digit of (9^t) will also not be 9. This implies that (9^{t-1}) having 9 as its leftmost digit is not feasible.Given (9^{4000}) has the leftmost digit as 9 and spans 3817 digits, and (10^{4000}) has 4001 digits, the decrease in the number of digits from (10^k) to (9^k) results in:[ 4001 - 3817 = 184. ]Thus, we calculate that there are 184 digits in the transition from (10^k) to (9^k) which occur without increment in numbers of digits and retaining the first digit of 9. Conclusion:The total number of elements in (T) which have 9 as their leftmost digit is therefore:[boxed{184}]

🧠:Okay, so I need to find the maximum value of the function ( 2^{x(1 - x)} ) where ( x ) is a real number. Hmm, let's start by understanding what this function looks like. First, I remember that exponential functions like ( 2^y ) are always increasing if the base is greater than 1, which it is here since the base is 2. That means the maximum value of ( 2^{x(1 - x)} ) will occur when the exponent ( x(1 - x) ) is maximized. So, instead of dealing with the exponential function directly, I can focus on maximizing the quadratic expression in the exponent, right?Let me write that down. Let ( y = x(1 - x) ). Then the problem reduces to finding the maximum value of ( y ) and then computing ( 2^y ). So, first step: maximize ( y = x(1 - x) ).Expanding that, ( y = x - x^2 ), which is a quadratic equation in terms of ( x ). Since the coefficient of ( x^2 ) is negative (-1), the parabola opens downward, which means the vertex is the maximum point. The vertex of a parabola given by ( ax^2 + bx + c ) is at ( x = -frac{b}{2a} ). In this case, ( a = -1 ) and ( b = 1 ), so plugging those into the formula: ( x = -frac{1}{2(-1)} = frac{1}{2} ). Therefore, the maximum value of ( y = x(1 - x) ) occurs at ( x = frac{1}{2} ). Now, substituting ( x = frac{1}{2} ) back into the expression for ( y ):( y = frac{1}{2}(1 - frac{1}{2}) = frac{1}{2} times frac{1}{2} = frac{1}{4} ).So the maximum value of the exponent ( y ) is ( frac{1}{4} ). Therefore, the maximum value of ( 2^{x(1 - x)} ) is ( 2^{frac{1}{4}} ).But wait, let me verify this. Maybe I should check the second derivative to ensure that it's indeed a maximum. Although since the quadratic opens downward, we know it's a maximum, but just to be thorough.First, let's consider the original function ( f(x) = 2^{x(1 - x)} ). To find its maximum, we can take the derivative and set it to zero. Let me try that approach.First, take the natural logarithm of ( f(x) ) to make differentiation easier. Let ( ln(f(x)) = x(1 - x) ln 2 ). Then, the derivative of ( ln(f(x)) ) is ( f'(x)/f(x) ). So:( frac{d}{dx} ln(f(x)) = frac{d}{dx} [x(1 - x) ln 2] ).Calculating the right-hand side:( frac{d}{dx} [x(1 - x) ln 2] = ln 2 times frac{d}{dx} [x - x^2] = ln 2 times (1 - 2x) ).Therefore, ( frac{f'(x)}{f(x)} = ln 2 times (1 - 2x) ).Setting this equal to zero to find critical points:( ln 2 times (1 - 2x) = 0 ).Since ( ln 2 ) is not zero, we can divide both sides by ( ln 2 ):( 1 - 2x = 0 Rightarrow x = frac{1}{2} ).So, this confirms that the critical point is at ( x = frac{1}{2} ), which matches our earlier result. To ensure it's a maximum, we can check the second derivative or observe the behavior around that point.Calculating the second derivative might be a bit involved, but let's try. Starting from the first derivative:( f'(x) = f(x) times ln 2 times (1 - 2x) ).Taking the second derivative:( f''(x) = frac{d}{dx} [f(x) times ln 2 times (1 - 2x)] ).Using the product rule:( f''(x) = f'(x) times ln 2 times (1 - 2x) + f(x) times ln 2 times (-2) ).At the critical point ( x = frac{1}{2} ), the first term becomes zero because ( 1 - 2x = 0 ). So:( f''(frac{1}{2}) = 0 + f(frac{1}{2}) times ln 2 times (-2) ).Since ( f(frac{1}{2}) = 2^{frac{1}{4}} ) is positive and ( ln 2 ) is positive, multiplying by -2 gives a negative result. Therefore, the second derivative is negative at ( x = frac{1}{2} ), confirming that this critical point is indeed a local maximum. Hence, the function attains its maximum value at ( x = frac{1}{2} ).So, the maximum value of ( 2^{x(1 - x)} ) is ( 2^{frac{1}{4}} ). But just to express this in another form, ( 2^{frac{1}{4}} ) is the fourth root of 2, which can also be written as ( sqrt[4]{2} ).Alternatively, since ( 2^{frac{1}{4}} = (2^{frac{1}{2}})^{frac{1}{2}} = sqrt{sqrt{2}} approx sqrt{1.4142} approx 1.1892 ), but the exact value is ( 2^{1/4} ).Let me check if there's another way to approach this problem. For example, using substitution or maybe completing the square for the exponent. The exponent ( x(1 - x) = -x^2 + x ), which is a quadratic. Completing the square:( -x^2 + x = -left(x^2 - xright) ).To complete the square inside the parentheses:( x^2 - x = x^2 - x + frac{1}{4} - frac{1}{4} = left(x - frac{1}{2}right)^2 - frac{1}{4} ).Therefore,( -x^2 + x = -left[left(x - frac{1}{2}right)^2 - frac{1}{4}right] = -left(x - frac{1}{2}right)^2 + frac{1}{4} ).So the exponent becomes ( frac{1}{4} - left(x - frac{1}{2}right)^2 ). Therefore, the original function is:( 2^{frac{1}{4} - left(x - frac{1}{2}right)^2} = 2^{frac{1}{4}} times 2^{- left(x - frac{1}{2}right)^2} ).Since ( 2^{- left(x - frac{1}{2}right)^2} ) is always less than or equal to 1 (because the exponent is non-positive), the maximum value occurs when ( left(x - frac{1}{2}right)^2 = 0 ), i.e., when ( x = frac{1}{2} ). Therefore, the maximum value is ( 2^{frac{1}{4}} ), which confirms our previous result.Another way to think about this is that the function ( 2^{- left(x - frac{1}{2}right)^2} ) is a Gaussian-like function centered at ( x = frac{1}{2} ), which has its peak there. Multiplying by ( 2^{frac{1}{4}} ) just scales it, so the peak remains at ( 2^{frac{1}{4}} ).So all methods point to the maximum value being ( 2^{1/4} ). Let me just verify with a quick example. If I plug in ( x = 0 ), the function is ( 2^{0} = 1 ). At ( x = 1 ), same thing. At ( x = 1/2 ), it's ( 2^{1/4} approx 1.189 ), which is indeed larger than 1. If I try ( x = 1/4 ), then ( x(1 - x) = 1/4 * 3/4 = 3/16 approx 0.1875 ), so ( 2^{0.1875} approx 2^{3/16} approx 1.139 ), which is less than 1.189. Similarly, if I try ( x = 1/3 ), then ( x(1 - x) = 1/3 * 2/3 = 2/9 approx 0.222 ), so ( 2^{0.222} approx 1.171 ), still less than 1.189. So that seems consistent.Therefore, after multiple approaches and verification, I'm confident that the maximum value of ( 2^{x(1 - x)} ) is ( 2^{1/4} ).Final AnswerThe maximum value is boxed{sqrt[4]{2}}.

✅:To find the maximum value of ( 2^{x(1-x)} ) where ( x ) is a real number, we start by analyzing the exponent ( x(1 - x) ). 1. Quadratic Analysis: - The expression ( x(1 - x) = -x^2 + x ) is a quadratic function opening downwards (since the coefficient of ( x^2 ) is negative). - The maximum value of this quadratic occurs at the vertex. For a quadratic ( ax^2 + bx + c ), the vertex is at ( x = -frac{b}{2a} ). - Here, ( a = -1 ) and ( b = 1 ), so the vertex is at ( x = frac{1}{2} ).2. Calculating the Maximum Exponent: - Substituting ( x = frac{1}{2} ) into ( x(1 - x) ): [ left( frac{1}{2} right) left( 1 - frac{1}{2} right) = frac{1}{2} times frac{1}{2} = frac{1}{4} ] - Therefore, the maximum value of the exponent is ( frac{1}{4} ).3. Exponential Function: - The maximum value of ( 2^{x(1 - x)} ) is ( 2^{frac{1}{4}} ).4. Verification via Derivatives: - Taking the natural logarithm of the function ( f(x) = 2^{x(1 - x)} ) and differentiating: [ frac{d}{dx} left[ x(1 - x) ln 2 right] = ln 2 times (1 - 2x) ] - Setting the derivative to zero confirms the critical point at ( x = frac{1}{2} ). - The second derivative test confirms this is a maximum.5. Completing the Square: - Rewriting the exponent ( -x^2 + x ) by completing the square: [ -left( left( x - frac{1}{2} right)^2 - frac{1}{4} right) = frac{1}{4} - left( x - frac{1}{2} right)^2 ] - This shows the maximum value of the exponent is ( frac{1}{4} ), leading to the maximum value of the function as ( 2^{frac{1}{4}} ).Thus, the maximum value of ( 2^{x(1-x)} ) is (boxed{sqrt[4]{2}}).

🔑:To find the maximum value of (2^{x(1-x)}), we can break it down step-by-step.1. Express the exponent as a function: Let ( y = x(1 - x) ). So, we need to maximize ( y ).2. Rewrite the function in terms of ( y ): We have the function ( y = -x^2 + x ).3. Identify the vertex of the quadratic function: The quadratic function ( y = -x^2 + x ) opens downward because the coefficient of ( x^2 ) is negative. The vertex of a parabola ( ax^2 + bx + c ) occurs at ( x = -frac{b}{2a} ).4. Find the vertex: For our function, ( a = -1 ) and ( b = 1 ). Thus, [ x = -frac{1}{2(-1)} = frac{1}{2}. ]5. Calculate the maximum value of ( y ): Plug the ( x ) value of the vertex back into the function to determine the maximum ( y ): [ y = - left( frac{1}{2} right)^2 + frac{1}{2} = -frac{1}{4} + frac{1}{2} = frac{1}{4}. ]6. Substitute ( y ) back into the original expression: Since ( y = x(1-x) ), the maximum value of ( 2^{x(1-x)} ) occurs when ( y = frac{1}{4} ).7. Compute ( 2^y ) at the maximum ( y ): Finally, we substitute ( y = frac{1}{4} ) into ( 2^y ): [ 2^{frac{1}{4}} = sqrt[4]{2}. ] Conclusion:The maximum value of ( 2^{x(1-x)} ) is ( boxed{sqrt[4]{2}} ).

🧠:Okay, so I have these parametric equations for a line: x equals R cosine t and y equals R sine t. The task is to convert them into an equation with just two variables, which I assume means eliminating the parameter t to get a relationship between x and y. Let me think about how to approach this.First, parametric equations use a parameter, in this case t, to define both x and y. To eliminate the parameter, I need to find a way to express t in terms of x or y and then substitute that into the other equation. Alternatively, maybe I can use a trigonometric identity since both x and y are expressed in terms of cosine and sine of the same parameter. Looking at the equations: x = R cos t and y = R sin t. If I can relate cos t and sin t, perhaps using the Pythagorean identity cos²t + sin²t = 1. That seems promising. Let me try that.Starting with x = R cos t. If I solve for cos t, I get cos t = x/R. Similarly, from y = R sin t, sin t = y/R. Then, substituting these into the identity:cos²t + sin²t = (x/R)² + (y/R)² = 1.Simplifying that equation: (x² + y²)/R² = 1. Multiplying both sides by R² gives x² + y² = R². Wait, that's the equation of a circle with radius R centered at the origin. But the problem says it's a line. Hmm, that's confusing. Did I misinterpret the original equations? Let me check again.The parametric equations are x = R cos t and y = R sin t. As t varies, these equations trace out a circle of radius R, not a line. So maybe there's a mistake in the problem statement? Or perhaps I need to reconsider.Wait, the problem mentions converting the parametric equations of a line. But according to the equations given, it's a circle. Unless there's a typo, maybe R is a constant and t is the parameter. But regardless, following the steps, eliminating t gives x² + y² = R², which is a circle equation. So either the original problem has a mistake, referring to a line instead of a circle, or maybe I misunderstood the question. Let me read the problem again: "Given the parametric equations of a line: x = R cos t, y = R sin t. Convert them to an equation with two variables." Hmm. The equations provided are for a circle, not a line. A line would typically have parametric equations like x = x0 + at and y = y0 + bt, where a and b are constants. So unless R is a function of t, but here R is probably a constant. Alternatively, maybe in some context, these equations can represent a line. Let me think. If R is fixed and t varies, then (x, y) moves around a circle. If R is varying as a function of t, but the problem states R as a constant, right? The equations are written as x = R cos t, y = R sin t, which is the standard parametrization of a circle. So perhaps the problem has an error, and they meant to say a circle instead of a line. But assuming the equations are correct as given, even though they represent a circle, the answer would still be x² + y² = R². Maybe the user intended to provide parametric equations of a line but made a mistake. However, based on the given information, I should proceed with the equations as they are.Alternatively, maybe there's another way to interpret the parametric equations as a line. Wait, if R is a parameter and t is fixed? But no, usually in parametric equations, the parameter is the variable that changes, so t is the parameter here. So unless R is also a variable, which isn't indicated. The problem says "the parametric equations of a line," so maybe there's a misunderstanding here. Alternatively, could this represent a line in polar coordinates? Because in polar coordinates, r = R is a circle, but if theta is fixed, then it's a line (a radial line). Wait, no. If theta is fixed, say theta = t, then r varies, but here we have x = R cos t, y = R sin t, where R is a constant. That would mean that as t changes, the point (x, y) moves around the circle of radius R. So it's definitely a circle. Therefore, unless there's a different interpretation, the conclusion is that the problem statement might have a typo, and "line" should be "circle." However, given the equations, the Cartesian equation is x² + y² = R². Alternatively, if we consider R as a variable and t as a fixed angle, but that's not standard. Usually, parametric equations use a single parameter (here, t) and express x and y in terms of that parameter. So R is likely a constant. Therefore, I think the correct answer is x² + y² = R², even though the problem mentions a line. Unless there's additional context or information missing, this seems to be the only possible conclusion. Just to double-check, if we consider t as a parameter, and for each t, we get a point (R cos t, R sin t). As t increases from 0 to 2π, the points trace a circle. Therefore, the Cartesian equation is indeed a circle. So despite the problem's mention of a line, the parametric equations given correspond to a circle. Therefore, converting them to an equation in x and y gives the circle equation. Maybe the user intended to write parametric equations for a line but made a mistake. But given the information provided, the answer must be x² + y² = R².

✅:The given parametric equations are ( x = R cos t ) and ( y = R sin t ). To eliminate the parameter ( t ) and derive a relationship between ( x ) and ( y ):1. Solve for ( cos t ) and ( sin t ): [ cos t = frac{x}{R}, quad sin t = frac{y}{R} ]2. Apply the Pythagorean identity ( cos^2 t + sin^2 t = 1 ): [ left(frac{x}{R}right)^2 + left(frac{y}{R}right)^2 = 1 ]3. Simplify the equation: [ frac{x^2 + y^2}{R^2} = 1 implies x^2 + y^2 = R^2 ]Conclusion: The parametric equations describe a circle of radius ( R ) centered at the origin. The final Cartesian equation is:[boxed{x^2 + y^2 = R^2}]Note: The original problem mentions a "line," but the equations provided parametrize a circle. The solution assumes the equations are correct as given. If the goal was to represent a line, there may be an error in the parametric equations.

🔑:1. The given parametric equations are: [ x = R cos t, quad y = R sin t ]2. From the second equation ( y = R sin t ), solve for ( sin t ): [ sin t = frac{y}{R} ]3. Use the Pythagorean identity ( cos^2 t + sin^2 t = 1 ) to find ( cos t ): [ cos t = sqrt{1 - sin^2 t} ]4. Substitute ( sin t = frac{y}{R} ) into the identity: [ cos t = sqrt{1 - left(frac{y}{R}right)^2} ]5. Simplify the expression for ( cos t ): [ cos t = sqrt{1 - frac{y^2}{R^2}} = sqrt{frac{R^2 - y^2}{R^2}} = frac{sqrt{R^2 - y^2}}{R} ]6. Substitute this expression for ( cos t ) back into the first parametric equation: [ x = R cos t = R cdot frac{sqrt{R^2 - y^2}}{R} = sqrt{R^2 - y^2} ]7. Square both sides to eliminate the square root: [ x^2 = R^2 - y^2 ]8. Rearrange the equation to the standard form of a circle equation: [ x^2 + y^2 = R^2 ]Conclusion. We have arrived at the Cartesian equation of the circle with radius ( R ) and center at the origin (0,0).[ boxed{x^2 + y^2 = R^2} ]

🧠:Okay, so I need to prove that the numbers from 1 to 16 can be written in a row where each pair of adjacent numbers adds up to a perfect square, but this isn't possible if we try to arrange them in a circle with the same condition. Let me start by understanding the problem and breaking it down.First, for the row (a linear arrangement), we need to find a permutation of numbers 1 to 16 such that every consecutive pair sums to a square number. For the circle (a cyclic arrangement), the same condition applies, but also the first and last numbers must sum to a square since they become adjacent in the circle. The task is to show that such a row exists, but a circle doesn't.Let me start with the row. I remember that this is a known problem, sometimes called the "square-sum problem." There are solutions for sequences up to certain numbers. Maybe for 16, someone has already found such an arrangement. But since I need to prove it, I should think about how to construct such a sequence or at least argue that it's possible.One approach could be to model this as a graph problem. Each number from 1 to 16 is a node, and there's an edge between two numbers if their sum is a perfect square. Then, finding such a row is equivalent to finding a Hamiltonian path in this graph. Similarly, arranging them in a circle would require a Hamiltonian circuit. So, if I can show that the graph has a Hamiltonian path but no Hamiltonian circuit, that would prove the statement.First, let me construct this graph. Let's list all possible pairs of numbers between 1 and 16 that add up to a square. The possible squares between 1+2=3 and 16+15=31 are 4, 9, 16, 25, and 36. Wait, 16 is 4², 25 is 5², 36 is 6². So possible square sums are 4, 9, 16, 25, 36. Let's check the maximum possible sum: 16+15=31, which is less than 36. Wait, 16+15=31, so the largest square less than or equal to 31 is 25 (5²=25, 6²=36). Wait, no, 25 is 5², 16 is 4². Wait, 25 is 25, 16 is 16. So possible squares are 4, 9, 16, 25. Wait, 1+3=4, which is 2². 16+9=25, which is 5². 16+20=36, but 20 isn't in the range. So the possible squares we can get are 4, 9, 16, 25.Wait, let's verify. The smallest possible sum is 1+2=3, which is not a square. The next possible sum is 1+3=4 (2²). The next square is 9 (3²), which can be achieved by pairs like 1+8, 2+7, 3+6, 4+5. Then 16 (4²) would require pairs like 1+15, 2+14, 3+13, 4+12, 5+11, 6+10, 7+9, 8+8. But since numbers are distinct, 8+8 is invalid. Similarly, 25 (5²) would be 9+16, 10+15, 11+14, 12+13. And 36 (6²) would need 16+20, but 20 isn't in our set. So the possible square sums are 4, 9, 16, 25.So each number from 1 to 16 can connect to others if their sum is in {4,9,16,25}. Let's note down for each number which others it can connect to.Let me list each number and its possible neighbors:1: Can pair with 3 (1+3=4), 8 (1+8=9), 15 (1+15=16), but 1+24=25 is too big. Wait, 25-1=24, which is beyond 16. So 1's possible neighbors are 3, 8, 15.2: 2+2=4, but same number, so invalid. 2+7=9, 2+14=16, 2+23=25 (invalid). So neighbors:7,14.3: 1 (3+1=4), 6 (3+6=9), 13 (3+13=16), 22 (too big). So neighbors:1,6,13.4: 5 (4+5=9), 12 (4+12=16), 21 (invalid). So neighbors:5,12.5: 4 (5+4=9), 11 (5+11=16), 20 (invalid). So neighbors:4,11.6: 3 (6+3=9), 10 (6+10=16), 19 (invalid). So neighbors:3,10.7: 2 (7+2=9), 9 (7+9=16), 18 (invalid). So neighbors:2,9.8: 1 (8+1=9), 8 (invalid), 17 (invalid). Wait, 8+ something =16? 8+8=16, but same number. 8+17=25, but 17 is too big. So neighbors:1.Wait, 8 can pair with 1 (8+1=9) and also with 17 for 25, but 17 is out of range. So 8 only connects to 1?Wait, 8+16=24, not a square. Wait, 8+? Let's check again. For sum=16: 16-8=8, but same number. For sum=25: 25-8=17, which is out of range. So yes, 8 can only connect to 1. That seems restrictive.9: 7 (9+7=16), 16 (9+16=25). So neighbors:7,16.10: 6 (10+6=16), 15 (10+15=25). So neighbors:6,15.11: 5 (11+5=16), 14 (11+14=25). So neighbors:5,14.12: 4 (12+4=16), 13 (12+13=25). So neighbors:4,13.13: 3 (13+3=16), 12 (13+12=25). So neighbors:3,12.14: 2 (14+2=16), 11 (14+11=25). So neighbors:2,11.15: 1 (15+1=16), 10 (15+10=25). So neighbors:1,10.16: 9 (16+9=25). Because 16+something=25, so 25-16=9. So neighbors:9.Wait, that's important. Let me check again for each number:1: 3,8,152:7,143:1,6,134:5,125:4,116:3,107:2,98:19:7,1610:6,1511:5,1412:4,1313:3,1214:2,1115:1,1016:9Hmm. So let's note the degrees (number of neighbors) for each node:1: 3 neighbors (3,8,15)2: 2 (7,14)3: 3 (1,6,13)4:2 (5,12)5:2 (4,11)6:2 (3,10)7:2 (2,9)8:1 (1)9:2 (7,16)10:2 (6,15)11:2 (5,14)12:2 (4,13)13:2 (3,12)14:2 (2,11)15:2 (1,10)16:1 (9)So in this graph, nodes 8 and 16 have degree 1. Nodes 1 and 3 have degree 3, others have degree 2.Now, in a Hamiltonian path (which is the row), we can have two nodes of degree 1 as the endpoints. But in a Hamiltonian cycle (circle), all nodes must have even degrees because it's a cycle, which is an Eulerian condition. Wait, but Hamiltonian cycle doesn't require Eulerian conditions. Wait, no, Hamiltonian cycle is a cycle that visits every node exactly once, so the degrees aren't directly related to Eulerian trails. But in any case, for a Hamiltonian cycle, the graph must be connected, and there's no requirement on node degrees, except that certain degree conditions can prevent a Hamiltonian cycle. For example, if a node has degree 1, then it's impossible to have a cycle including it because it can't have two edges. But in our case, nodes 8 and 16 have degree 1. Therefore, in any cycle, each node must have degree at least 2. But here, 8 and 16 have degree 1, so they cannot be part of any cycle. Therefore, the graph cannot have a Hamiltonian cycle because those nodes cannot be included in a cycle. Wait, but in the problem statement, we need to arrange all numbers from 1 to 16 in a circle, which would require all nodes to be part of the cycle. But nodes 8 and 16 have degree 1, meaning they can only be connected to one node each. In a cycle, each node must have exactly two neighbors (the previous and next in the cycle). Therefore, nodes with degree 1 cannot be part of a cycle. Hence, the graph as constructed cannot have a Hamiltonian cycle because nodes 8 and 16 can't be included properly. Therefore, arranging them in a circle is impossible. That seems like a solid argument.But wait, maybe there's a mistake here. Let me check again. If nodes 8 and 16 have degree 1, then in any cycle, each node must have two neighbors. But 8 only connects to 1, so 8 can't have two neighbors. Similarly, 16 only connects to 9. Therefore, they can't be part of a cycle. Thus, it's impossible to have a Hamiltonian cycle including all 16 nodes. Therefore, the circular arrangement is impossible.But the problem says to prove that such a circle cannot exist. So this would be the reason. Therefore, the key point is that nodes 8 and 16 have only one possible neighbor each, making a cycle impossible because they can't have two neighbors in the cycle.For the row, the Hamiltonian path, it's possible because we can start at 8, go to 1, then proceed through other nodes, ending at 16 via 9, or start at 16, go to 9, etc. Since 8 and 16 are the only nodes with degree 1, they must be the endpoints of the path. So if there exists a Hamiltonian path from 8 to 16, then the row is possible.Now, the next step is to confirm that such a Hamiltonian path exists. This might require constructing it or using some theorem. Since constructing it manually would be tedious, perhaps there's a known path or a way to argue existence.Alternatively, I can try to construct the path step by step. Let me attempt that.Starting from 8, which can only go to 1. So 8-1. From 1, possible next nodes are 3,8,15. But 8 is already used, so next is 3 or 15.Let's try 8-1-3. From 3, options are 1,6,13. 1 is used, so 6 or 13.Take 6 next: 8-1-3-6. From 6, options are 3,10. 3 used, so 10.8-1-3-6-10. From 10, options are 6,15. 6 used, so 15.8-1-3-6-10-15. From 15, options are 1,10. Both used, dead end. Hmm.Alternatively, at 3, go to 13 instead of 6: 8-1-3-13. From 13, options are 3,12. 3 used, so 12.8-1-3-13-12. From 12, options are 4,13. 13 used, go to 4.8-1-3-13-12-4. From 4, options are 5,12. 12 used, go to 5.8-1-3-13-12-4-5. From 5, options are 4,11. 4 used, go to 11.8-1-3-13-12-4-5-11. From 11, options are 5,14. 5 used, go to 14.8-1-3-13-12-4-5-11-14. From 14, options are 2,11. 11 used, go to 2.8-1-3-13-12-4-5-11-14-2. From 2, options are 7,14. 14 used, go to 7.8-1-3-13-12-4-5-11-14-2-7. From 7, options are 2,9. 2 used, go to 9.8-1-3-13-12-4-5-11-14-2-7-9. From 9, options are 7,16. 7 used, go to 16.Now, we have 8-1-3-13-12-4-5-11-14-2-7-9-16. But we still have numbers 6,10,15, and maybe others missing. Wait, let's check which numbers are missing:Original list: 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16.Used so far:8,1,3,13,12,4,5,11,14,2,7,9,16. Missing:6,10,15.Wait, how did we skip those? Let's see. After 8-1-3-13-12-4-5-11-14-2-7-9-16, we have 13 nodes. The path is 13 nodes long, so missing 3 numbers:6,10,15.But in this path, after 9-16, the path ends at 16, which is a degree 1 node. So we can't go further. But we missed 6,10,15. So this attempt didn't work. Maybe choosing a different path earlier on.Let's backtrack. At 8-1-3-13-12-4-5-11-14-2-7-9, maybe instead of going to 16, can we go somewhere else? But 9's only connections are 7 and 16. 7 is already used, so must go to 16. Thus, that path ends there. So perhaps the initial choices led to missing some nodes. Let's try a different route.Starting again from 8-1. From 1, instead of going to 3, go to 15.So 8-1-15. From 15, options are 1,10. 1 used, so go to 10.8-1-15-10. From 10, options are 6,15. 15 used, so go to 6.8-1-15-10-6. From 6, options are 3,10. 10 used, go to 3.8-1-15-10-6-3. From 3, options are 1,6,13. 1 and 6 used, go to 13.8-1-15-10-6-3-13. From 13, options are 3,12. 3 used, go to 12.8-1-15-10-6-3-13-12. From 12, options are 4,13. 13 used, go to 4.8-1-15-10-6-3-13-12-4. From 4, options are 5,12. 12 used, go to 5.8-1-15-10-6-3-13-12-4-5. From 5, options are 4,11. 4 used, go to 11.8-1-15-10-6-3-13-12-4-5-11. From 11, options are 5,14. 5 used, go to 14.8-1-15-10-6-3-13-12-4-5-11-14. From 14, options are 2,11. 11 used, go to 2.8-1-15-10-6-3-13-12-4-5-11-14-2. From 2, options are 7,14. 14 used, go to 7.8-1-15-10-6-3-13-12-4-5-11-14-2-7. From 7, options are 2,9. 2 used, go to 9.8-1-15-10-6-3-13-12-4-5-11-14-2-7-9. From 9, go to 16.8-1-15-10-6-3-13-12-4-5-11-14-2-7-9-16. Now check which numbers are missing: 8,1,15,10,6,3,13,12,4,5,11,14,2,7,9,16. Wait, that's all 16 numbers! Wait, did I count correctly? Let me list them in order:1. 82. 13. 154.105.66.37.138.129.410.511.1112.1413.214.715.916.16Yes, all numbers from 1 to 16 are there. So this is a valid Hamiltonian path from 8 to 16. Therefore, the row arrangement is possible.Therefore, the row exists, and the circle doesn't because nodes 8 and 16 have degree 1, making a cycle impossible.But let me confirm that this path works. Let's check each adjacent sum:8+1=9 (3²)1+15=16 (4²)15+10=25 (5²)10+6=16 (4²)6+3=9 (3²)3+13=16 (4²)13+12=25 (5²)12+4=16 (4²)4+5=9 (3²)5+11=16 (4²)11+14=25 (5²)14+2=16 (4²)2+7=9 (3²)7+9=16 (4²)9+16=25 (5²)All sums are perfect squares. Great! So this path works. Therefore, the row is possible.For the circle, as established, nodes 8 and 16 have only one connection each, so they can't be part of a cycle that requires two neighbors. Hence, arranging all 16 numbers in a circle with the given condition is impossible.Therefore, the proof is complete.

✅:To determine whether numbers 1 to 16 can be arranged in a row or a circle such that adjacent pairs sum to perfect squares, we analyze the problem using graph theory. Key Steps:1. Graph Construction: - Represent each number (1–16) as a node. - Connect nodes with an edge if their sum is a perfect square (4, 9, 16, 25).2. Degree Analysis: - Nodes with degree 1: 8 (connects only to 1) and 16 (connects only to 9). - Nodes with degree 3: 1 (connects to 3, 8, 15) and 3 (connects to 1, 6, 13). - All other nodes have degree 2.3. Row Arrangement (Hamiltonian Path): - A Hamiltonian path requires two endpoints with odd degrees (here, nodes 8 and 16, both degree 1). - Constructing such a path: - Example Path: (8 to 1 to 15 to 10 to 6 to 3 to 13 to 12 to 4 to 5 to 11 to 14 to 2 to 7 to 9 to 16) Each adjacent pair sums to a square (verified below).4. Circle Arrangement (Hamiltonian Cycle): - A cycle requires all nodes to have even degrees (each node must have two neighbors). - Nodes 8 and 16 have degree 1, making it impossible to form a cycle that includes them. Thus, no Hamiltonian cycle exists.--- Verification for the Row:Adjacency sums in the constructed path:- (8+1=9), (1+15=16), (15+10=25), (10+6=16), (6+3=9), (3+13=16), (13+12=25), (12+4=16), (4+5=9), (5+11=16), (11+14=25), (14+2=16), (2+7=9), (7+9=16), (9+16=25). All sums are perfect squares.--- Conclusion:- Row Possible: A Hamiltonian path exists (as constructed). - Circle Impossible: Nodes 8 and 16 cannot satisfy the degree requirement for a cycle. Final Answer: Numbers 1 to 16 can be arranged in a row but not in a circle under the given condition. Row example: (8, 1, 15, 10, 6, 3, 13, 12, 4, 5, 11, 14, 2, 7, 9, 16). Circle impossible due to nodes 8 and 16 having insufficient connections. boxed{text{Such a row exists, but a circular arrangement is impossible.}}

🔑:1. Initial Setup: We need to show that the numbers from 1 to 16 can be arranged in a line such that the sum of any two adjacent numbers is a perfect square, but such an arrangement is not possible in a circular manner.2. Analysis with the Number 16: Let's focus on the number 16 first. Given that two numbers (a) and (b) must be adjacent if (a + b) is a perfect square let’s start analyzing how 16 can be positioned. [ text{If } x text{ is adjacent to } 16, text{ then } 16 + x = a^2 ] We know (16 + x) must be a perfect square. It implies: [ 16 + 1 leq 16 + x = a^2 leq 16 + 15 ] Simplifying further, we get: [ 17 leq a^2 leq 31 ] The only perfect square in this range is 25, meaning: [ a^2 = 25 implies 16 + x = 25 implies x = 9 ] 3. Implication for 16 and 9: This implies that the only number that can be adjacent to 16 is 9. Hence, 16 can have only one neighbor if we need the sum to be a perfect square.4. Circular Arrangement Contradiction: - In a circular arrangement, every number must have precisely two neighbors. - Since 16 can only have one neighbor which is 9, it violates the requirement for a circular arrangement.5. Linear Arrangement Example: Although arranging in a circle is impossible, we can still demonstrate a possible linear arrangement of numbers from 1 to 16 where the sum of any two adjacent numbers is a perfect square. One possible arrangement is: [ 16, 9, 7, 2, 14, 11, 5, 4, 12, 13, 3, 6, 10, 15, 1, 8 ]6. Verification: - (16 + 9 = 25) (which is (5^2)) - (9 + 7 = 16) (which is (4^2)) - (7 + 2 = 9) (which is (3^2)) - (2 + 14 = 16) (which is (4^2)) - (14 + 11 = 25) (which is (5^2)) - (11 + 5 = 16) (which is (4^2)) - (5 + 4 = 9) (which is (3^2)) - (4 + 12 = 16) (which is (4^2)) - (12 + 13 = 25) (which is (5^2)) - (13 + 3 = 16) (which is (4^2)) - (3 + 6 = 9) (which is (3^2)) - (6 + 10 = 16) (which is (4^2)) - (10 + 15 = 25) (which is (5^2)) - (15 + 1 = 16) (which is (4^2)) - (1 + 8 = 9) (which is (3^2))Thus, the numbers from 1 to 16 can indeed be arranged in a straight line satisfying the given condition, but such an arrangement in a circular manner is impossible.(boxed{text{16, 9, 7, 2, 14, 11, 5, 4, 12, 13, 3, 6, 10, 15, 1, 8}})