🧠:Okay, so I need to find the smallest positive number λ such that for any complex numbers z₁, z₂, z₃ inside the unit disk (|z| < 1), if they satisfy z₁ + z₂ + z₃ = 0, then the expression |z₁z₂ + z₂z₃ + z₃z₁|² + |z₁z₂z₃|² is less than λ. Hmm, interesting problem. Let me start breaking it down.First, since z₁, z₂, z₃ are complex numbers inside the unit disk, their magnitudes are all less than 1. Also, they sum to zero. This seems like a problem involving symmetric functions of the roots of a polynomial. Maybe I can relate z₁, z₂, z₃ to the roots of a cubic equation. Let me recall that if z₁, z₂, z₃ are roots of a cubic polynomial, then the polynomial can be written as z³ - (z₁ + z₂ + z₃)z² + (z₁z₂ + z₂z₃ + z₃z₁)z - z₁z₂z₃. But in our case, z₁ + z₂ + z₃ = 0, so the polynomial simplifies to z³ + (z₁z₂ + z₂z₃ + z₃z₁)z - z₁z₂z₃. So maybe the coefficients of this polynomial are related to the expressions we have in the problem.The problem wants the sum of the squares of the magnitudes of these coefficients, except the leading coefficient which is 1. Wait, the coefficients are (z₁z₂ + z₂z₃ + z₃z₁) and (-z₁z₂z₃). So the expression given in the problem is |(z₁z₂ + z₂z₃ + z₃z₁)|² + |z₁z₂z₃|². So we need to find the maximum value of this expression over all possible triples z₁, z₂, z₃ inside the unit disk summing to zero, and then λ would be the smallest number greater than this maximum.Therefore, our task reduces to maximizing |S|² + |P|² where S = z₁z₂ + z₂z₃ + z₃z₁ and P = z₁z₂z₃, under the constraints |z₁|, |z₂|, |z₃| < 1 and z₁ + z₂ + z₃ = 0.Hmm, how do I approach maximizing this? Maybe using Lagrange multipliers, but since we're dealing with complex numbers and magnitudes, it might get complicated. Alternatively, perhaps we can use some inequalities or known results about complex numbers with given sums.Another idea is to parametrize the variables. Since z₁ + z₂ + z₃ = 0, we can express one variable in terms of the other two. Let's say z₃ = - (z₁ + z₂). Then substitute z₃ into S and P.Let's try that. Let z₃ = -z₁ - z₂. Then S = z₁z₂ + z₂(-z₁ - z₂) + (-z₁ - z₂)z₁.Calculating S:z₁z₂ + z₂(-z₁) + z₂(-z₂) + (-z₁)z₁ + (-z₂)z₁Simplify term by term:z₁z₂ - z₁z₂ - z₂² - z₁² - z₂z₁Wait, let me compute step by step:First term: z₁z₂Second term: z₂ * (-z₁ - z₂) = -z₁z₂ - z₂²Third term: (-z₁ - z₂) * z₁ = -z₁² - z₂z₁So S = z₁z₂ - z₁z₂ - z₂² - z₁² - z₂z₁Combine like terms:z₁z₂ - z₁z₂ cancels out. Then we have -z₂² - z₁² - z₂z₁So S = - (z₁² + z₂² + z₁z₂)Similarly, P = z₁z₂z₃ = z₁z₂(-z₁ - z₂) = -z₁²z₂ - z₁z₂²So S = - (z₁² + z₂² + z₁z₂) and P = - z₁z₂(z₁ + z₂)Our expression to maximize becomes |S|² + |P|² = |z₁² + z₂² + z₁z₂|² + |z₁z₂(z₁ + z₂)|².But we also have the constraint that |z₁| < 1, |z₂| < 1, and |z₃| = |z₁ + z₂| < 1. Wait, but z₃ must also lie inside the unit disk. So |z₁ + z₂| < 1. Therefore, the variables z₁ and z₂ must satisfy |z₁ + z₂| < 1 in addition to |z₁| < 1 and |z₂| < 1.So the problem now reduces to maximizing |z₁² + z₂² + z₁z₂|² + |z₁z₂(z₁ + z₂)|² with |z₁| < 1, |z₂| < 1, and |z₁ + z₂| < 1. Hmm. This seems more manageable with two variables instead of three.But this is still a complex optimization problem. Maybe we can parameterize z₁ and z₂ in terms of polar coordinates. Let me set z₁ = r₁ e^{iθ₁}, z₂ = r₂ e^{iθ₂}, where 0 ≤ r₁, r₂ < 1, and θ₁, θ₂ are real numbers. Then z₁ + z₂ = r₁ e^{iθ₁} + r₂ e^{iθ₂}, whose magnitude must be less than 1. However, this seems complicated due to the angles.Alternatively, perhaps we can assume some symmetry. For example, maybe the maximum occurs when z₁ and z₂ are real numbers, or when they are equally spaced in some way. Let me test some specific cases.First, consider the case where all three numbers are real. Let’s set z₁ = a, z₂ = b, z₃ = -a - b, where a, b are real numbers in (-1, 1). Then the constraints are |a| < 1, |b| < 1, |a + b| < 1.Then S = -(a² + b² + ab)P = -ab(a + b)So the expression becomes (a² + b² + ab)² + (ab(a + b))².We need to maximize this over a, b real numbers with |a|, |b| < 1 and |a + b| < 1.This is a two-variable optimization problem. Maybe we can set variables such that a and b are non-negative (since the expression is symmetric in a and b, and if they are negative, maybe squaring makes it positive). Let's assume a, b ≥ 0. Then |a + b| < 1 implies a + b < 1. So a, b ∈ [0, 1) and a + b < 1.So let's set variables in this domain. Let’s let a and b be non-negative with a + b < 1.Let me define x = a, y = b. Then x, y ≥ 0, x + y < 1. Then the expression becomes (x² + y² + xy)² + (xy(x + y))².Let me compute this. Let me denote E = (x² + y² + xy)² + (xy(x + y))².To maximize E over x, y ≥ 0 with x + y < 1.Perhaps we can set t = x + y, s = xy. Then since x, y ≥ 0, s ≤ t²/4 by AM-GM. But x + y < 1, so t < 1.But x² + y² + xy = (x + y)^2 - xy = t² - s.So E = (t² - s)^2 + (s t)^2.So E(t, s) = (t² - s)^2 + (s t)^2, where 0 < t < 1 and 0 ≤ s ≤ t²/4.Hmm, perhaps we can express this as a function of s for a fixed t, then find the maximum over s and t.For fixed t, E(t, s) = (t² - s)^2 + (s t)^2.Let’s expand this:= t^4 - 2 t² s + s² + s² t²= t^4 - 2 t² s + s² (1 + t²)To find the maximum over s ∈ [0, t²/4].This is a quadratic in s. Let’s treat it as a quadratic function in s:E(t, s) = (1 + t²) s² - 2 t² s + t^4The quadratic in s has its vertex at s = [2 t²] / [2(1 + t²)] = t² / (1 + t²). The vertex is a minimum or maximum? Since the coefficient of s² is (1 + t²) > 0, it's a convex function, so the vertex is a minimum. Therefore, the maximum of E(t, s) over s ∈ [0, t²/4] occurs at one of the endpoints: either s = 0 or s = t²/4.Compute E(t, 0) = t^4Compute E(t, t²/4) = (1 + t²)(t^4 / 16) - 2 t²*(t² / 4) + t^4Wait, let's compute step by step:First term: (t² - s)^2 = (t² - t²/4)^2 = (3 t² / 4)^2 = 9 t^4 / 16Second term: (s t)^2 = (t²/4 * t)^2 = (t^3 / 4)^2 = t^6 / 16So E(t, t²/4) = 9 t^4 / 16 + t^6 / 16Therefore, for each t, the maximum of E(t, s) is the maximum between t^4 and 9 t^4 / 16 + t^6 / 16.Compare t^4 and 9 t^4 /16 + t^6 /16.Set t^4 = 9 t^4 /16 + t^6 /16Multiply both sides by 16: 16 t^4 = 9 t^4 + t^6Simplify: 7 t^4 = t^6 => t^6 - 7 t^4 = 0 => t^4(t² -7)=0. So solutions t=0 or t=±√7, but since t is in (0,1), the only solution is t=0.Therefore, for t >0 in (0,1), which is larger: t^4 or 9 t^4 /16 + t^6 /16?Let’s compute the difference: t^4 - [9 t^4 /16 + t^6 /16] = (16 t^4 -9 t^4 - t^6)/16 = (7 t^4 - t^6)/16 = t^4(7 - t²)/16.Since t ∈ (0,1), t² <1, so 7 - t² >6 >0. Therefore, t^4 > 9 t^4 /16 + t^6 /16 for all t ∈ (0,1). Therefore, the maximum over s is at s=0, which gives E(t, s) = t^4. Wait, but that contradicts. Wait, if s can be up to t²/4, but when s=0, E is t^4, and when s=t²/4, E is 9 t^4 /16 + t^6 /16. Since t ∈ (0,1), t^6 < t^4, so 9 t^4 /16 + t^6 /16 < (9/16 + 1/16) t^4 = 10/16 t^4 = 5/8 t^4 < t^4. So indeed, E(t, s) is maximized at s=0 for each t. Hmm, that seems odd. Wait, but if s=0, that means either x=0 or y=0. Because s=xy=0.So in the case when either x=0 or y=0, then E = t^4. For example, if x=0, then y = t, and E becomes (0 + y² +0)^2 +0 = y^4. Similarly, if y=0, E = x^4. But when s= t²/4, which is achieved when x=y=t/2. So in that case, the expression becomes [t² - t²/4]^2 + [t²/4 * t]^2 = (3 t² /4)^2 + (t^3 /4)^2 = 9 t^4 /16 + t^6 /16. But this is less than t^4 as we saw.So this suggests that in the real case, the maximum of E is achieved when either x or y is zero, giving E = t^4. But since t = x + y <1, the maximum possible t is approaching 1. So as t approaches 1, E approaches 1. Therefore, in the real case, the maximum is 1. But wait, let's verify with actual numbers.Suppose x=1-ε, y=0, where ε approaches 0. Then z₃ = -x - y = - (1 - ε). But |z₃| must be less than 1. But |z₃| = 1 - ε <1, which is okay. Then E = x⁴ +0 = (1 - ε)^4, which approaches 1 as ε→0. Similarly, if x and y are both approaching 0.5, say x = y = 0.5 - ε, then t =1 - 2ε, s=(0.5 - ε)^2. Then E would be [t² -s]^2 + [s t]^2. Let’s compute t=1 - 2ε, s=(0.5 - ε)^2 ≈0.25 - ε. Then t² ≈1 -4ε, s≈0.25 - ε. So t² -s ≈1 -4ε -0.25 + ε=0.75 -3ε. Then (t² -s)^2 ≈0.75²=0.5625. And s t ≈(0.25)(1)=0.25, so (s t)^2≈0.0625. So total E≈0.5625 +0.0625=0.625, which is less than 1. So indeed, when x and y are both 0.5, E≈0.625, which is less than 1. So the maximum in the real case is indeed approaching 1 when one variable approaches 1 and the other approaches 0.But wait, in that case, z₃ = -x - y = - (1 - ε + 0) = - (1 - ε). So |z₃| =1 - ε <1. So this is allowed. But in the limit as ε→0, z₁ approaches 1, z₂ approaches 0, z₃ approaches -1. But |z₁|, |z₂|, |z₃| must all be less than 1. So in the limit, they approach the boundary, but are still inside. Therefore, the supremum of E in the real case is 1, but it's never actually reached because |z_i| must be less than 1. Therefore, in the real case, the maximum is approaching 1, but never actually reaches it.But the problem states that z_i are in the open unit disk, so strictly less than 1. Therefore, the maximum in the real case is less than 1. However, maybe when considering complex numbers, we can get higher?Wait, the problem asks for the smallest λ such that the expression is less than λ for all such triples. So even if in the real case the supremum is 1, but since it's never reached, λ must be at least 1. However, maybe in the complex case, we can get higher values. So perhaps 1 is not the answer.Wait, let me check with complex numbers. Let's try an example where z₁, z₂, z₃ are complex.Suppose z₁ = a, z₂ = a, z₃ = -2a, with |a| <1/2 so that |z₃| =2|a| <1. Then S = z₁z₂ + z₂z₃ + z₃z₁ = a² + a*(-2a) + (-2a)*a = a² -2a² -2a² = -3a².P = z₁z₂z₃ = a*a*(-2a) = -2a³.So |S|² + |P|² = | -3a² |² + | -2a³ |² =9|a|⁴ +4|a|⁶.Since |a| <1/2, let's compute the maximum of 9r⁴ +4r⁶ for r <1/2. Let's set r approaches 1/2. Then 9*(1/2)^4 +4*(1/2)^6=9/16 +4/64=9/16 +1/16=10/16=5/8=0.625. So in this case, the maximum approaches 0.625, which is less than 1. Hmm, so even with complex numbers arranged this way, we get a lower value than the real case.Wait, maybe another configuration. Let's take z₁, z₂, z₃ as the cube roots of some number, but adjusted so that they sum to zero. For example, the cube roots of unity sum to zero, but their magnitudes are 1, which is on the boundary. But if we scale them down by a factor r <1, then z₁ = r, z₂ = r e^{2πi/3}, z₃ = r e^{4πi/3}. Then z₁ + z₂ + z₃ =0, as they are the scaled cube roots of unity.Let's compute S and P for these.z₁ = r, z₂ = r e^{2πi/3}, z₃ = r e^{4πi/3}Compute S = z₁z₂ + z₂z₃ + z₃z₁.First, z₁z₂ = r * r e^{2πi/3} = r² e^{2πi/3}z₂z₃ = r e^{2πi/3} * r e^{4πi/3} = r² e^{6πi/3} = r² e^{2πi} = r² *1 = r²z₃z₁ = r e^{4πi/3} * r = r² e^{4πi/3}Therefore, S = r² e^{2πi/3} + r² + r² e^{4πi/3}= r² [ e^{2πi/3} +1 + e^{4πi/3} ]But e^{2πi/3} + e^{4πi/3} = -1 (since they are the non-real cube roots of unity), so S = r² [ -1 +1 ] =0. Interesting, S=0.Then P = z₁z₂z₃ = r * r e^{2πi/3} * r e^{4πi/3} = r³ e^{6πi/3} = r³ e^{2πi} = r³ *1 = r³.Therefore, |S|² + |P|² = 0 + |r³|² = r⁶. Since r <1, the maximum approaches 1 as r approaches 1. But wait, if r approaches 1, then |z₁|=|z₂|=|z₃|=r approaches 1, but they are still inside the unit disk. So as r→1-, |S|² + |P|² approaches 1. However, in this case, S=0, so the entire expression is just |P|²= r⁶. Wait, but when r approaches 1, r⁶ approaches 1. So in this configuration, the expression approaches 1. But in this case, the variables are approaching the boundary of the unit disk. However, the problem states that z_i are strictly inside the unit disk, so we can't have them equal to 1. Therefore, the supremum in this case is also approaching 1. But this is a different configuration where S=0 and P approaches 1 in magnitude.But in the previous real case, S approaches -1 and P approaches 0. So depending on the configuration, different parts of the expression contribute. Wait, in the real case when z₁ approaches 1, z₂ approaches 0, z₃ approaches -1, then S = z₁z₂ + z₂z₃ + z₃z₁ approaches (1*0) + (0*(-1)) + (-1)*1 = -1. So |S|² approaches 1, and |P|² approaches |1*0*(-1)|² =0. So in that case, the expression approaches 1.But in the cube roots case, when r approaches 1, |S|² + |P|² approaches 0 +1=1. So both configurations approach 1, but from different terms. So in both real and complex cases, the supremum is approaching 1.But the problem asks for the smallest λ such that the expression is strictly less than λ for all z₁, z₂, z₃ inside the unit disk. Since in some cases, the expression can approach 1, but never actually reach it. Therefore, λ must be at least 1. But is 1 sufficient? Or is there a configuration where the expression can exceed 1?Wait, let's test another case. Suppose two variables are equal in magnitude and opposite in direction, and the third adjusts to make the sum zero. For example, let z₁ = a, z₂ = -a, then z₃ =0. Then S = z₁z₂ + z₂z₃ + z₃z₁ = a*(-a) + (-a)*0 +0*a = -a². P = z₁z₂z₃ =0. So |S|² + |P|² = | -a² |² +0 = a⁴. Since |a| <1, the maximum approaches 1 as |a| approaches 1. So again, approaches 1.Another case: take z₁ = a, z₂ = a, z₃ = -2a, but with a complex. Let a = re^{iθ}. Then |z₃| =2|a| <1, so |a| <1/2. Then S = -3a², P = -2a³. Then |S|² + |P|² =9|a|⁴ +4|a|⁶. Since |a| <1/2, the maximum when |a| approaches 1/2 is 9*(1/2)^4 +4*(1/2)^6 =9/16 +4/64=9/16 +1/16=10/16=5/8, which is less than 1.Hmm, so in this case, even with complex numbers, the maximum is lower.Wait, but earlier, with the cube roots scaled by r approaching 1, we get |P|² approaching 1, and S=0. Similarly, in the real case with one variable approaching 1, we get |S|² approaching 1 and P=0. So in different configurations, different parts contribute to the expression approaching 1. But is there a case where the sum exceeds 1? For instance, can we have both |S|² and |P|² contributing significantly?Suppose we take a combination where both S and P are non-zero. Let’s try to construct such a case.Let’s take z₁ = a, z₂ = a, z₃ = -2a, but with a complex. Wait, but earlier, when a is real, we saw that the expression becomes (9a⁴) + (4a⁶). If a is complex, maybe we can get a different result? Wait, let me check.Suppose a is complex, say a = re^{iθ}. Then S = -3a² = -3r² e^{i2θ}, so |S|² =9r⁴. P = -2a³ = -2r³ e^{i3θ}, so |P|²=4r⁶. So regardless of θ, |S|² + |P|²=9r⁴ +4r⁶. Since |a| <1/2, as before, the maximum is 5/8. So no, even with complex a, it doesn't help. So maybe in this configuration, the maximum is still low.Alternatively, let's take z₁, z₂, z₃ such that two are complex conjugates. For example, let z₁ = a + bi, z₂ = a - bi, z₃ = -2a. Then z₁ + z₂ + z₃ =0. The magnitudes |z₁| = |z₂| = sqrt(a² + b²) <1, and |z₃| = | -2a | <1, so |a| <1/2.Compute S = z₁z₂ + z₂z₃ + z₃z₁z₁z₂ = (a + bi)(a - bi) = a² + b²z₂z₃ = (a - bi)(-2a) = -2a² + 2abiz₃z₁ = (-2a)(a + bi) = -2a² -2abiSo S = (a² + b²) + (-2a² + 2abi) + (-2a² -2abi) = a² + b² -2a² -2a² + (2abi -2abi) = (a² -4a² + b²) +0 = (-3a² + b²)Similarly, P = z₁z₂z₃ = (a² + b²)(-2a) = -2a(a² + b²)Therefore, |S|² + |P|² = (-3a² + b²)^2 + [2a(a² + b²)]^2Let me compute this expression:First term: 9a⁴ -6a²b² + b⁴Second term:4a²(a² + b²)^2 =4a²(a⁴ +2a²b² + b⁴)=4a⁶ +8a⁴b² +4a²b⁴So total expression:9a⁴ -6a²b² + b⁴ +4a⁶ +8a⁴b² +4a²b⁴=4a⁶ + (9a⁴ +8a⁴b²) + (-6a²b² +4a²b⁴) + b⁴Hmm, this seems complicated, but perhaps we can set variables to simplify. Let’s set t = a², s = b². Then since |z₁|=sqrt(t + s) <1, and |z₃|=2|a|=2sqrt(t) <1 => t <1/4.So t ∈ (0,1/4), s ∈ (0,1 - t). Wait, but |z₁| <1 => t + s <1. So s <1 - t.But given that t <1/4, then s <1 - t <3/4.So the expression becomes:4t³ +9t² +8t² s -6t s +4t s² + s² + s⁴Wait, perhaps not helpful. Let me try specific values. Let’s set a =1/2 - ε, so t approaches 1/4. Then b can be up to sqrt(1 - t) - a. Wait, maybe this is getting too involved. Alternatively, set a=0. Then z₃=0, z₁ and z₂ are conjugates with a=0, so z₁=bi, z₂=-bi, z₃=0. Then S = (bi)(-bi) + (-bi)(0) +0(bi) = b². P =0. So |S|² + |P|² =b⁴. With |b| <1, so maximum approaching1. But in this case, z₃=0, which is allowed. Wait, but if a=0, then z₃=0, and z₁=bi, z₂=-bi, then indeed z₁ + z₂ + z₃=0. Then S = z₁z₂ + z₂z₃ + z₃z₁ = (bi)(-bi) +0 +0= b². So |S|² + |P|² =b⁴ +0. As b approaches1, this approaches1. So again, this approaches1.But here, a=0, so z₃=0, and z₁ and z₂ are purely imaginary with magnitudes approaching1. So in this case, the expression approaches1. Therefore, again, the supremum is1.But wait, let's see if in some configuration, we can have both |S| and |P| contributing to make the total sum exceed1. For example, if |S|² is close to1 and |P|² is also significant, then the total might exceed1. But from the previous examples, in the cases where |S|² approaches1, |P|² is approaching0, and vice versa.Wait, let me test a case where both |S| and |P| are non-zero. For example, take z₁ = a, z₂ = a, z₃ = -2a with a complex, but a is not purely real or purely imaginary. Let's say a = re^{iθ}, with θ ≠0, π/2.Then S = -3a², so |S|² =9r⁴, and P = -2a³, so |P|²=4r⁶. Then |S|² + |P|²=9r⁴ +4r⁶. For this to exceed1, we need 9r⁴ +4r⁶ >1. Let’s solve for r:9r⁴ +4r⁶ >1Let’s set r=0.9, which is less than1. Then 9*(0.9)^4 +4*(0.9)^6 ≈9*0.6561 +4*0.531441≈5.9049 +2.1258≈8.0307>1. Wait, but hold on, if r=0.9, then |z₃|=2r=1.8>1, which is not allowed. So in this case, z₃ would be outside the unit disk. Therefore, invalid.So in the configuration z₁ = z₂ =a, z₃=-2a, we must have |z₃|=2|a|<1, so |a|<1/2. Therefore, r<1/2. Let's set r=0.49. Then |z₃|=0.98<1. Then 9*(0.49)^4 +4*(0.49)^6≈9*(0.0576) +4*(0.0138)=0.5184 +0.0552≈0.5736<1. So even with maximum allowed |a|=0.5, then 9*(0.5)^4 +4*(0.5)^6=9*(0.0625)+4*(0.015625)=0.5625+0.0625=0.625<1. So no, in this configuration, we can't reach1.But in the case where one variable approaches1 and another approaches0, then |S|² approaches1, while |P|² approaches0, so their sum approaches1. Similarly, in the cube roots case, |P|² approaches1 while |S|² approaches0. So in both cases, the total is approaching1.Is there a case where both |S|² and |P|² are non-negligible, leading to a total exceeding1? Let's try another configuration. Suppose we take z₁ = a, z₂ = b, z₃ = -a -b, with a and b chosen such that S and P are both non-zero.Alternatively, perhaps take z₁, z₂, z₃ as vertices of an equilateral triangle on the circle of radius r <1, centered at the origin. Wait, but if they are vertices of an equilateral triangle, their sum might not be zero unless they are the cube roots of unity scaled by r. Then, as before, their sum is zero, S=0, P = r³. So again, |S|² + |P|² =r⁶, approaching1 as r approaches1.Alternatively, take two points and the third balancing them. For example, z₁ = re^{iθ}, z₂=re^{-iθ}, z₃= -2r cosθ. Then z₁ + z₂ + z₃=0. Let's compute S and P.First, z₁ = re^{iθ}, z₂=re^{-iθ}, z₃=-2r cosθ.Compute S = z₁z₂ + z₂z₃ + z₃z₁.z₁z₂ = r².z₂z₃ = re^{-iθ}*(-2r cosθ) = -2r² cosθ e^{-iθ}z₃z₁ = -2r cosθ * re^{iθ} = -2r² cosθ e^{iθ}Therefore, S = r² -2r² cosθ e^{-iθ} -2r² cosθ e^{iθ}= r² -2r² cosθ (e^{iθ} + e^{-iθ})= r² -2r² cosθ *2 cosθ= r² -4r² cos²θSimilarly, P = z₁z₂z₃ = r² * (-2r cosθ) = -2r³ cosθTherefore, |S|² + |P|² = |r²(1 -4 cos²θ)|² + | -2r³ cosθ |²= r⁴ (1 -4 cos²θ)² +4r⁶ cos²θLet’s denote this expression as E(r, θ) = r⁴ (1 -4 cos²θ)² +4r⁶ cos²θ.We need to maximize this over r <1 and θ.First, note that |z₃| = | -2r cosθ | <1. So 2r |cosθ| <1. Therefore, |cosθ| <1/(2r). Since r <1, the maximum possible |cosθ| is less than1/(2r). But since |cosθ| ≤1, this constraint is active only when1/(2r) <1, i.e., r >1/2. For r ≤1/2, |cosθ| can be up to1.Let’s consider two cases: r ≤1/2 and r >1/2.Case1: r ≤1/2. Then |cosθ| can be up to1. So θ can vary freely.E(r, θ)= r⁴ (1 -4 cos²θ)^2 +4r⁶ cos²θ.Let’s set x = cos²θ. Then x ∈[0,1]. Then E(r, x)= r⁴(1 -4x)^2 +4r⁶x.We can treat E as a function of x for fixed r.E(r, x)=r⁴(1 -8x +16x²) +4r⁶x.Take derivative with respect to x:dE/dx= r⁴(-8 +32x) +4r⁶.Set derivative to zero:r⁴(-8 +32x) +4r⁶=0Divide both sides by r⁴ (since r≠0):-8 +32x +4r²=032x=8 -4r²x=(8 -4r²)/32=(2 -r²)/8.So critical point at x=(2 -r²)/8.We need to check if this x is in [0,1].Since r ≤1/2, r² ≤1/4, so 2 -r² ≥2 -1/4=7/4, so x=(7/4)/8=7/32≈0.21875. Which is in [0,1]. So the maximum occurs either at this critical point or at the endpoints x=0 or x=1.Compute E at x=(2 - r²)/8:E(r, x)=r⁴[1 -4*(2 -r²)/8]^2 +4r⁶*(2 -r²)/8.Simplify:First term inside [ ]:1 - (8 -4r²)/8=1 -1 + (4r²)/8= r²/2.So first term: r⁴*(r²/2)^2 =r⁴*r⁴/4= r⁸/4.Second term:4r⁶*(2 -r²)/8= (4/8)r⁶*(2 -r²)=0.5 r⁶*(2 -r²)= r⁶ -0.5 r⁸.Therefore, total E= r⁸/4 + r⁶ -0.5 r⁸= r⁶ -0.25 r⁸.Compare this with endpoints:At x=0: E= r⁴*(1)^2 +0= r⁴.At x=1: E= r⁴*(1 -4)^2 +4r⁶*1=9r⁴ +4r⁶.So for each r ≤1/2, we need to compare E at critical point and endpoints.For example, take r=1/2. Then critical x=(2 - (1/2)^2)/8=(2 -1/4)/8=7/4/8=7/32≈0.21875.E= (1/2)^6 -0.25*(1/2)^8=1/64 -0.25/256=1/64 -1/1024= (16 -1)/1024=15/1024≈0.0146.Compare with endpoints:At x=0: E=(1/2)^4=1/16≈0.0625.At x=1:E=9*(1/2)^4 +4*(1/2)^6=9/16 +4/64=9/16 +1/16=10/16=5/8≈0.625.So maximum at x=1.Therefore, for r=1/2, maximum E is5/8≈0.625.Similarly, for smaller r, let's say r=1/4:Critical x=(2 - (1/4)^2)/8=(2 -1/16)/8=31/16/8=31/128≈0.242.E=(1/4)^6 -0.25*(1/4)^8=1/4096 -0.25/65536≈0.000244 -0.0000038≈0.000240.At x=0:E=(1/4)^4=1/256≈0.0039.At x=1:E=9*(1/4)^4 +4*(1/4)^6=9/256 +4/4096≈0.0352 +0.00098≈0.0361.So maximum at x=1.Therefore, for r ≤1/2, the maximum of E occurs at x=1, which corresponds to cos²θ=1, i.e., θ=0 or π. Therefore, in this case, the variables are real numbers. So we end up back in the real case, which approaches1 as r approaches1/2.Wait, but when r=1/2, and x=1 (θ=0), then cosθ=1, so z₁ = z₂ =1/2, z₃=-1. But |z₃|=1, which is not allowed. Wait, but our constraint was |z₃|<1. So actually, when r=1/2 and x=1 (θ=0), then z₃=-2*(1/2)*1=-1, which is on the boundary. Therefore, to satisfy |z₃|<1, we need r <1/2. So as r approaches1/2 from below, z₃ approaches-1. Therefore, the maximum E approaches9*(1/2)^4 +4*(1/2)^6=9/16 +4/64=10/16=5/8=0.625.Wait, but earlier in the real case, when z₁ approaches1, z₂ approaches0, z₃ approaches-1, we had |S|² + |P|² approaches1. So why the discrepancy?Ah, because in this specific configuration (z₁ =z₂ =a, z₃=-2a), we have to keep |z₃|=2|a|<1, so |a|<1/2. Therefore, in this case, even as |a| approaches1/2, the maximum expression is5/8. But in the other real case, where z₁ approaches1, z₂ approaches0, z₃ approaches-1, that's a different configuration where two variables are not equal. So different configurations give different maximums.This suggests that the supremum of the expression |S|² + |P|² is1, approached as one variable approaches the boundary (|z₁|→1), another approaches0, and the third approaches-1 (on the boundary but not included). Similarly, in the cube roots case, when scaled to approach the boundary.However, the problem states that z_i are strictly inside the unit disk, so the expression |S|² + |P|² never actually reaches1, but can get arbitrarily close. Therefore, the smallest λ satisfying the condition would be1, since the expression can approach1 but never exceed it.But wait, the problem says "Find the smallest positive number λ such that for any complex numbers z₁,z₂,z₃∈{z∈ℂ | |z|<1}, if z₁+z₂+z₃=0, then |z₁z₂ +z₂z₃ +z₃z₁|² + |z₁z₂z₃|² <λ."If the supremum is1, then λ must be at least1. However, if there's no actual maximum (since the supremum is1 but not achieved), then any λ>1 would satisfy the inequality, but the smallest such λ is1. But the problem says "<λ", so if the expression can approach1, then λ must be at least1, and since the inequality is strict, λ can be1 if the expression never actually reaches1. But wait, in real analysis, the supremum is the least upper bound. If the set of values of the expression has supremum1, but never actually attains1, then the smallest λ for which the expression is less thanλ is1. Because for any λ' <1, there exists values of the expression greater thanλ', but for λ=1, the expression is always less than1. However, this is only true if the expression never actually reaches1. But wait, the problem states that z_i are in the open unit disk, so |z_i| <1. Therefore, in the case where one variable approaches1, another approaches0, and the third approaches-1, the expression approaches1, but since all |z_i| are strictly less than1, the expression never actually equals1. Therefore, the expression is always strictly less than1, so λ=1 is the smallest such number. But is this correct?Wait, let me verify with an explicit example. Let’s take z₁ =1 -ε, z₂=ε, z₃= -1, where ε approaches0. But wait, z₃ must have |z₃|<1. So z₃= -1 +δ, but then z₁ +z₂ +z₃ = (1 -ε) +ε +(-1 +δ)=δ=0 =>δ=0. Therefore, z₃=-1. But |z₃|=1 is not allowed. Therefore, in reality, to keep |z₃|<1, we need to have z₃= -z₁ -z₂, so if z₁ approaches1 and z₂ approaches0, then z₃ approaches-1, but must stay inside the disk. Therefore, z₁ cannot actually reach1. Let’s make this precise.Let’s set z₁ =1 -ε, z₂=ε/2, then z₃= -z₁ -z₂= -1 +ε -ε/2= -1 +ε/2. Then |z₃|=| -1 +ε/2 | =√(1 -ε + (ε/2)^2 ). Wait, no. The magnitude of a real number is its absolute value. So |z₃|=| -1 +ε/2 |=1 -ε/2. Which is less than1 if ε/2 >0, which it is. Therefore, if we set z₁=1 -ε, z₂=ε/2, then z₃= -1 +ε/2. All three have magnitudes less than1 for ε>0. Then let’s compute the expression:S = z₁z₂ + z₂z₃ + z₃z₁= (1 -ε)(ε/2) + (ε/2)(-1 +ε/2) + (-1 +ε/2)(1 -ε)Compute each term:First term: (1 -ε)(ε/2)= ε/2 - ε²/2Second term: (ε/2)(-1 +ε/2)= -ε/2 + ε²/4Third term: (-1 +ε/2)(1 -ε)= -1*(1 -ε) + (ε/2)(1 -ε)= -1 +ε +ε/2 -ε²/2= -1 + (3ε)/2 -ε²/2Sum all terms:First term + Second term + Third term:(ε/2 - ε²/2) + (-ε/2 + ε²/4) + (-1 +3ε/2 -ε²/2)= [ε/2 - ε²/2 -ε/2 + ε²/4] + [-1 +3ε/2 -ε²/2]= [0 - ε²/2 + ε²/4] + [-1 +3ε/2 -ε²/2]= (-ε²/4) + (-1 +3ε/2 -ε²/2)= -1 +3ε/2 - (3ε²)/4Similarly, P =z₁z₂z₃=(1 -ε)(ε/2)(-1 +ε/2)= (ε/2)(1 -ε)(-1 +ε/2)Let’s expand this:= (ε/2)[ -1 +ε/2 +ε -ε²/2 ]= (ε/2)[ -1 + (3ε)/2 -ε²/2 ]= -ε/2 + (3ε²)/4 -ε³/4So |S|² + |P|²= | -1 +3ε/2 -3ε²/4 |² + | -ε/2 +3ε²/4 -ε³/4 |²As ε approaches0, the leading terms are:S ≈ -1 + (3ε)/2P ≈ -ε/2Therefore, |S|² ≈1 -3ε + (9ε²)/4|P|²≈ε²/4So |S|² + |P|²≈1 -3ε + (9ε²)/4 +ε²/4=1 -3ε + (10ε²)/4=1 -3ε + (5/2)ε²As ε approaches0 from the positive side, this expression approaches1 from below. Therefore, even though we take ε approaching0, making z₁ approach1 and z₃ approach-1 from within the disk, the expression |S|² + |P|² approaches1 from below, but never reaches1. Therefore, in this configuration, the expression is always less than1.Similarly, in the cube roots case, when scaling the roots by r approaching1, then |P|² approaches1 from below, and |S|² is0, so the total approaches1 from below.Therefore, in all cases, the expression |S|² + |P|² remains strictly less than1 for z_i inside the unit disk. Hence, the supremum is1, but it's never attained. Therefore, the smallest λ satisfying the condition is1.But wait, I need to confirm if there isn't some other configuration where the expression could exceed1. Let's think of another approach.Suppose we consider symmetric configurations. Let’s assume that z₁, z₂, z₃ are equally spaced on a circle of radius r <1, but rotated such that their sum is zero. This would be the case of the cube roots of unity scaled by r. In this case, as before, S=0 and P = r³. So |S|² + |P|² = r⁶. Since r <1, this is less than1. Therefore, no problem.Another configuration: two points close to1 and the third close to-2, but wait, |z₃|=| -2a |<1 implies |a|<1/2, so a can't be close to1.Alternatively, take z₁= a, z₂= a, z₃= -2a with a complex. As before, even with a complex, the expression is9|a|⁴ +4|a|⁶, which is less than9*(1/2)^4 +4*(1/2)^6=9/16 +4/64=10/16=5/8<1.Alternatively, consider three points forming a triangle with angles between them, but I think such cases either reduce to the cases we've already considered or give lower expressions.Another approach: use Lagrange multipliers in the complex plane. However, since we're dealing with magnitudes, it's often useful to use polar coordinates or consider variables on the real axis due to rotational symmetry.Alternatively, we can use the maximum modulus principle, which states that a non-constant holomorphic function on an open set cannot attain its maximum modulus in the set. However, I'm not sure how to apply this here.Alternatively, consider that the expression |S|² + |P|² is a real-valued function on the domain {(z₁,z₂,z₃) ∈ D³ | z₁ +z₂ +z₃=0}, where D is the unit disk. The function is continuous, and the domain is open (since D is open and the condition z₁ +z₂ +z₃=0 defines a closed subset, but within the open set D³, the intersection is relatively closed). However, since we are considering open disks, the supremum might not be attained.Given all the test cases and configurations, it seems that the supremum of the expression is1, approached but not reached. Therefore, the smallest λ satisfying the condition is λ=1.But wait, let me check another source or similar problem. I recall that in problems involving symmetric sums of complex numbers with constraints on their magnitudes, sometimes the maximum is achieved at specific symmetric points. For three roots summing to zero, maybe the maximum of |z₁z₂ + z₂z₃ + z₃z₁|² + |z₁z₂z₃|² is indeed1, but I need to verify.Alternatively, consider the following: for any z₁, z₂, z₃ inside the unit disk with z₁ + z₂ + z₃ =0, we can apply the triangle inequality or other inequalities.Note that |z₁z₂ + z₂z₃ + z₃z₁| ≤ |z₁z₂| + |z₂z₃| + |z₃z₁| ≤ |z₁||z₂| + |z₂||z₃| + |z₃||z₁|.Since each |z_i| <1, then each term is less than1. But this gives |S| <3, which is not helpful.Alternatively, note that since z₁ + z₂ + z₃ =0, we can relate S and P to the roots of the cubic polynomial as before: z³ + Sz - P =0. The roots are inside the unit disk, so perhaps we can use properties of polynomials with roots in the unit disk. However, I'm not sure of specific theorems that would directly apply here.Alternatively, use the fact that for any three complex numbers summing to zero, we have |z₁|² + |z₂|² + |z₃|² = |z₁ + z₂ + z₃|² + 2 Re(z₁z₂ + z₂z₃ + z₃z₁). But since z₁ + z₂ + z₃=0, this gives |z₁|² + |z₂|² + |z₃|² = 2 Re(S). Therefore, Re(S)= (|z₁|² + |z₂|² + |z₃|²)/2. But this relates the real part of S to the magnitudes of the variables. However, S is a complex number, and we are interested in |S|².But perhaps we can bound |S| in terms of the magnitudes.Using the identity above, Re(S) = (|z₁|² + |z₂|² + |z₃|²)/2.But since Re(S) ≤ |S|, we have (|z₁|² + |z₂|² + |z₃|²)/2 ≤ |S|. Therefore, |S| ≥ (|z₁|² + |z₂|² + |z₃|²)/2.But not sure if this helps.Alternatively, consider the inequality between arithmetic and geometric means. For three numbers, but not sure.Alternatively, use the Cauchy-Schwarz inequality. For instance, |z₁z₂ + z₂z₃ + z₃z₁| ≤ |z₁||z₂| + |z₂||z₃| + |z₃||z₁|. But each |z_i| <1, so this sum is less than3. Again, not helpful.Alternatively, think of the variables as vectors in the complex plane summing to zero, forming a triangle. The product terms might relate to the area or other properties.But perhaps this is overcomplicating. Since all test cases and configurations approach1 but don't exceed it, and given the problem's requirement for the smallest λ such that the expression is strictly less than λ, I think the answer is λ=1.However, I have a doubt because in some references, similar problems yield λ=27/16 or other constants. Wait, maybe I need to check a specific example where the expression might exceed1.Wait, let me try z₁= r, z₂= r ω, z₃= r ω², where ω is a primitive cube root of unity and r approaches1. Wait, but then z₁ +z₂ +z₃= r(1 + ω + ω²)=0, so they sum to zero. Then S= z₁z₂ +z₂z₃ +z₃z₁= r²(ω + ω³ + ω²)= but ω³=1, so ω³=1. Wait, no, ω is a primitive cube root, so ω³=1, but 1 + ω + ω²=0.Wait, compute S:z₁z₂ = r * r ω = r² ωz₂z₃ = r ω * r ω² = r² ω³ = r² *1= r²z₃z₁ = r ω² * r = r² ω²Therefore, S= r² ω + r² + r² ω² = r²(1 + ω + ω²)=0. So S=0.P= z₁z₂z₃= r * r ω * r ω²= r³ ω³= r³ *1= r³.Therefore, |S|² + |P|²=0 + r⁶. As r approaches1, this approaches1. But again, strictly less than1.But is there any case where the expression exceeds1? If not, then λ=1 is the answer.Alternatively, consider three equal points forming an equilateral triangle inside the unit disk but scaled such that their sum is zero. As above, this gives S=0 and P=r³, leading to the expression approaching1.Another case: let z₁= a, z₂= b, z₃= -a -b, with a and b chosen such that S and P are both large.Let’s attempt to maximize |S|² + |P|². Let’s suppose that a and b are such that z₁, z₂ are near the boundary. For example, set a approaching1, b approaching1, but then z₃ approaching-2, which is outside the disk. Therefore, invalid.Alternatively, set a approaching1, b approaching-1, then z₃=0. Then S= z₁z₂ +z₂z₃ +z₃z₁= z₁z₂= (1)(-1)=-1. P= z₁z₂z₃=0. So |S|² + |P|²=1 +0=1. But again, z₁ and z₂ approach1 and-1, but they must stay within the open unit disk. So the expression approaches1.Another example: z₁= a, z₂= -a, z₃=0, with a approaching1. Then S= z₁z₂ +0 +0= -a², so |S|²= a⁴, and P=0. So the expression approaches1 as a approaches1.So regardless of the configuration, the expression seems to approach1 but never exceed it. Therefore, the minimal λ is1.But wait, let me check one last case. Suppose z₁=0.9, z₂=0.9e^{iθ}, z₃=-0.9 -0.9e^{iθ}. Compute |z₃|=| -0.9(1 + e^{iθ}) |. Let's compute this magnitude.|1 + e^{iθ}|=2|cos(θ/2)|. So |z₃|=0.9*2|cos(θ/2)|=1.8|cos(θ/2)|. To ensure |z₃|<1, we need 1.8|cos(θ/2)| <1 => |cos(θ/2)| <5/9≈0.5556. Therefore, θ/2 >acos(5/9)≈56.25 degrees, so θ>112.5 degrees.So θ must be between112.5 and247.5 degrees to keep |z₃|<1.Let’s choose θ=120 degrees (2π/3 radians). Then e^{iθ}=e^{i2π/3}= -1/2 +i√3/2.Compute z₁=0.9, z₂=0.9*(-1/2 +i√3/2)= -0.45 +i0.9√3/2≈-0.45 +i0.7785, z₃=-0.9 -0.9*(-1/2 +i√3/2)= -0.9 +0.45 -i0.9√3/2= -0.45 -i0.7785.Compute S= z₁z₂ + z₂z₃ + z₃z₁.First term:0.9*(-0.45 +i0.7785)= -0.405 +i0.70065Second term: (-0.45 +i0.7785)*(-0.45 -i0.7785)= (-0.45)^2 + (0.7785)^2=0.2025 +0.606≈0.8085Third term: (-0.45 -i0.7785)*0.9= -0.405 -i0.70065Therefore, S= (-0.405 +i0.70065) +0.8085 + (-0.405 -i0.70065)= (-0.405 -0.405 +0.8085) +i(0.70065 -0.70065)= (-0.81 +0.8085) +0≈-0.0015≈0.Wait, nearly zero. Then P= z₁z₂z₃=0.9*(-0.45 +i0.7785)*(-0.45 -i0.7785)=0.9*(0.2025 +0.606)=0.9*0.8085≈0.72765.So |S|² + |P|²≈0 +0.72765²≈0.529, which is less than1. But if we increase r slightly, say r=0.95, but then |z₃|=1.8*0.95*cos(θ/2). For θ=120 degrees, cos(60 degrees)=0.5, so |z₃|=1.8*0.95*0.5=0.855<1. Wait, actually, with r=0.95 and θ=120 degrees:z₁=0.95, z₂=0.95*(-0.5 +i√3/2), z₃=-0.95 -0.95*(-0.5 +i√3/2)= -0.95 +0.475 -i0.95*√3/2≈-0.475 -i0.823.Compute |z₃|=sqrt((-0.475)^2 + (0.823)^2)≈sqrt(0.2256 +0.6773)=sqrt(0.9029)≈0.9502<1. Okay.Compute S= z₁z₂ + z₂z₃ + z₃z₁.First term:0.95*(-0.475 +i0.823)= -0.45125 +i0.78185Second term: (-0.475 +i0.823)*(-0.475 -i0.823)=0.475² +0.823²≈0.2256 +0.6773≈0.9029Third term: (-0.475 -i0.823)*0.95≈-0.45125 -i0.78185Sum all terms:(-0.45125 +i0.78185) +0.9029 + (-0.45125 -i0.78185)= (-0.45125 -0.45125 +0.9029) +i(0.78185 -0.78185)=0.0004≈0.So S≈0, P= z₁z₂z₃=0.95*0.9029≈0.8578. Then |S|² + |P|²≈0 +0.8578²≈0.735<1. Still less than1.Therefore, even with larger r, the expression |S|² + |P|² remains less than1.Thus, after checking various configurations and approaching the boundary cases, I conclude that the supremum of the expression is1, but it's never attained. Therefore, the smallest λ satisfying the condition is λ=1.Final Answerboxed{1}

✅:To find the smallest positive number (lambda) such that for any complex numbers (z_1, z_2, z_3 in {z in mathbb{C} mid |z| < 1}) with (z_1 + z_2 + z_3 = 0), the expression (left|z_1 z_2 + z_2 z_3 + z_3 z_1right|^2 + left|z_1 z_2 z_3right|^2 < lambda), we analyze various configurations and cases.1. Real Case Analysis: - Consider (z_1 = a), (z_2 = b), (z_3 = -a - b) where (a, b) are real numbers. - The expression simplifies to ((a^2 + b^2 + ab)^2 + (ab(a + b))^2). - As (a) approaches 1 and (b) approaches 0, the expression approaches 1 but remains strictly less than 1.2. Complex Case Analysis: - Consider (z_1, z_2, z_3) as scaled cube roots of unity, (z_1 = r), (z_2 = r e^{2pi i/3}), (z_3 = r e^{4pi i/3}). - The expression simplifies to (|0|^2 + |r^3|^2 = r^6), which approaches 1 as (r) approaches 1 but remains strictly less than 1.3. Other Configurations: - Testing configurations such as (z_1 = a), (z_2 = -a), (z_3 = 0) shows the expression approaches 1 as (|a|) approaches 1. - Considering symmetric configurations and other complex arrangements, the expression remains bounded below 1.4. Supremum Analysis: - In all cases, the supremum of the expression (left|z_1 z_2 + z_2 z_3 + z_3 z_1right|^2 + left|z_1 z_2 z_3right|^2) approaches 1 but never exceeds it due to the constraints (|z_1|, |z_2|, |z_3| < 1).Thus, the smallest (lambda) such that the expression is always less than (lambda) is (lambda = 1).[boxed{1}]

🔑:1. Claim: We claim that (lambda = 1).2. Proof that (lambda ge 1): - Consider (z_1 = 1 - epsilon), (z_2 = 0), and (z_3 = epsilon - 1), where (epsilon) is a small positive real number. - Given (z_1 + z_2 + z_3 = 0), we have: [ text{L.H.S} = |z_1 z_3|^2 = |(1 - epsilon)(epsilon - 1)|^2 = (1 - epsilon^2)^2 ] - As (epsilon to 0), ((1 - epsilon^2)^2 to 1). Therefore, (lambda ge 1).3. Proof that (lambda = 1) works: - Let (z_k = r_k (cos theta_k + i sin theta_k)) for (k = 1, 2, 3). - Given (z_1 + z_2 + z_3 = 0), we have: [ sum r_i cos theta_i = 0 quad text{and} quad sum r_i sin theta_i = 0 ] - This implies: [ r_1 cos theta_1 + r_2 cos theta_2 = -r_3 cos theta_3 quad text{and} quad r_1 sin theta_1 + r_2 sin theta_2 = -r_3 sin theta_3 ] - Squaring both equations and adding them, we get: [ r_1^2 + r_2^2 + 2r_1 r_2 cos (theta_2 - theta_1) = r_3^2 ] - Solving for (cos (theta_2 - theta_1)): [ cos (theta_2 - theta_1) = frac{r_3^2 - r_1^2 - r_2^2}{2r_1 r_2} ] - Now consider: [ 2r_1^2 r_2^2 cos (2theta_2 - 2theta_1) = 2r_1^2 r_2^2 (2 cos^2 (theta_2 - theta_1) - 1) ] [ = (r_3^2 - r_1^2 - r_2^2)^2 - 2r_1^2 r_2^2 ] [ = r_1^4 + r_2^4 + r_3^4 - 2r_1^2 r_3^2 - 2r_2^2 r_3^2 ] - Adding cyclic permutations, we get: [ sum_{1 le i < j le 3} 2r_i^2 r_j^2 cos (2theta_j - 2theta_i) = 3(r_1^4 + r_2^4 + r_3^4) - 4(r_1^2 r_2^2 + r_2^2 r_3^2 + r_3^2 r_1^2) ] - Since (z_1 + z_2 + z_3 = 0), we can swap (z_1 z_2 + z_2 z_3 + z_3 z_1) with (frac{1}{2}(z_1^2 + z_2^2 + z_3^2)). - This gives us: [ text{L.H.S} = frac{1}{4} |z_1^2 + z_2^2 + z_3^2|^2 + |z_1 z_2 z_3|^2 ] [ = frac{1}{4} left| sum r_i^2 cos 2theta_i + i sum r_i^2 sin 2theta_i right|^2 + r_1^2 r_2^2 r_3^2 ] [ = frac{1}{4} left( left( sum r_i^2 cos 2theta_i right)^2 + left( sum r_i^2 sin 2theta_i right)^2 right) + r_1^2 r_2^2 r_3^2 ] [ = frac{1}{4} left( r_1^4 + r_2^4 + r_3^4 + 2 sum_{1 le i < j le 3} r_i^2 r_j^2 cos (2theta_j - 2theta_i) right) + r_1^2 r_2^2 r_3^2 ] [ = frac{1}{4} left( r_1^4 + r_2^4 + r_3^4 + 3(r_1^4 + r_2^4 + r_3^4) - 4(r_1^2 r_2^2 + r_2^2 r_3^2 + r_3^2 r_1^2) right) + r_1^2 r_2^2 r_3^2 ] [ = r_1^4 + r_2^4 + r_3^4 - (r_1^2 r_2^2 + r_2^2 r_3^2 + r_3^2 r_1^2) + r_1^2 r_2^2 r_3^2 ] [ le r_1^2 + r_2^2 + r_3^2 - (r_1^2 r_2^2 + r_2^2 r_3^2 + r_3^2 r_1^2) + r_1^2 r_2^2 r_3^2 ] [ = 1 - (1 - r_1^2)(1 - r_2^2)(1 - r_3^2) le 1 ] - This implies that (lambda = 1) works. Done. (blacksquare)The final answer is ( boxed{ lambda = 1 } )

🧠:Okay, so I need to find the least possible value of ABCD minus AB times CD, where ABCD is a 4-digit number, and AB and CD are 2-digit numbers. Also, A and C can't be zero. Let me break this down step by step.First, let's clarify the notation. ABCD is a 4-digit number, which means A is in the thousands place, B in the hundreds, C in the tens, and D in the ones place. So the value of ABCD is 1000A + 100B + 10C + D. Similarly, AB is a 2-digit number, which is 10A + B, and CD is another 2-digit number, 10C + D. The problem is asking for the minimum value of ABCD - (AB × CD).So, mathematically, the expression we need to minimize is:(1000A + 100B + 10C + D) - ( (10A + B) × (10C + D) )Our goal is to find the smallest possible value of this expression, given that A and C are digits from 1 to 9, and B and D are digits from 0 to 9.Let me start by expanding the product (10A + B)(10C + D) to see if I can simplify the expression. Expanding the product:(10A + B)(10C + D) = 10A × 10C + 10A × D + B × 10C + B × D = 100AC + 10AD + 10BC + BDSo, subtracting this from ABCD:ABCD - (AB × CD) = (1000A + 100B + 10C + D) - (100AC + 10AD + 10BC + BD)Let me rewrite this expression term by term:1000A - 100AC - 10AD + 100B - 10BC - BD + 10C + D - BDWait, perhaps that's messy. Let me compute each term separately.Original expression:= 1000A + 100B + 10C + D - [100AC + 10AD + 10BC + BD]Combine like terms:1000A - 100AC - 10AD + 100B - 10BC + 10C + D - BDHmm. Maybe factor some terms:Factor A terms:1000A - 100AC -10AD = A(1000 - 100C -10D)Factor B terms:100B -10BC - BD = B(100 -10C - D)Then the remaining terms:10C + DSo the entire expression can be written as:A(1000 - 100C -10D) + B(100 -10C - D) + 10C + DHmm, not sure if that helps yet, but maybe.Alternatively, perhaps reorganize the original expression:= 1000A + 100B + 10C + D - 100AC -10AD -10BC -BD= 1000A -100AC -10AD + 100B -10BC +10C - BD + DWait, maybe group terms with A, B, C, D:A terms: 1000A -100AC -10ADB terms: 100B -10BC - BDC terms: 10CD terms: D - BDWait, that might not be helpful. Alternatively, perhaps factor variables:Let's see:1000A -100AC -10AD = A(1000 - 100C -10D)Similarly, 100B -10BC - BD = B(100 -10C - D)Then 10C + D remains.So putting it all together:Expression = A(1000 - 100C -10D) + B(100 -10C - D) +10C + DHmm. Maybe this form can be useful. Let's see.Our goal is to minimize this expression. Let's note that all variables A, B, C, D are digits with A, C ∈ {1,...,9}, B, D ∈ {0,...,9}.Since we need the minimal value, we need to make the expression as small as possible. So, maybe we can analyze each part.First, note that A, C are from 1 to 9, B, D from 0 to 9.Let me think: To minimize the expression, since it's ABCD - AB×CD, we need to make ABCD as small as possible and AB×CD as large as possible. So, perhaps the minimal value occurs when ABCD is small and AB×CD is large.But it's a trade-off because ABCD and AB×CD are related through the digits. So maybe we can't just minimize ABCD and maximize AB×CD independently because they share the same digits.Alternatively, perhaps some combination where ABCD is not too small, but AB×CD is significantly larger, leading to a smaller result. Since we are subtracting a product, the larger the product, the smaller the result. So our goal is to find ABCD such that AB×CD is as large as possible relative to ABCD.So, perhaps if AB and CD are both large numbers, their product will be large, but ABCD as a 4-digit number will also be large. So the key is to have AB×CD be as large as possible compared to ABCD.Alternatively, maybe the difference ABCD - AB×CD can be negative? Let's check.If AB×CD > ABCD, then the difference would be negative, which could be a candidate for the minimum. So maybe we can have negative values here.So, the question is, can AB×CD be greater than ABCD? Let's see.Let's take an example. Suppose ABCD is 9999, then AB=99, CD=99, so AB×CD=9801, which is less than 9999. So the difference is 9999 - 9801 = 198.But if we take a smaller ABCD, maybe AB×CD can be larger than ABCD.Wait, let's test with a smaller number. Suppose ABCD is 1000. Then AB=10, CD=00, but CD can't be 00 because C can't be zero. So CD must be at least 10. So ABCD=1000, AB=10, CD=10. Then AB×CD=10×10=100. So 1000 - 100 = 900.But maybe take ABCD=1111, AB=11, CD=11, product=121, difference=1111-121=990.Still positive. Hmm. Wait, perhaps if AB and CD are larger. Let's take ABCD=9999, AB=99, CD=99, product=9801, difference=198 as before. But if we take ABCD=9900, AB=99, CD=00, but CD can't be 00. So CD=01, then product=99×1=99, difference=9900 -99=9801. So that's still positive.Wait, maybe if AB and CD are in such a way that their product is larger. Wait, suppose ABCD=1212, AB=12, CD=12. Then product is 144, difference=1212 -144=1068. Still positive. Hmm.Wait, perhaps when AB and CD are both large, but ABCD is a number where the digits are arranged such that ABCD is not too large. For example, ABCD= 1999, AB=19, CD=99. Then product=19×99=1881. Difference=1999 -1881=118. Still positive.Wait, another example: ABCD= 9991, AB=99, CD=91. Product=99×91=9009. Difference=9991 -9009= -18. Oh, here we have a negative value.So that's possible. So the difference can be negative. So our job is to find the least possible value, which could be negative.So in this case, ABCD=9991, AB=99, CD=91, product=9009, so difference is 9991 -9009= -18.But maybe we can find a more negative difference. Let's check.Let me try ABCD= 9900, but CD can't be 00. Let's try ABCD= 9909, AB=99, CD=09. Then product=99×09=891, difference=9909 -891=9018. Not good.Wait, perhaps ABCD= 9199, AB=91, CD=99. Then product=91×99=9009, difference=9199 -9009=190. Still positive.Wait, another example: ABCD= 8999, AB=89, CD=99. Product=89×99=8811, difference=8999 -8811=188.Wait, maybe a different approach. Let's consider that ABCD - AB×CD = (1000A + 100B + 10C + D) - (10A + B)(10C + D).We can write this as:= 1000A + 100B + 10C + D - [100AC + 10AD + 10BC + BD]= 1000A - 100AC + 100B -10BC +10C - BD + D -10ADLet me group terms:= 1000A - 100AC -10AD + 100B -10BC - BD + 10C + DHmm. Maybe factor A, B, C, D:For A:1000A -100AC -10AD = A(1000 -100C -10D)For B:100B -10BC - BD = B(100 -10C - D)For C:10CFor D:D - BD -10AD? Wait, no. Wait, D is only in the last two terms: + D - BD.Wait, so the D terms: D(1 - B)Wait, so let's re-express:Expression = A(1000 -100C -10D) + B(100 -10C - D) + 10C + D(1 - B)Hmm, not sure if that helps, but maybe.Alternatively, let's consider that to minimize the expression, we need to maximize (AB × CD) - ABCD. Since we are looking for ABCD - AB×CD minimal, it's equivalent to maximizing AB×CD - ABCD. So the problem reduces to maximizing AB×CD - ABCD, which is equivalent to finding the maximum of (AB×CD - ABCD). Therefore, the minimal value of ABCD - AB×CD is equal to - (maximum of (AB×CD - ABCD)).So, if we can find the maximum value of (AB×CD - ABCD), then the minimal value of the original expression is the negative of that.So, maybe this perspective helps. Let's try to maximize AB×CD - ABCD.AB×CD - ABCD = (10A + B)(10C + D) - (1000A + 100B + 10C + D)Let's compute this:= 100AC + 10AD + 10BC + BD - 1000A - 100B -10C - D= 100AC -1000A + 10AD -100B + 10BC -10C + BD - DFactor terms:For A:100AC -1000A +10AD = A(100C -1000 +10D)For B:-100B +10BC + BD = B(-100 +10C + D)For C:-10CFor D:BD - D -10AD? Wait, no. Wait, BD - D is D(B -1). Wait, let's check:Original terms after expansion:100AC +10AD +10BC +BD -1000A -100B -10C -DSo:100AC -1000A +10AD -100B +10BC -10C +BD - DSo grouping:A(100C +10D -1000) + B(10C + D -100) + (-10C) + (-D)Hmm, maybe.So the expression to maximize is:A(100C +10D -1000) + B(10C + D -100) -10C -DSo, since A and B are digits (A,C:1-9; B,D:0-9), we need to choose A,B,C,D such that this expression is maximized.Each term:First term: A(100C +10D -1000). For this term to be positive, 100C +10D -1000 >0 => 100C +10D >1000 => 10C + D >100. But 10C + D is CD, a 2-digit number. The maximum CD is 99. So 10C + D ≤99 <100. Therefore, 100C +10D -1000 =10*(10C + D) -1000 ≤10*99 -1000=990-1000=-10. Therefore, 100C +10D -1000 is always negative. Therefore, the first term is A multiplied by a negative number. Therefore, to maximize the first term (since A is positive), we need to minimize (100C +10D -1000). Wait, but since it's negative, the smaller (more negative) it is, the larger the term (since A is multiplied by a negative number). Wait, no. Let's think.If we have A*(negative number), to maximize this term, since A is positive, we need the negative number to be as close to zero as possible (i.e., least negative). So, for the first term, to maximize it, we need 100C +10D -1000 to be as close to zero as possible (i.e., the least negative). Therefore, maximize 100C +10D. Since 100C +10D =10*(10C + D)=10*CD. So maximizing CD. Since CD is a 2-digit number, maximum CD is 99. Therefore, 100C +10D=10*99=990. Then 990 -1000= -10. So the first term is A*(-10). To maximize this term (i.e., make it less negative), we choose A as small as possible. So A=1. Then first term is 1*(-10)= -10. If A=9, then first term is 9*(-10)= -90. So to maximize the first term, set A=1 and CD=99.Similarly, the second term is B*(10C + D -100). Let's analyze 10C + D -100. 10C + D is CD, which is at most 99. So 10C + D -100 is negative. Therefore, the second term is B*(negative). To maximize this term, we need to minimize the negative value (i.e., make it as close to zero as possible) and set B as small as possible. So maximum CD=99, so 99 -100= -1. Then the second term is B*(-1). To maximize this term (i.e., make it less negative), set B=0. Then the term is 0*(-1)=0. If B=9, term=9*(-1)= -9. So to maximize the second term, set CD=99 and B=0.Third term: -10C. To maximize the expression, we need to minimize 10C, so set C as small as possible. C=1.Fourth term: -D. Similarly, to maximize the expression, set D as small as possible. D=0.Wait, but we need to consider the dependencies. Since CD is part of the 4-digit number ABCD, if we set CD=99 (max CD), then C=9 and D=9. But if we set C=1 (to minimize -10C), then CD=19, which is not the maximum CD.Therefore, there is a conflict here. The first term suggests maximizing CD (to make 100C +10D -1000 as large as possible, i.e., less negative), which would mean CD=99. However, the third term -10C suggests minimizing C, which would be C=1. Similarly, the fourth term suggests D=0, but CD=10 in that case, which is not maximum CD.So, these terms are in conflict. Therefore, we need to find a balance between these terms.Alternatively, let's consider that to maximize AB×CD - ABCD, we can look for AB and CD such that their product is as large as possible, while ABCD is as small as possible.But ABCD is a 4-digit number, so making ABCD small would mean A=1, B as small as possible, but AB is part of the product. So if AB is small, but CD is large, maybe their product is still large.Wait, maybe if AB is small but CD is very large. For example, AB=10 and CD=99, product=10×99=990, ABCD=1099, so difference=990 -1099= -109.Alternatively, AB=19 and CD=99, product=19×99=1881, ABCD=1999, difference=1881 -1999= -118.But this is more negative than the previous example. Wait, but is this possible?Wait, in the first case, ABCD=1099, AB=10, CD=99. Then ABCD - AB×CD=1099 -990=109. But if we compute AB×CD - ABCD=990 -1099= -109.But if we take AB=19, CD=99, ABCD=1999, then AB×CD=1881, so difference=1999 -1881=118. So AB×CD - ABCD=1881 -1999= -118.But if we take ABCD= 1899, AB=18, CD=99. Then AB×CD=18×99=1782, difference=1899 -1782=117. AB×CD - ABCD=1782 -1899= -117.Wait, so in these examples, the difference is negative but the magnitude varies. So maybe the more AB and CD are large, the more the product can be large compared to ABCD.But how do we get the maximum of AB×CD - ABCD? Let's think of it as AB×CD - ABCD = AB×CD - (1000A +100B +10C +D). To make this as large as possible, AB×CD needs to be as large as possible, and ABCD as small as possible.So, ideally, we want AB and CD to be as large as possible, and A as small as possible (since A contributes 1000A to ABCD, which is a large term). However, AB is a 2-digit number starting with A. So if A is small, say A=1, then AB can be at most 19. But CD can be 99. Then AB×CD=19×99=1881, and ABCD=1B99. To minimize ABCD, set B=0: ABCD=1099. Then AB×CD - ABCD=1881 -1099=782.Wait, that's positive. Wait, but earlier when I took AB=19 and CD=99, ABCD=1999, the difference was negative. Wait, there's confusion here. Wait, if ABCD is 1099, AB=10, CD=99. So AB×CD=10×99=990, and ABCD=1099. So 990 -1099= -109. But if AB=19, CD=99, ABCD=1999, then 19×99=1881, 1881 -1999= -118.But if ABCD=1899, AB=18, CD=99, product=18×99=1782, difference=1782 -1899= -117.Wait, so when AB is 19 and CD=99, the difference is -118. If AB is 10 and CD=99, the difference is -109. So the more AB increases, even though ABCD increases, the product increases more, leading to a more negative difference.Wait, let's check with AB=50, CD=99. Then ABCD=5099, AB×CD=50×99=4950, difference=4950 -5099= -149. So that's even more negative.Wait, but AB=50 implies A=5, which makes ABCD start with 5, which is larger. But AB×CD is 50×99=4950. So 4950 -5099= -149.Similarly, AB=99, CD=99: ABCD=9999, product=9801, difference=9801 -9999= -198.So in this case, as AB and CD both increase, ABCD increases, but the product increases even more? Wait, no. Wait, when AB=99 and CD=99, ABCD=9999, product=99×99=9801, so difference=9801 -9999= -198. So that's the most negative so far.But earlier with AB=50, CD=99, difference= -149.Wait, but if we take AB=99 and CD=99, the difference is -198.Is there a way to get even more negative?Wait, let's try AB=89, CD=99. Then ABCD=8999, product=89×99=8811, difference=8811 -8999= -188.Still, the difference is -188. So less negative than -198.Wait, so AB=99, CD=99 gives the most negative difference. But ABCD=9999, which is a large number, but product is 99×99=9801, so difference= -198.But maybe if we take AB=91 and CD=99. Then ABCD=9199, product=91×99=9009, difference=9009 -9199= -190. Hmm, more negative than -188 but less than -198.Alternatively, AB=99, CD=95. Then ABCD=9995, product=99×95=9405, difference=9405 -9995= -590. Wait, wait, that can't be. 99×95=9405, ABCD=9995. 9995 -9405=590. So the difference is 590. Wait, but if we compute AB×CD - ABCD=9405 -9995= -590. But that's a very large negative number. Wait, but that seems conflicting with previous results.Wait, hold on: If ABCD=9995, AB=99, CD=95. Then AB×CD=99×95=9405. So ABCD - AB×CD=9995 -9405=590. Therefore, AB×CD - ABCD= -590. So this is a very negative number.Wait, but why is this so different? Because CD is 95, which is less than 99, but AB is still 99.Wait, but in this case, CD=95 is smaller, but ABCD=9995 is very large. So AB×CD=99×95=9405, which is much less than ABCD=9995, so the difference is 9995 -9405=590. Therefore, AB×CD - ABCD= -590. Wait, but this is more negative than previous examples. So why?But in this case, even though CD is not maximum, ABCD is large, so the difference is more negative? Wait, no. Wait, AB×CD - ABCD=9405 -9995= -590. So it's more negative. But why? Because ABCD is large, but AB×CD is not as large. So in this case, even though AB is maximum, CD is not, leading to a larger ABCD but not a proportionally larger product. Hence, the difference becomes more negative. But this seems contradictory.Wait, maybe I made a mistake here. Let me check again. ABCD=9995, AB=99, CD=95. So AB×CD=99×95=9405. Then ABCD - AB×CD=9995 -9405=590. So the difference is positive. Therefore, AB×CD - ABCD= -590. So this is a very negative number. But we were supposed to maximize AB×CD - ABCD. So in this case, AB×CD - ABCD= -590. But if we take AB=99 and CD=99, then AB×CD - ABCD= -198. So which one is larger? Since -590 is less than -198, the maximum of AB×CD - ABCD is -198 (the higher value between -590 and -198 is -198). Therefore, even though the difference is more negative, it's not the maximum. Wait, perhaps I need to clarify.We are trying to find the maximum of AB×CD - ABCD. So if AB×CD - ABCD= -590, that's a lower value than -198. Therefore, the maximum is the least negative value. So in this context, the maximum of AB×CD - ABCD would be -198, achieved when AB=99 and CD=99. Whereas when AB=99 and CD=95, it's -590, which is smaller. Therefore, the maximum is -198. Hence, the minimal value of ABCD - AB×CD is 198. Wait, but that contradicts the previous example where ABCD=9991, AB=99, CD=91, difference= -18. Wait, ABCD=9991, AB=99, CD=91. Then AB×CD=99×91=9009. Then ABCD - AB×CD=9991 -9009=982. Wait, no, 9991-9009=982? Wait, 9009 + 982=9991? Let's check 9009 + 900=9909, 9909 +82=9991. Yes. So the difference is 982. So AB×CD - ABCD= -982. But that's even worse. So why is this happening?Wait, maybe there's confusion here. Let me clarify.The problem asks for the minimal value of ABCD - AB×CD.To find the minimal value, which can be negative, we need to find the case where ABCD - AB×CD is as small as possible (most negative).But when I thought about maximizing AB×CD - ABCD, that is equivalent to finding the maximum of (AB×CD - ABCD), which would correspond to the minimal value of ABCD - AB×CD = - (AB×CD - ABCD). So the minimal value is the negative of the maximum of (AB×CD - ABCD). However, if AB×CD - ABCD can be made very large negative, then ABCD - AB×CD can be very large positive. Wait, no. Wait, if AB×CD - ABCD is negative, then ABCD - AB×CD is positive. If AB×CD - ABCD is positive, then ABCD - AB×CD is negative.Wait, this is getting confusing. Let's rephrase.We need to compute ABCD - AB×CD. We need the minimal value. So if ABCD - AB×CD can be negative, that would be smaller (more minimal) than a positive value.So to get the minimal value, we need cases where ABCD - AB×CD is as negative as possible, i.e., AB×CD is as large as possible compared to ABCD.Therefore, we need to maximize AB×CD while keeping ABCD as small as possible.Alternatively, find AB and CD such that their product is large, but ABCD is a number that is not too large. For example, if AB is large and CD is large, but ABCD is arranged such that the number ABCD is not the maximum possible. For instance, AB=99, CD=99, ABCD=9999. Here, the product is 9801, difference=9999 -9801=198. But if AB=91, CD=99, ABCD=9199, product=9009, difference=9199 -9009=190. So difference is 190. But wait, 9199 -9009=190, which is still positive. Wait, so how can we get a negative difference?Wait, in my earlier example, ABCD=9991, AB=99, CD=91. Then ABCD=9991, product=99×91=9009, difference=9991 -9009=982. Positive. Wait, but if AB=99, CD=95, ABCD=9995, product=9405, difference=9995 -9405=590. Still positive. Wait, so when does ABCD - AB×CD become negative?Wait, maybe when AB is small and CD is large. Let's take AB=10, CD=99. Then ABCD=1099. Product=10×99=990. Difference=1099 -990=109. Positive. AB=12, CD=99, ABCD=1299. Product=12×99=1188. Difference=1299 -1188=111. Still positive.Wait, how about AB=50, CD=99. ABCD=5099. Product=50×99=4950. Difference=5099 -4950=149. Positive.AB=99, CD=99, difference=198. AB=99, CD=00 is invalid. AB=99, CD=01, product=99×1=99, difference=9901 -99=9802. Very positive.Wait, so all these examples result in a positive difference. Wait, but earlier I thought there was a negative difference. Wait, maybe I made a miscalculation before.Wait, let's recast.Let me compute ABCD - AB×CD with AB=81, CD=19. So ABCD=8119. Then AB×CD=81×19=1539. Difference=8119 -1539=6580. That's positive.Wait, maybe if CD is small. Let's take AB=99, CD=10. Then ABCD=9910. Product=99×10=990. Difference=9910 -990=8920. Positive.Wait, maybe take AB=19, CD=99. ABCD=1999. Product=19×99=1881. Difference=1999 -1881=118. Positive.Wait, perhaps there is no case where ABCD - AB×CD is negative. Then the minimal value would be the smallest positive value, which in the examples above is 118.But wait, when I thought of ABCD=9991, AB=99, CD=91, difference=9991 -9009=982.Wait, but when I considered AB=99 and CD=99, the difference is 198.Wait, but how about AB=50, CD=50. ABCD=5050. Product=50×50=2500. Difference=5050 -2500=2550.Wait, so perhaps the minimal value is 118? Or maybe even smaller?Wait, let's check another example. Let's take AB=25, CD=97. Then ABCD=2597. Product=25×97=2425. Difference=2597 -2425=172.Another example: AB=39, CD=61. ABCD=3961. Product=39×61=2379. Difference=3961 -2379=1582.Wait, this is larger. Maybe AB=49, CD=51. ABCD=4951. Product=49×51=2499. Difference=4951 -2499=2452.Wait, not helpful. Let's think of AB and CD such that AB×CD is as large as possible compared to ABCD.Wait, for example, if AB=89, CD=99, ABCD=8999. Product=89×99=8811. Difference=8999 -8811=188. So 188.AB=89, CD=98, ABCD=8998. Product=89×98=8722. Difference=8998 -8722=276.AB=88, CD=99, ABCD=8899. Product=88×99=8712. Difference=8899 -8712=187.AB=87, CD=99, ABCD=8799. Product=87×99=8613. Difference=8799 -8613=186.AB=79, CD=99, ABCD=7999. Product=79×99=7821. Difference=7999 -7821=178.AB=78, CD=99, ABCD=7899. Product=78×99=7722. Difference=7899 -7722=177.AB=69, CD=99, ABCD=6999. Product=69×99=6831. Difference=6999 -6831=168.AB=59, CD=99, ABCD=5999. Product=59×99=5841. Difference=5999 -5841=158.AB=49, CD=99, ABCD=4999. Product=49×99=4851. Difference=4999 -4851=148.AB=39, CD=99, ABCD=3999. Product=39×99=3861. Difference=3999 -3861=138.AB=29, CD=99, ABCD=2999. Product=29×99=2871. Difference=2999 -2871=128.AB=19, CD=99, ABCD=1999. Product=19×99=1881. Difference=1999 -1881=118.AB=18, CD=99, ABCD=1899. Product=18×99=1782. Difference=1899 -1782=117.AB=17, CD=99, ABCD=1799. Product=17×99=1683. Difference=1799 -1683=116.AB=16, CD=99, ABCD=1699. Product=16×99=1584. Difference=1699 -1584=115.AB=15, CD=99, ABCD=1599. Product=15×99=1485. Difference=1599 -1485=114.AB=14, CD=99, ABCD=1499. Product=14×99=1386. Difference=1499 -1386=113.AB=13, CD=99, ABCD=1399. Product=13×99=1287. Difference=1399 -1287=112.AB=12, CD=99, ABCD=1299. Product=12×99=1188. Difference=1299 -1188=111.AB=11, CD=99, ABCD=1199. Product=11×99=1089. Difference=1199 -1089=110.AB=10, CD=99, ABCD=1099. Product=10×99=990. Difference=1099 -990=109.So, as we can see, when we decrease AB from 19 to 10, keeping CD=99, the difference reduces from 118 down to 109. Wait, so the minimal difference in these examples is 109.Wait, but earlier when AB=19, CD=99, difference=118; AB=10, CD=99, difference=109. So 109 is smaller.But is there a way to get even lower?Let's try AB=10, CD=99, difference=109.What if we vary CD? For example, AB=10, CD=98. Then ABCD=1098. Product=10×98=980. Difference=1098 -980=118. That's higher.AB=10, CD=97. ABCD=1097. Product=10×97=970. Difference=1097 -970=127. Higher.Similarly, AB=10, CD=96. Difference=1096 -960=136.So, when CD decreases from 99, the difference increases. Therefore, to minimize the difference, CD should be as large as possible.Therefore, for AB=10, CD=99 gives the minimal difference in that category.Similarly, if we take AB=11, CD=99, difference=110.So, as AB increases from 10 to 19, the difference increases from 109 to 118. Wait, no, when AB=10, difference=109; AB=11, difference=110; up to AB=19, difference=118. So increasing AB while keeping CD=99 results in increasing difference.Therefore, the minimal difference when CD=99 occurs at AB=10, difference=109.But perhaps we can get a lower difference by choosing CD not equal to 99.Wait, let's check AB=19, CD=89. ABCD=1989. Product=19×89=1691. Difference=1989 -1691=298. Higher.AB=19, CD=98. Product=19×98=1862. ABCD=1998. Difference=1998 -1862=136. Higher than 118.AB=19, CD=79. Product=19×79=1501. ABCD=1979. Difference=1979 -1501=478.No, worse.Alternatively, let's try AB=12, CD=98. ABCD=1298. Product=12×98=1176. Difference=1298 -1176=122. Which is higher than when CD=99.So, seems like CD=99 gives the minimal difference for each AB.Now, let's check for AB=10, CD=99, difference=109.Is there a way to get a smaller difference by changing AB and CD?Wait, what if CD is not 99 but another number, but AB is different. For example, AB=21, CD=93. ABCD=2193. Product=21×93=1953. Difference=2193 -1953=240. Not helpful.Alternatively, AB=13, CD=95. Product=13×95=1235. ABCD=1395. Difference=1395 -1235=160.Still higher.Alternatively, AB=15, CD=95. ABCD=1595. Product=15×95=1425. Difference=1595 -1425=170.Hmm.What if AB=22, CD=97. ABCD=2297. Product=22×97=2134. Difference=2297 -2134=163.Not better.Wait, perhaps another approach. Let's consider that ABCD - AB×CD = 1000A + 100B +10C +D - (10A + B)(10C + D).We can write this as:= 1000A + 100B +10C + D - [100AC +10AD +10BC +BD]= 1000A -100AC -10AD +100B -10BC -BD +10C +DLet me factor terms:= A(1000 -100C -10D) + B(100 -10C - D) +10C + DNow, to minimize this expression, we need to assign values to A, B, C, D such that each term is minimized.Let's analyze each component:1. A(1000 -100C -10D): To minimize this term, since A is at least 1, and (1000 -100C -10D) can be positive or negative. Wait, 100C +10D is the value of the 3-digit number CD0, but since C is a digit from 1-9 and D is 0-9, 100C +10D ranges from 100*1 +10*0=100 to 100*9 +10*9=990.Therefore, 1000 -100C -10D ranges from 1000 -990=10 to 1000 -100=900. So 1000 -100C -10D is always positive. Therefore, the term A(1000 -100C -10D) is positive, and to minimize it, we need to minimize A and maximize 100C +10D.Since A is at least 1, to minimize this term, set A=1, and maximize 100C +10D, which would be C=9, D=9. So 100*9 +10*9=990, so the term becomes 1*(1000 -990)=10.2. B(100 -10C - D): Let's look at 100 -10C -D. Since C is from 1-9 and D is 0-9, 10C + D ranges from 10*1 +0=10 to 10*9 +9=99. Therefore, 100 -10C -D ranges from 100 -99=1 to 100 -10=90. Therefore, this term is positive. To minimize the entire expression, since B is from 0-9, we can set B=0, which would make this term zero. If B>0, this term would add a positive value. Therefore, to minimize the expression, set B=0.3. 10C + D: This term is simply the 2-digit number CD. To minimize the entire expression, since this term is positive, we need to minimize CD. Since C is at least 1, the smallest CD is 10, with C=1, D=0.So, putting it all together, to minimize the expression:- Set A=1 (minimizes the first term)- Set B=0 (minimizes the second term)- Set CD as small as possible, which is 10 (C=1, D=0) to minimize the third term.Therefore, the minimal value occurs at ABCD=1 0 1 0=1010, AB=10, CD=10.Compute the expression:ABCD - AB×CD =1010 - (10×10)=1010 -100=910.But wait, is this the minimal value? Earlier examples gave us lower values like 109.Wait, but according to the mathematical analysis, the minimal value should be 910. However, in practice, when we set A=1, B=0, C=1, D=0, the difference is 910, but earlier examples with A=10, B=0, C=9, D=9 gave difference=1099 -990=109, which is lower.This discrepancy arises because the mathematical analysis assumed that minimizing each term would lead to the minimal total. However, the terms are not independent. When we set CD=10 (minimizing the third term), we might be increasing other terms.Wait, let's substitute the values:If A=1, B=0, C=1, D=0:Expression=1*(1000 -100*1 -10*0) +0*(100 -10*1 -0) +10*1 +0=1*(1000 -100) +0 +10 +0=1*900 +10=910.If A=1, B=0, C=9, D=9:Expression=1*(1000 -100*9 -10*9) +0*(100 -10*9 -9) +10*9 +9=1*(1000 -900 -90) +0 +90 +9=1*10 +99=10 +99=109.Therefore, the latter gives a much lower result. Therefore, the mathematical approach of individually minimizing each term does not work because the terms are interdependent. For example, choosing a larger CD (which increases the third term) can lead to a decrease in the first term (due to the subtraction in 1000 -100C -10D), which can overall result in a lower total.Therefore, the correct approach must consider the interdependencies between variables.Hence, perhaps the initial approach of testing examples where AB is small and CD is large is more effective. From the examples, when AB=10, CD=99, the difference is 109. When AB=10, CD=99, ABCD=1099, product=990, difference=1099 -990=109. When AB=10, CD=99, the first term is 1*(1000 -990)=10, second term is 0*(100 -99)=0, third term is 99. So total=10 +0 +99=109. That aligns.Alternatively, if we take AB=11, CD=99, ABCD=1199, product=1089, difference=1199 -1089=110. Similarly, the first term would be A*(1000 -100C -10D)=1*(1000 -990)=10, B*(100 -10C -D)=1*(100 -99)=1, and CD=99. So total=10 +1 +99=110. Which matches.Therefore, the minimal value so far is 109, achieved when A=1, B=0, C=9, D=9.Is there a way to get even lower?Let's see. Let's see if changing B affects the result.Suppose A=1, B=1, C=9, D=9. Then ABCD=1199. Product AB=11, CD=99. Product=1089. Difference=1199 -1089=110.If A=1, B=0, C=9, D=9. Difference=109.If B=0, it's better.What if A=1, B=0, C=9, D=8. Then CD=98. Then ABCD=1098. Product=10×98=980. Difference=1098 -980=118. Which is higher than 109.If A=1, B=0, C=8, D=9. CD=89. ABCD=1089. Product=10×89=890. Difference=1089 -890=199.Even higher.Therefore, keeping CD as 99 gives the minimal difference for A=1, B=0.If we try A=1, B=0, C=9, D=9, difference=109.Is there a way to get a lower difference by changing A?For example, A=2, B=0, C=9, D=9. Then ABCD=2099. Product=20×99=1980. Difference=2099 -1980=119. Higher than 109.A=3, B=0, C=9, D=9. Difference=3099 -30×99=3099 -2970=129.Similarly, increasing A increases the difference.If A=1, B=0, C=9, D=9 gives the minimal difference so far.What if we try A=1, B=0, C=9, D=9. Difference=109.Is there a way to have a smaller difference by adjusting B and C?Wait, if we set B=0, but lower C. For example, A=1, B=0, C=8, D=9. Then ABCD=1089. Product=10×89=890. Difference=1089 -890=199. Worse.If A=1, B=0, C=9, D=8. ABCD=1098. Product=10×98=980. Difference=1098 -980=118.Still worse.If we set B=0, C=9, D=9: difference=109.If we set B=0, C=9, D=9. If we set D=9, C=9. Is there a way to have C or D higher? No, since they are already at 9.Therefore, the minimal difference found so far is 109.But wait, in the example I thought of earlier with ABCD=9991, AB=99, CD=91. Difference=9991 -99×91=9991 -9009=982. But that's a positive difference. Not helpful.Wait, but maybe if AB and CD are arranged such that AB is large and CD is large, but ABCD is not the maximum. For example, AB=99, CD=99 gives ABCD=9999, difference=198. But if AB=99, CD=89, ABCD=9989. Product=99×89=8811. Difference=9989 -8811=1178. Still positive.Wait, all examples where AB and CD are large result in positive differences. The only way to get a lower difference is by setting AB to be small and CD to be large, as in AB=10, CD=99, difference=109.But wait, earlier I thought that difference could be negative, but all my calculations resulted in positive differences. Did I make a mistake in previous thinking?Wait, let's recast the original problem: Compute the least possible value of ABCD - AB×CD.All examples so far give positive results. So is it possible that the minimal value is positive? But the problem says "least possible value", so it could be negative. But according to the examples, it's positive.Wait, perhaps there's a case where AB×CD is larger than ABCD. Let me find such a case.Suppose ABCD=1000. Then AB=10, CD=00, but CD cannot be 00. So CD=01. Then product=10×01=10. Difference=1000 -10=990.ABCD=1001, AB=10, CD=01. Product=10×1=10. Difference=1001 -10=991.If AB=50, CD=20. ABCD=5020. Product=50×20=1000. Difference=5020 -1000=4020.If AB=99, CD=10. ABCD=9910. Product=99×10=990. Difference=9910 -990=8920.If AB=80, CD=15. ABCD=8015. Product=80×15=1200. Difference=8015 -1200=6815.Wait, all these examples give positive differences. But when I take AB=50, CD=20, product=1000, ABCD=5020. Difference=4020. Still positive.Wait, what if AB=25, CD=40. ABCD=2540. Product=25×40=1000. Difference=2540 -1000=1540.Still positive.Wait, how about AB=99, CD=99. Product=9801, ABCD=9999. Difference=198.Another example: AB=91, CD=99. Product=9009, ABCD=9199. Difference=9199 -9009=190.AB=90, CD=99. Product=90×99=8910, ABCD=9099. Difference=9099 -8910=189.AB=85, CD=99. Product=85×99=8415, ABCD=8599. Difference=8599 -8415=184.AB=80, CD=99. Product=80×99=7920, ABCD=8099. Difference=8099 -7920=179.AB=75, CD=99. Product=75×99=7425, ABCD=7599. Difference=7599 -7425=174.AB=70, CD=99. Product=70×99=6930, ABCD=7099. Difference=7099 -6930=169.AB=65, CD=99. Product=65×99=6435, ABCD=6599. Difference=6599 -6435=164.AB=60, CD=99. Product=60×99=5940, ABCD=6099. Difference=6099 -5940=159.AB=55, CD=99. Product=55×99=5445, ABCD=5599. Difference=5599 -5445=154.AB=50, CD=99. Product=50×99=4950, ABCD=5099. Difference=5099 -4950=149.AB=49, CD=99. Product=49×99=4851, ABCD=4999. Difference=4999 -4851=148.AB=40, CD=99. Product=40×99=3960, ABCD=4099. Difference=4099 -3960=139.AB=30, CD=99. Product=30×99=2970, ABCD=3099. Difference=3099 -2970=129.AB=20, CD=99. Product=20×99=1980, ABCD=2099. Difference=2099 -1980=119.AB=10, CD=99. Product=10×99=990, ABCD=1099. Difference=1099 -990=109.AB=09, CD=99. Wait, AB=09 is allowed? A=0, but A can't be zero. So AB must be at least 10.Therefore, the minimal difference found in this sequence is 109.Is there a case where the difference is lower than 109?Let's check AB=10, CD=99. Difference=109.If we try AB=10, CD=98. Difference=1098 -10×98=1098 -980=118.AB=10, CD=99. Difference=1099 -990=109.AB=10, CD=99, with ABCD=1099.Alternatively, AB=10, CD=99, ABCD=1099.What if we try AB=11, CD=99. ABCD=1199. Product=11×99=1089. Difference=1199 -1089=110.AB=12, CD=99. Difference=1299 -12×99=1299 -1188=111.So increasing AB from 10 onward increases the difference.What if we try AB=10, CD=89. ABCD=1089. Product=10×89=890. Difference=1089 -890=199.Higher.AB=10, CD=99. Difference=109.Is there a way to set AB and CD such that the product is larger than ABCD? It seems not, because AB is the first two digits of ABCD, and CD is the last two digits. Since AB is at least 10 and CD is at least 10, the product AB×CD is at least 10×10=100, while ABCD is at least 1000. So ABCD is at least 1000, and the product is at least 100. Therefore, the difference ABCD - AB×CD is at least 900. But in our examples, we have differences as low as 109, which contradicts this reasoning. Wait, there's a mistake here.Wait, ABCD is a 4-digit number, which ranges from 1000 to 9999. AB is the first two digits, so ranges from 10 to 99. CD is the last two digits, from 00 to 99 (but C can't be 0). So CD ranges from 10 to 99. Therefore, the minimal AB×CD is 10×10=100, and maximal AB×CD is 99×99=9801.But ABCD ranges from 1000 to 9999. Therefore, the difference ABCD - AB×CD can range from as low as 1000 - 9801= -8801 to as high as 9999 -100=9899.Wait, but this is a miscalculation. ABCD is a 4-digit number, so it's at least 1000. AB×CD can be as high as 99×99=9801. Therefore, the difference ABCD - AB×CD can be as low as 1000 -9801= -8801. However, in reality, ABCD and AB×CD are related, because AB and CD are parts of ABCD. Therefore, ABCD is not independent of AB and CD.For example, if AB=99 and CD=99, then ABCD=9999, and AB×CD=9801, so difference=9999 -9801=198.If AB=99 and CD=00, ABCD=9900, but CD cannot be 00.But if AB and CD are independent variables, the minimal difference would be -8801, but since AB and CD are parts of ABCD, the actual minimal difference is constrained.So, in order to find the minimal value of ABCD - AB×CD, we need to maximize AB×CD while keeping ABCD as small as possible.But how can AB×CD be large while ABCD is small?For example, if AB is large and CD is large, but ABCD is not too large.Wait, for instance, if AB=99 and CD=99, ABCD=9999. But that's the largest 4-digit number. However, if AB=99 and CD=98, ABCD=9998, product=99×98=9702, difference=9998 -9702=296.If AB=98, CD=99, ABCD=9899, product=98×99=9702, difference=9899 -9702=197.Wait, which is lower than when AB=99, CD=99.AB=97, CD=99, ABCD=9799, product=97×99=9603, difference=9799 -9603=196.AB=96, CD=99, difference=9699 -9504=195.Continuing this pattern:AB=99 -k, CD=99, ABCD=(99 -k)99, product=(99 -k)*99=99*(99 -k), difference= ( (99 -k)*100 +99 ) -99*(99 -k)= ( (99 -k)*100 +99 ) -99*(99 -k)= (99 -k)*100 +99 -99*(99 -k)= (99 -k)*(100 -99) +99= (99 -k)*1 +99= 99 -k +99=198 -k.So when AB=99 -k, CD=99, the difference is 198 -k. Therefore, as k increases from 0 to 89 (AB from 99 down to 10), the difference decreases from 198 to 198 -89=109.So when AB=10, CD=99, difference=109.This aligns with the previous examples.Therefore, the minimal difference in this sequence is 109, achieved when AB=10, CD=99, ABCD=1099.But is there another combination where AB×CD - ABCD is larger, leading to a smaller difference?Wait, consider AB=21, CD=93. Then ABCD=2193, product=21×93=1953. Difference=2193 -1953=240.But this is higher than 109.What if we take AB=15, CD=95. ABCD=1595. Product=15×95=1425. Difference=1595 -1425=170.Still higher.Alternatively, AB=25, CD=75. ABCD=2575. Product=25×75=1875. Difference=2575 -1875=700.Not helpful.Alternatively, AB=10, CD=95. ABCD=1095. Product=10×95=950. Difference=1095 -950=145.Higher than 109.AB=10, CD=99. Difference=109.What if AB=10, CD=99. Is there a way to make ABCD smaller than 1099 while keeping CD=99?The smallest ABCD with CD=99 is 1099 (A=1, B=0, C=9, D=9). If A=1, B can be 0, giving ABCD=1099.If A=1, B=0, C=9, D=9: ABCD=1099.If we try A=1, B=0, C=9, D=9. That's the smallest possible ABCD with CD=99. Any other combination with CD=99 would have a higher ABCD, like 1199, 1299, etc.Therefore, ABCD=1099, AB=10, CD=99, difference=109.Is this the minimal possible?Wait, another case: CD=99, AB=10, ABCD=1099. If we try to have CD=99 and AB=10, but ABCD=1099. If we try to reduce ABCD further, but CD must be 99. Therefore, the thousands digit A must be 1, and the hundreds digit B must be 0. So ABCD=1099 is the minimal.Alternatively, is there a case where CD is not 99 but the product AB×CD is larger relative to ABCD?For example, AB=85, CD=13. ABCD=8513. Product=85×13=1105. Difference=8513 -1105=7408. Not helpful.AB=12, CD=83. ABCD=1283. Product=12×83=996. Difference=1283 -996=287.Higher than 109.AB=13, CD=76. Product=13×76=988. ABCD=1376. Difference=1376 -988=388.Still higher.AB=14, CD=71. Product=14×71=994. ABCD=1471. Difference=1471 -994=477.Not helpful.AB=15, CD=66. Product=15×66=990. ABCD=1566. Difference=1566 -990=576.Hmm.AB=16, CD=62. Product=16×62=992. ABCD=1662. Difference=1662 -992=670.No improvement.AB=17, CD=58. Product=17×58=986. ABCD=1758. Difference=1758 -986=772.Still higher.AB=18, CD=55. Product=18×55=990. ABCD=1855. Difference=1855 -990=865.No.AB=19, CD=52. Product=19×52=988. ABCD=1952. Difference=1952 -988=964.No.So none of these examples give a lower difference than 109.Alternatively, if we take AB=99, CD=10. ABCD=9910. Product=99×10=990. Difference=9910 -990=8920.Not helpful.Therefore, based on exhaustive testing, the minimal difference occurs when AB=10, CD=99, ABCD=1099, giving a difference of 109.But wait, in the previous analysis, when AB=99 -k and CD=99, the difference is 198 -k, where k ranges from 0 to 89. So when k=89, AB=10, difference=109. This seems to be the minimal value.Is there a way to get a lower difference by setting CD to something other than 99?For example, let's try AB=11, CD=9A. Wait, CD must be a 2-digit number with C=9. So CD can be 90 to 99.Take CD=99: difference=11×99=1089, ABCD=1199. Difference=110.CD=98: product=11×98=1078, ABCD=1198. Difference=1198 -1078=120. Higher.CD=97: product=11×97=1067, ABCD=1197. Difference=130.So, the difference increases as CD decreases.Similarly, if AB=10, CD=99: difference=109.If AB=10, CD=98: difference=1098 -10×98=1098 -980=118.Similarly, higher.Thus, for AB=10, the minimal difference is achieved when CD=99.Therefore, the minimal possible value of ABCD - AB×CD is 109.But wait, let's check if there's another combination where CD is not 99 but AB is slightly larger, leading to a higher product that might offset the increase in ABCD.For example, AB=11, CD=90. Product=11×90=990. ABCD=1190. Difference=1190 -990=200.Higher than 109.AB=12, CD=85. Product=12×85=1020. ABCD=1285. Difference=1285 -1020=265.Higher.AB=20, CD=50. Product=20×50=1000. ABCD=2050. Difference=2050 -1000=1050.No.AB=25, CD=40. Product=25×40=1000. ABCD=2540. Difference=1540.No.Alternatively, AB=50, CD=20. Product=50×20=1000. ABCD=5020. Difference=5020 -1000=4020.No.It seems that any other combination doesn't yield a lower difference than 109.Therefore, after thorough analysis and testing various cases, the minimal value of ABCD - AB×CD is 109, achieved when ABCD=1099, AB=10, CD=99.However, to ensure this is the minimal possible, let's consider another approach.Suppose we treat the expression ABCD - AB×CD as:1000A + 100B +10C + D - (10A + B)(10C + D)Expand the product:1000A + 100B +10C + D - (100AC + 10AD + 10BC + BD)= 1000A - 100AC -10AD + 100B -10BC - BD +10C + DGroup terms:= A(1000 -100C -10D) + B(100 -10C -D) + C(10) + D(1)To minimize this expression, we need to minimize each component.Let's analyze each term:1. A(1000 -100C -10D): Since A ≥ 1 and 100C +10D ≤ 990 (when C=9, D=9), 1000 -100C -10D ≥ 10. Therefore, this term is positive. To minimize it, set A=1 and maximize 100C +10D. Max 100C +10D=990 (C=9, D=9). So this term becomes 1*(1000 -990)=10.2. B(100 -10C -D): Since 10C +D ≥10 (C≥1), 100 -10C -D ≤90. To minimize this term, set B=0, making the entire term zero. If B>0, the term increases.3. C(10) + D(1): This is the value of CD, which is minimized when C=1 and D=0 (CD=10). However, we already fixed C=9 and D=9 to minimize the first term. Here, CD=99, which is the maximum.Thus, there is a conflict between minimizing the first term and the third term. To minimize the first term, we need CD=99, which maximizes the third term. However, the third term is added, so maximizing it increases the total. Therefore, we need to balance these two.If we set CD=99 (maximizing 100C +10D), the first term becomes 10 (minimal possible for A=1), the second term becomes zero (B=0), and the third term is 99. Total=10 +0 +99=109.If we set CD=10 (minimizing the third term), then 100C +10D=100*1 +10*0=100. The first term becomes A*(1000 -100)=A*900. With A=1, this term is 900. The second term with B=0 is zero. The third term is 10. Total=900 +0 +10=910. This is much higher.Thus, the balance between minimizing the first term and the third term leads us to choose CD=99, A=1, B=0, resulting in a total of 109.Therefore, after careful analysis, the least possible value is 109.Final AnswerThe least possible value is boxed{109}.

✅:To find the least possible value of (ABCD - AB times CD), where (ABCD) is a 4-digit number, and (AB) and (CD) are 2-digit numbers, we need to minimize the expression:[ABCD - (AB times CD)]This can be expanded as:[(1000A + 100B + 10C + D) - (10A + B)(10C + D)]Expanding the product ((10A + B)(10C + D)) gives:[100AC + 10AD + 10BC + BD]Subtracting this from (ABCD) results in:[1000A + 100B + 10C + D - (100AC + 10AD + 10BC + BD)]Grouping and factoring terms, we get:[A(1000 - 100C - 10D) + B(100 - 10C - D) + 10C + D]To minimize this expression, we need to consider the interplay between the terms. The key is to balance the contributions from each term. After thorough analysis and testing various combinations, the minimal value is found when (A = 1), (B = 0), (C = 9), and (D = 9). This results in:- (ABCD = 1099)- (AB = 10)- (CD = 99)- The product (AB times CD = 10 times 99 = 990)Thus, the expression evaluates to:[1099 - 990 = 109]Therefore, the least possible value is (boxed{109}).

🔑:1. Let ( x ) be the two-digit number ( AB ), and let ( y ) be the two-digit number ( CD ). We want to minimize the quantity ( ABCD - AB times CD ).2. Express ( ABCD ) in terms of ( x ) and ( y ): [ ABCD = 100x + y ] where ( x = 10A + B ) and ( y = 10C + D ).3. The expression we need to minimize is: [ 100x + y - xy ]4. Rewrite the expression: [ 100x + y - xy = 100x + y - (10A + B)(10C + D) ]5. To simplify, let’s consider the expression ( 100x + y - xy ) directly: [ 100x + y - xy = 100x + y - 10A cdot 10C - 10A cdot D - B cdot 10C - B cdot D ]6. Notice that ( 100x + y - xy ) can be rewritten as: [ 100x + y - xy = 100x + y - 100AC - 10AD - 10BC - BD ]7. To find the minimum value, we need to maximize the negative part ( xy ). Since ( x ) and ( y ) are two-digit numbers, the maximum value for ( x ) is 99 and for ( y ) is 99.8. However, we need to find the least possible value of ( 100x + y - xy ). Let’s consider the boundary values for ( x ) and ( y ): - If ( x = 10 ) and ( y = 99 ): [ 100 cdot 10 + 99 - 10 cdot 99 = 1000 + 99 - 990 = 109 ]9. Verify if this is indeed the minimum: - If ( x = 11 ) and ( y = 98 ): [ 100 cdot 11 + 98 - 11 cdot 98 = 1100 + 98 - 1078 = 120 ] - If ( x = 12 ) and ( y = 97 ): [ 100 cdot 12 + 97 - 12 cdot 97 = 1200 + 97 - 1164 = 133 ]10. From the calculations, the minimum value is indeed achieved when ( x = 10 ) and ( y = 99 ).The final answer is ( boxed{109} ).

🧠:Alright, let me try to work through this problem step by step. The problem is about a function F(n) defined as the sum of f(d) over all divisors d of n, where f(n) is (-1) raised to the number of prime factors of n, counting multiplicities. We need to show that F(n) is either 0 or 1 for any positive integer n, and determine for which n it is 1.First, let's parse the definitions again to make sure I understand them correctly. The function Ω(n) counts the number of prime factors of n, including multiplicities. For example, Ω(12) = Ω(2^2 * 3) = 3. Then f(n) is defined as (-1)^{Ω(n)}, so f(1) is 1 since Ω(1) is 0 (since 1 has no prime factors), and for any prime power, say p^k, Ω(p^k) = k, so f(p^k) = (-1)^k. For a general number n with prime factorization n = p1^{k1} p2^{k2} ... pr^{kr}, Ω(n) = k1 + k2 + ... + kr, so f(n) = (-1)^{k1 + k2 + ... + kr}.Then F(n) is the sum of f(d) over all divisors d of n. So, for example, if n is a prime number p, then its divisors are 1 and p. Then F(p) = f(1) + f(p) = 1 + (-1)^1 = 1 - 1 = 0. If n is p^2, divisors are 1, p, p^2. Then F(p^2) = f(1) + f(p) + f(p^2) = 1 + (-1) + 1 = 1. Similarly, for n = p^3, divisors are 1, p, p^2, p^3. So F(p^3) = 1 -1 +1 -1 = 0. So for prime powers, it seems like F(p^k) alternates between 1 and 0 depending on whether k is even or odd? Wait, let's check.Wait, if n = p^k, the number of divisors is k+1. Each divisor is p^m for m from 0 to k. Then f(p^m) = (-1)^m. So the sum F(p^k) is the sum from m=0 to m=k of (-1)^m. That is a geometric series with ratio -1. The sum is (1 - (-1)^{k+1}) / (1 - (-1)) ) = (1 - (-1)^{k+1}) / 2. So if k is even, then k+1 is odd, so (-1)^{k+1} = -1, so the sum is (1 - (-1)) / 2 = 2/2 = 1. If k is odd, then (-1)^{k+1} = 1, so the sum is (1 - 1)/2 = 0. Therefore, F(p^k) is 1 if k is even and 0 if k is odd. So for prime powers, F(n) is 1 if the exponent is even, 0 if odd. Got that.Now, the problem states that F(n) is either 0 or 1 for any positive integer n. So the first part is to prove that. Then, we have to determine for which n we have F(n)=1.Since F(n) is multiplicative? Wait, let's check if F(n) is multiplicative. If n and m are coprime, is F(nm) = F(n)F(m)? Let's see. Let's take n and m coprime. Then the divisors of nm are products of divisors of n and divisors of m. So each divisor d of nm can be written uniquely as d1*d2 where d1|n and d2|m. Then, F(nm) = sum_{d|nm} f(d) = sum_{d1|n, d2|m} f(d1 d2). Now, since d1 and d2 are coprime, Ω(d1 d2) = Ω(d1) + Ω(d2), so f(d1 d2) = (-1)^{Ω(d1) + Ω(d2)} = (-1)^{Ω(d1)} * (-1)^{Ω(d2)} = f(d1) f(d2). Therefore, F(nm) = sum_{d1|n, d2|m} f(d1) f(d2) = (sum_{d1|n} f(d1)) * (sum_{d2|m} f(d2)) = F(n) F(m). Hence, F is multiplicative.Therefore, since F is multiplicative, to compute F(n) for any n, we can factor n into primes, n = p1^{k1} p2^{k2} ... pr^{kr}, and then F(n) = F(p1^{k1}) F(p2^{k2}) ... F(pr^{kr}).But we already found that F(p^k) is 1 if k is even, 0 if k is odd. Therefore, F(n) is the product over each prime power factor of F(p_i^{k_i}). So each F(p_i^{k_i}) is 1 if k_i is even, 0 if k_i is odd. Therefore, the entire product F(n) is 1 if and only if all k_i are even (i.e., n is a perfect square), and 0 otherwise. Wait, but hold on, that would mean F(n)=1 if and only if n is a square, but earlier, when n is a prime squared, F(p^2)=1, but when n is a product of two primes, say n = p*q, which is not a square, then F(n) = F(p) F(q) = 0 * 0 = 0. Wait, but hold on, n = p*q, which is square-free. Let's compute F(n) directly.n = p*q, divisors are 1, p, q, pq. Then f(1) = 1, f(p) = -1, f(q) = -1, f(pq) = (-1)^2 = 1. So sum is 1 -1 -1 +1 = 0. So indeed F(n) = 0. If n is square-free, then in its prime factorization, all exponents are 1 (odd), so F(n) = product of 0s, hence 0. If n is a square, say n = p^2, then F(n) = 1. If n is a product of squares, like n = p^2 q^2, then F(n) = 1 * 1 = 1. But wait, n = p^2 q^2 is a square. So in general, n is a square if and only if all exponents in its prime factorization are even. So in that case, F(n) = 1. Otherwise, at least one exponent is odd, so F(n) = 0.But wait, that seems to contradict the problem statement. The problem says "Prove that F(n)=0,1 for all positive integers n. For which integers n is F(n)=1?" So according to my reasoning, F(n) is 1 if n is a square, 0 otherwise. Therefore, the answer is that F(n)=1 if and only if n is a perfect square. But let me check with some examples.Take n=1. 1 is a square. F(1) = sum_{d|1} f(d) = f(1) =1, which is correct.Take n=4. Divisors are 1,2,4. f(1)=1, f(2)=(-1)^1=-1, f(4)=(-1)^2=1. Sum:1 -1 +1=1. Correct, since 4 is a square.Take n=6. Divisors 1,2,3,6. f(1)=1, f(2)=f(3)=-1, f(6)=(-1)^2=1. Sum:1 -1 -1 +1=0. Since 6 is not a square.Take n=9. Divisors 1,3,9. f(1)=1, f(3)=-1, f(9)=1. Sum:1 -1 +1=1. 9 is a square.Take n=12. Prime factors 2^2 *3^1. So exponents 2 and 1. Therefore, F(n)=F(2^2)*F(3^1)=1*0=0. Let's compute directly. Divisors of 12:1,2,3,4,6,12.f(1)=1, f(2)=f(3)=-1, f(4)=1, f(6)=1, f(12)=(-1)^3=-1. Sum:1 -1 -1 +1 +1 -1 = (1 -1) + (-1 +1) + (1 -1) = 0 +0 +0=0. Correct.Take n=16. Divisors 1,2,4,8,16. f(1)=1, f(2)=-1, f(4)=1, f(8)=-1, f(16)=1. Sum:1 -1 +1 -1 +1=1. Correct, since 16 is a square.Take n=25. Divisors 1,5,25. f(1)=1, f(5)=-1, f(25)=1. Sum:1 -1 +1=1. Correct.Take n=36=2^2 *3^2. F(n)=1*1=1. Divisors of 36:1,2,3,4,6,9,12,18,36. Compute f(d) for each:1:1, 2:-1,3:-1,4:1,6:1,9:1,12:-1,18:-1,36:1. Sum:1 -1 -1 +1 +1 +1 -1 -1 +1 = Let's group:(1 -1) + (-1 +1) + (1 +1) + (-1 -1) +1= 0 +0 +2 + (-2) +1= 0+0+0+1=1. Correct.So all these examples support the conclusion that F(n)=1 if and only if n is a square. Therefore, the answer is that F(n) is 1 exactly when n is a perfect square, and 0 otherwise. Therefore, F(n) is always 0 or 1, as required.But wait, let me check another non-square with multiple exponents. Let's take n=8=2^3. Divisors:1,2,4,8. f(1)=1, f(2)=-1, f(4)=1, f(8)=(-1)^3=-1. Sum:1 -1 +1 -1=0. Correct, since 8 is not a square.Another example: n=100=2^2*5^2. F(n)=1*1=1. Divisors:1,2,4,5,10,20,25,50,100. Compute f(d):1:1, 2:-1,4:1,5:-1,10:1,20:-1,25:1,50:-1,100:1. Sum:1 -1 +1 -1 +1 -1 +1 -1 +1 = (1 -1) + (1 -1) + (1 -1) + (1 -1) +1= 0 +0 +0 +0 +1=1. Correct.So all the examples work out, so my reasoning seems to hold. Therefore, the conclusion is that F(n) is multiplicative, and since F(p^k) is 1 if k is even and 0 if k is odd, then F(n) is 1 if and only if all exponents in the prime factorization of n are even, i.e., n is a perfect square, and 0 otherwise. Therefore, F(n) is always 0 or 1, and it is 1 exactly when n is a perfect square.But let me make sure there isn't a mistake in the multiplicative part. Suppose n and m are coprime, then F(nm) = F(n)F(m). If either F(n) or F(m) is 0, then F(nm) is 0. So to have F(nm)=1, both F(n) and F(m) need to be 1. Therefore, n and m must both be squares. Therefore, the product of squares is a square. So yes, if n is a square, then all exponents in its prime factorization are even, so each F(p^k) is 1, hence product is 1. If n is not a square, at least one exponent is odd, so F(n)=0. Therefore, the reasoning is correct.Therefore, the answer is that F(n) is 1 if and only if n is a perfect square, and 0 otherwise. Hence, F(n) is always 0 or 1.Final AnswerThe function ( F(n) ) equals 1 if and only if ( n ) is a perfect square. Therefore, ( F(n) = 1 ) for all square numbers and ( F(n) = 0 ) otherwise. The final answer is (boxed{F(n) = 1 text{ if and only if } n text{ is a perfect square}}).Final AnswerThe integers ( n ) for which ( F(n) = 1 ) are exactly the perfect squares. Thus, the solution is (boxed{n text{ is a perfect square}}).

✅:To solve the problem, we need to analyze the function ( F(n) = sum_{d|n} f(d) ), where ( f(n) = (-1)^{Omega(n)} ) and ( Omega(n) ) is the number of prime factors of ( n ) counting multiplicities. We aim to show that ( F(n) ) is either 0 or 1 for all positive integers ( n ) and determine for which ( n ) it is 1.1. Understanding ( f(n) ): - ( f(n) = (-1)^{Omega(n)} ). - For a prime power ( p^k ), ( Omega(p^k) = k ), so ( f(p^k) = (-1)^k ).2. Prime Power Case: - For ( n = p^k ), divisors are ( 1, p, p^2, ldots, p^k ). - Summing ( f ) over divisors: ( F(p^k) = sum_{m=0}^{k} (-1)^m ). - This is a geometric series: ( F(p^k) = frac{1 - (-1)^{k+1}}{2} ). - Result: ( F(p^k) = 1 ) if ( k ) is even, ( 0 ) if ( k ) is odd.3. Multiplicativity of ( F(n) ): - For coprime ( n ) and ( m ), ( F(nm) = F(n)F(m) ). - This follows because ( f(d_1d_2) = f(d_1)f(d_2) ) when ( d_1 ) and ( d_2 ) are coprime.4. General Case: - Factor ( n ) as ( n = p_1^{k_1} p_2^{k_2} cdots p_r^{k_r} ). - ( F(n) = prod_{i=1}^r F(p_i^{k_i}) ). - ( F(p_i^{k_i}) = 1 ) if ( k_i ) is even, ( 0 ) if ( k_i ) is odd. - Therefore, ( F(n) = 1 ) if all ( k_i ) are even (i.e., ( n ) is a perfect square), otherwise ( F(n) = 0 ).5. Verification with Examples: - Examples such as ( n = 1, 4, 6, 9, 12, 16, 25, 36, 100 ) confirm the result. ConclusionThe function ( F(n) ) is 1 if and only if ( n ) is a perfect square, and 0 otherwise. This is because ( F(n) ) is multiplicative and each prime power component must have an even exponent for the product to be 1.[boxed{n text{ is a perfect square}}]

🔑:1. Define the function ( f(n) ): - For ( n = 1 ), ( f(1) = 1 ). - For ( n > 1 ), ( f(n) = (-1)^{Omega(n)} ), where ( Omega(n) ) is the number of prime factors of ( n ) (counted with multiplicity).2. Define the function ( F(n) ): - ( F(n) = sum_{d|n} f(d) ), where the sum is over all divisors ( d ) of ( n ).3. Show that ( f ) is multiplicative: - A function ( f ) is multiplicative if ( f(mn) = f(m)f(n) ) for all coprime ( m ) and ( n ). - Since ( Omega(mn) = Omega(m) + Omega(n) ) for coprime ( m ) and ( n ), we have: [ f(mn) = (-1)^{Omega(mn)} = (-1)^{Omega(m) + Omega(n)} = (-1)^{Omega(m)} cdot (-1)^{Omega(n)} = f(m)f(n) ] - Therefore, ( f ) is multiplicative.4. Show that ( F ) is multiplicative: - Since ( f ) is multiplicative, ( F ) is also multiplicative by properties of Dirichlet convolution.5. Evaluate ( F(p^k) ) for a prime ( p ) and integer ( k ): - Consider ( n = p^k ). The divisors of ( p^k ) are ( 1, p, p^2, ldots, p^k ). - We need to compute: [ F(p^k) = sum_{j=0}^k f(p^j) ] - Since ( f(p^j) = (-1)^j ), we have: [ F(p^k) = sum_{j=0}^k (-1)^j ]6. Sum the series ( sum_{j=0}^k (-1)^j ): - If ( k ) is even, say ( k = 2m ), then: [ sum_{j=0}^{2m} (-1)^j = 1 - 1 + 1 - 1 + cdots + 1 = 1 ] - If ( k ) is odd, say ( k = 2m + 1 ), then: [ sum_{j=0}^{2m+1} (-1)^j = 1 - 1 + 1 - 1 + cdots + 1 - 1 = 0 ]7. Conclusion for ( F(p^k) ): - ( F(p^k) = 1 ) if ( k ) is even. - ( F(p^k) = 0 ) if ( k ) is odd.8. Generalize to any ( n ): - Since ( F ) is multiplicative, for any ( n ) with prime factorization ( n = p_1^{e_1} p_2^{e_2} cdots p_m^{e_m} ): [ F(n) = F(p_1^{e_1}) F(p_2^{e_2}) cdots F(p_m^{e_m}) ] - ( F(p_i^{e_i}) = 1 ) if ( e_i ) is even, and ( 0 ) if ( e_i ) is odd. - Therefore, ( F(n) = 1 ) if and only if all ( e_i ) are even, i.e., ( n ) is a perfect square.The final answer is ( boxed{ F(n) = 1 } ) if and only if ( n ) is a perfect square.

🧠:Okay, let's tackle this problem. So, we have a sequence of positive integers where x₁ is a fixed six-digit number. For each n ≥ 1, x_{n+1} is a prime divisor of x_n + 1. We need to find x₁₉ + x₂₀. Hmm, interesting. Let me try to figure out how this sequence behaves.First, let's parse the problem statement again to make sure I understand it correctly. The starting term x₁ is a six-digit positive integer, which means it's between 100,000 and 999,999 inclusive. Then, each subsequent term is a prime divisor of the previous term plus one. So, x_{n+1} is a prime factor of x_n + 1. Since primes are involved, the sequence could involve some interesting number theory concepts.Wait, the problem says "a prime divisor" but doesn't specify which one. So, does that mean we can choose any prime divisor each time, or is there a specific rule? The problem statement doesn't specify, so I guess we have to consider that the sequence is defined for any choice of prime divisors, and the answer should be the same regardless of the choices made. Otherwise, the problem would be ill-posed. So, perhaps regardless of the choices of prime divisors at each step, the sequence eventually stabilizes into some cycle or constant value, so that x₁₉ and x₂₀ are fixed. That must be the case here.Let me think. Let's consider how the sequence progresses. Starting with x₁, which is a six-digit number. Then x₂ is a prime divisor of x₁ + 1. Then x₃ is a prime divisor of x₂ + 1, and so on. Each term is a prime factor of the previous term plus one. Since each term is a prime, except possibly when the previous term plus one is a prime itself, then x_{n+1} would be that prime. Wait, no: x_{n+1} is a prime divisor of x_n + 1. Since x_{n+1} must be a prime, even if x_n + 1 is composite, we pick one of its prime factors. So, each term is a prime number, right? Because even if x_n + 1 is composite, we take a prime divisor of it. Therefore, starting from x₂ onwards, all terms are primes.But x₁ is a six-digit number, which is given as a positive integer. So x₁ could be prime or composite, but since it's six digits, it's more likely composite, but not necessarily. However, regardless of that, x₂ is a prime divisor of x₁ + 1. So even if x₁ is prime, x₁ + 1 is even (unless x₁ = 2, but x₁ is six-digit, so x₁ is at least 100,000, which is even). Therefore, x₁ + 1 is odd if x₁ is even, or even if x₁ is odd. Wait, but x₁ is a six-digit number. Let's check parity. If x₁ is even, then x₁ + 1 is odd. If x₁ is odd, x₁ + 1 is even. So x₂, being a prime divisor of x₁ + 1, would be 2 if x₁ is odd, because x₁ + 1 would be even, so 2 is a prime divisor. If x₁ is even, then x₁ + 1 is odd, so the prime divisors are odd primes.But since x₁ is a six-digit number, which can be either even or odd. However, the problem states that x₁ is a positive integer, but doesn't specify it's prime. So, x₁ could be either even or odd. Wait, but since it's a six-digit number, the parity of x₁ affects the next term. Let me think.Suppose x₁ is even. Then x₁ + 1 is odd, so x₂ is an odd prime divisor of x₁ + 1. Then x₃ is a prime divisor of x₂ + 1. Since x₂ is odd, x₂ + 1 is even, so 2 is a prime divisor. Therefore, x₃ could be 2. If x₃ is 2, then x₄ would be a prime divisor of 2 + 1 = 3. So x₄ is 3. Then x₅ is a prime divisor of 3 + 1 = 4, which is 2. Then x₆ is a prime divisor of 2 + 1 = 3. Then x₇ is 2 again, and so on. So we get a cycle between 2 and 3: 2 → 3 → 2 → 3 → ...Alternatively, if after x₃, instead of choosing 2, we could choose another prime divisor? Wait, but x₂ + 1 is even, so its prime divisors are 2 and possibly others if x₂ + 1 is a power of 2. But x₂ is an odd prime. So x₂ + 1 is even, so 2 is a prime divisor, but if x₂ + 1 is a power of 2, then the only prime divisor is 2. Otherwise, it could have other even prime divisors, but since 2 is the only even prime, all other prime divisors would be odd. But since x₂ + 1 is even, if it's not a power of 2, it would have 2 and another odd prime. However, x_{n+1} is defined as "a prime divisor", so we can choose any. But if we choose a different prime divisor, say an odd one, then the next term would be that odd prime. But then x₃ is odd, so x₄ is a prime divisor of x₃ + 1, which is even, so x₄ would be 2 again, and then x₅ is 3, etc. So maybe regardless of the choices, after a few terms, we end up in the 2-3 cycle.Similarly, if x₁ is odd, then x₁ + 1 is even, so x₂ must be 2 (since 2 is the only even prime). Then x₃ is 3, as above, and then we cycle between 2 and 3. Therefore, regardless of whether x₁ is even or odd, the sequence might eventually reach 2, then 3, then 2, 3, etc., creating a cycle. Therefore, if this is the case, then x₁₉ and x₂₀ would be 2 and 3 in some order, depending on where the cycle starts.But wait, let's verify this step by step. Let's take an example. Suppose x₁ is even. Let's pick x₁ = 100,000 (which is even). Then x₂ is a prime divisor of 100,001. Let's compute 100,001. Hmm, 100,001 divided by 11: 11*9090.999...? Wait, maybe 100,001 is prime? Probably not. Let's check. 100,001: Let's see, 100,001 divided by 101: 101*990.099... Not sure. Alternatively, 100,001 is 10^5 + 1. 10^5 +1 factors as (10 + 1)(10^4 - 10^3 + 10^2 -10 +1) = 11*9091. So 100,001 = 11 * 9091. Are both 11 and 9091 primes? 11 is prime. 9091: Let's check if it's prime. Let's try dividing by small primes: 9091 ÷ 13 = 699.307... Not an integer. 9091 ÷ 7 = 1298.714... Not integer. 9091 ÷ 17 = 534.764... Hmm. Let me check up to sqrt(9091) ≈ 95.35. So primes up to 97. 9091 ÷ 19 = 478.473... Not integer. 23: 9091 ÷23 ≈ 395.26... 29: 9091 ÷29 ≈313.48... 31: 9091 ÷31 ≈293.258... 37: 9091 ÷37 ≈245.7... 41: 9091 ÷41 ≈221.73... 43: 9091 ÷43 ≈211.418... 47: 9091 ÷47 ≈193.425... 53: 9091 ÷53 ≈171.52... 59: 9091 ÷59 ≈154.08... 61: 9091 ÷61 ≈149.03... 67: 9091 ÷67 ≈135.68... 71: 9091 ÷71 ≈128.0... Let's check 71*128 = 9088, which is 3 less than 9091, so not. 73: 9091 ÷73 ≈124.53... 79: 9091 ÷79 ≈115.07... 83: 9091 ÷83 ≈109.53... 89: 9091 ÷89 ≈102.14... 97: 9091 ÷97 ≈93.72... So 9091 seems to be prime. Therefore, 100,001 factors into 11 * 9091. So x₂ can be either 11 or 9091. If we choose 11, then x₃ is a prime divisor of 11 + 1 = 12. The prime divisors of 12 are 2 and 3. So x₃ can be 2 or 3. If we choose 2, then x₄ is a prime divisor of 3 (since 2 + 1 = 3), which is 3. Then x₅ is a prime divisor of 4, which is 2. Then x₆ is 3, x₇ is 2, etc. So cycle between 2 and 3. Alternatively, if we choose 3 for x₃, then x₄ is a prime divisor of 4, which is 2, and the cycle starts. So either way, after x₃, we get into 2 and 3 cycle.Alternatively, if x₂ is 9091, then x₃ is a prime divisor of 9091 + 1 = 9092. 9092 factors into 4 * 2273. So prime divisors are 2 and 2273. If we take 2, then x₄ is a prime divisor of 3, which is 3, and then cycle. If we take 2273, which is a prime (assuming), then x₄ is a prime divisor of 2274. 2274 divided by 2 is 1137, which is 3 * 379. So prime divisors are 2, 3, 379. If we choose 2, then x₅ is 3, cycle. If we choose 3, x₅ is a prime divisor of 4, which is 2, cycle. If we choose 379, then x₅ is a prime divisor of 380, which is 2^2 * 5 * 19. So prime divisors are 2, 5, 19. Choosing 2 leads to cycle, 5 would lead to x₆ being a prime divisor of 6 (2 or 3), cycle again. 19 would lead to x₆ being a prime divisor of 20 (2 or 5). Again, choosing 2 leads to cycle. So regardless of the path, eventually, we end up in the 2-3 cycle.Therefore, it seems that no matter the starting six-digit number, the sequence eventually reaches the cycle 2, 3, 2, 3, etc. So, since we have to compute x₁₉ + x₂₀, which are terms relatively far in the sequence, we can expect that by the 19th and 20th terms, the sequence has already entered this cycle. Therefore, x₁₉ and x₂₀ would be 2 and 3 in some order, so their sum would be 5.But to confirm this, let's check with another example. Let's take x₁ as an odd six-digit number. Let's say x₁ = 100,001. Then x₂ is a prime divisor of 100,002. Since 100,002 is even, x₂ can be 2 or another prime divisor. Let's factor 100,002: 100,002 = 2 * 50,001. 50,001 divided by 3: 50,001 / 3 = 16,667. So 50,001 = 3 * 16,667. 16,667: check if prime. Let's divide by small primes: 16,667 ÷ 7 = 2381, which is 7*2381. Is 2381 prime? Let's check. 2381 ÷ 13 ≈183.15, 2381 ÷ 17 ≈140.05, not integer. Up to sqrt(2381) ≈ 48.8. Check primes up to 47. 2381 ÷ 19 ≈125.315, 23: 2381 ÷23 ≈103.52, 29: 2381 ÷29 ≈82.1, 31: ≈76.8, 37: ≈64.35, 41: ≈58.07, 43: ≈55.37, 47: ≈50.65. So 2381 is prime. Therefore, 100,002 = 2 * 3 * 17 * 19609? Wait, no, 50,001 was 3 * 16,667. Wait, 16,667 divided by 17: 16,667 ÷17 ≈980.411... Not integer. Let's check if 16,667 is prime. Trying dividing by 7: 16,667 ÷7 ≈2381, which we already saw. So 16,667 seems prime. Wait, but 16,667 is 17*980.411, so not integer. Let me check 16,667 ÷ 13 = 1282.07... Not integer. So 16,667 is prime. Therefore, 100,002 factors into 2 * 3 * 16,667. So x₂ can be 2, 3, or 16,667. If we choose 2, then x₃ is prime divisor of 3, which is 3, then x₄ is 2, and so on. If we choose 3, then x₃ is prime divisor of 4, which is 2, then cycle. If we choose 16,667, then x₃ is prime divisor of 16,667 + 1 = 16,668. 16,668 is even, so 2 is a divisor. 16,668 /2 = 8,334, which is 2*4,167. 4,167 ÷3 = 1,389. So 4,167 = 3*1,389. 1,389 ÷3 = 463. So 16,668 factors into 2^2 * 3^2 * 463. So prime factors are 2, 3, 463. If we choose 2, x₄ is 3, cycle. If we choose 3, x₄ is 2, cycle. If we choose 463, then x₄ is a prime divisor of 464, which is 16*29, so prime factors 2 and 29. Choosing 2 leads to cycle, 29 leads to x₅ being a prime divisor of 30 (2, 3, 5). Then choosing any of these leads to cycle. So again, regardless of the path, we end up in the 2-3 cycle.Therefore, it's looking like regardless of the starting six-digit number x₁, the sequence x_n eventually enters the cycle between 2 and 3. Therefore, after some terms, all subsequent terms alternate between 2 and 3. So, even if the starting number is different, the sequence will eventually fall into this cycle.Now, the question is: how many steps does it take for the sequence to reach this cycle? If it's a six-digit number, even if we take the longest possible path, would it take more than 18 steps to reach the cycle? Because we need x₁₉ and x₂₀. Let's see. Let's consider the worst-case scenario where each time we choose the largest possible prime divisor, trying to delay entering the cycle.But even then, how long can it take? Let's think recursively. Suppose we have a term p (a prime), then the next term is a prime divisor of p + 1. If p is large, p + 1 could factor into smaller primes, but unless p + 1 is a power of 2, which would give us another prime. Wait, but primes are odd (except 2), so p is odd, so p + 1 is even, so 2 is always a divisor. So, if we choose 2 each time, then the next term is 2, followed by 3, then 2, etc. But if we choose another prime divisor, which is odd, then next term is odd, then the term after that would be 2, since odd + 1 is even.Wait, if we have a prime p, then p + 1 is even, so 2 is a divisor. Therefore, the next term can be 2 or another prime divisor. But if we choose another prime divisor q (odd), then the term after that would be a prime divisor of q + 1, which is even, so again 2 is a divisor. Therefore, unless q + 1 is a power of 2, which would make the next term 2. If q + 1 is a power of 2, then q is a Mersenne prime. For example, if q = 3, then 3 + 1 = 4 = 2^2, so next term is 2. If q = 7, 7 + 1 = 8 = 2^3, so next term is 2. So, primes like 3, 7, 31, etc., which are one less than a power of 2. But these are rare.Therefore, if we choose a prime q such that q + 1 is a power of 2, then next term is 2. Otherwise, we can choose another prime divisor, but it's again even, so we would have to choose 2 or another prime. But even if we choose another prime, the next term will be 2. So perhaps the maximum number of steps between a prime and 2 is two steps: prime q, then some other prime divisor (maybe another odd prime), but then the next term would be 2. However, this seems complicated.But considering the problem is asking for x₁₉ and x₂₀, which are the 19th and 20th terms. Given that starting from a six-digit number, which is enormous, but each subsequent term is a prime divisor of the previous term plus one. Each time, the number can decrease or increase. However, primes can be large, but in practice, because we are taking prime divisors, which are factors, so the terms can't be larger than the previous term plus one. But actually, if x_n is a prime, then x_{n+1} is a prime divisor of x_n + 1, which could be as big as x_n + 1 itself if x_n + 1 is prime. But x_n + 1 is even (since x_n is a prime, except when x_n = 2). Wait, if x_n is 2, then x_{n+1} is a prime divisor of 3, which is 3. So, if we have 2, next term is 3. Then 3 + 1 = 4, prime divisor is 2. So cycle.But if x_n is a large prime, say p, then p + 1 is even, so x_{n+1} could be 2, or another prime divisor. But if p + 1 is not a power of 2, then it will have other prime divisors. For example, if p = 7, then p + 1 = 8, which is 2^3, so only prime divisor is 2. If p = 5, then p + 1 = 6, which factors into 2 and 3. So, in that case, we can choose 2 or 3.But the key point is that unless p + 1 is a power of 2, we have a choice of primes. But even if we take the larger prime divisor, how does that affect the sequence? Let's take an example. Suppose we have p = 11. Then p + 1 = 12, which factors into 2, 3. If we take 3 as the next term, then 3 + 1 = 4, so next term is 2. Then cycle. If we take 2, next term is 3, cycle. So either way, two steps. Similarly, if p = 17, p + 1 = 18, factors into 2, 3. Choosing 3 leads to 4, then 2. Choosing 2 leads to 3. So again, two steps.But for larger primes. Let's take a prime like 127. Then 127 + 1 = 128 = 2^7. So the only prime divisor is 2. So next term is 2, then 3, cycle. Similarly, if p is a Mersenne prime, like 8191 (which is 2^13 - 1), then p + 1 = 8192 = 2^13, so next term is 2. So regardless, primes of the form 2^k - 1 will lead directly to 2. For other primes, p + 1 will have 2 as a prime factor and possibly others. But if we choose to take the non-2 prime factor, then next term is that prime, which is odd, then the term after is 2, and then cycle. So in the worst case, it takes two steps after encountering a prime to get back to 2. Unless the prime is 2 itself.But starting from a six-digit number, even if we take the worst path, how many steps would it take to get down to 2 or 3? Let's see. Let's simulate a possible worst-case scenario.Start with x₁ = 999,983 (a six-digit prime). Then x₂ is a prime divisor of 999,984. 999,984 is even, so factors: 2 * 499,992. 499,992 is also even, so 2^2 * 249,996, and so on. 999,984 = 2^4 * 62499. 62499 divided by 3: 62499 /3 = 20,833. 20,833: check if prime. Let's divide by 3: 20,833 ÷3 ≈6,944.333. Not divisible. 5: ends with 3, no. 7: 20,833 ÷7 ≈2,976.14. 11: 20,833 ÷11 ≈1,893. So not. Let's check 13: 13*1,603 = 20,839, which is over. So 13*1,602 = 20,826, remainder 7. So not. 17: 17*1,225 = 20,825, remainder 8. Not. 19: 19*1,096 = 20,824, remainder 9. Not. So 20,833 could be prime. Therefore, 999,984 factors into 2^4 *3 *20,833. So x₂ could be 2, 3, or 20,833. If we choose 20,833 (assuming it's prime), then x₃ is a prime divisor of 20,834. 20,834 is even, so factors: 2 *10,417. 10,417: check if prime. Divided by 3: 10,417 ÷3 ≈3,472.333. 5: ends with 7. 7: 10,417 ÷7 ≈1,488.14. 11: 10,417 ÷11 ≈947. So not. Maybe prime. So x₃ could be 2 or 10,417. Choosing 10,417 (if prime), then x₄ is a prime divisor of 10,418. 10,418 is even, so 2 *5,209. 5,209: check if prime. Divided by 3: 5,209 ÷3 ≈1,736.333. 5: ends with 9. 7: 5,209 ÷7 ≈744.14. 11: 5,209 ÷11 ≈473.545. 13: 5,209 ÷13 ≈400.69. 17: 5,209 ÷17 ≈306.41. 19: 5,209 ÷19 ≈274.15. 23: 5,209 ÷23 ≈226.48. 29: 5,209 ÷29 ≈179.62. 31: 5,209 ÷31 ≈168.03. 37: 5,209 ÷37 ≈140.78. So, 5,209 could be prime. Then x₄ could be 2 or 5,209. Choosing 5,209, then x₅ is a prime divisor of 5,210. 5,210 is even, so 2 *2,605. 2,605 is 5 *521. 521 is a prime. Therefore, x₅ could be 2, 5, or 521. Choosing 521, then x₆ is a prime divisor of 522. 522 is even, so 2 *261. 261 is 3*87, which is 3*3*29. So prime factors are 2, 3, 29. Choosing 29, x₇ is a prime divisor of 30, which is 2, 3, 5. Choosing 5, x₈ is a prime divisor of 6, which is 2 or 3. Then cycle begins.But even in this path, we only took 8 steps to get to 5, then a couple more to get to cycle. But this is just an example. However, starting from a six-digit number, each time we might have to factor large numbers, but even if we take the largest possible prime each time, the sequence seems to reduce to smaller primes relatively quickly. Because each time, when you take a large prime p, p + 1 will have factors, and if we choose a large factor, that factor is still smaller than p. For example, if p is around 10^6, then p + 1 is about 10^6 +1, which factors into primes that are at most 10^6 +1, but in reality, the largest prime factor would be much smaller. So each time, the primes are decreasing or fluctuating, but tending to decrease.Therefore, even in the worst case, the number of steps required to get down to 2 or 3 is logarithmic in the size of the original number. Since the original number is six digits (up to ~10^6), log2(10^6) ≈20. So maybe in 20 steps, it would reduce down. Therefore, x₁₉ and x₂₀ would almost certainly be in the 2-3 cycle.Alternatively, maybe there's a mathematical theorem that ensures that such sequences always end up in the 2-3 cycle. I recall that similar problems involve sequences reducing to small primes and eventually cycles. For example, the Collatz conjecture, although different, involves similar iterative processes. However, in this case, since we are dividing by primes each time, the sequence is forced to decrease in some sense.Another angle: Let's suppose that once the sequence reaches 2, it goes 2 →3 →2 →3… So if we can show that the sequence must eventually reach 2, then after that, it cycles. How can we guarantee that it reaches 2? Because every time we have an odd prime p, p +1 is even, so 2 is a prime divisor. But the problem allows choosing any prime divisor. However, if we choose 2 at any point, then the cycle begins. If we never choose 2, is that possible? Let's see.Suppose we have a prime p, then p +1 is even, so 2 is a prime divisor. If we choose to take another prime divisor (if exists), which is odd. Let's say p is an odd prime, then p +1 is even, so 2 divides it. If p +1 is not a power of two, then there exists another odd prime divisor. If we keep choosing the odd prime divisor each time, can we avoid ever choosing 2? For example, let's say we have a sequence where each term is an odd prime, and each term is a prime divisor of the previous term +1. Is such an infinite sequence possible?Suppose we have p₁, p₂, p₃, ... where each p_{n+1} is an odd prime divisor of p_n +1. Let's see if such a sequence can exist infinitely. For example, starting with p₁=3. Then p₂ divides 4, but the only prime divisor is 2, which is even. So we have to choose 2, can't choose an odd prime. Next example: p₁=5. p₂ divides 6, prime divisors are 2, 3. If we choose 3, which is odd. Then p₃ divides 4, which is 2. So again, forced to choose 2. Another example: p₁=7. p₂ divides 8, which is 2^3. Only prime divisor is 2. p₁=11. p₂ divides 12. Primes are 2, 3. Choosing 3. Then p₃ divides 4. Only 2. So again, forced to 2. Another example: p₁=13. p₂ divides 14: 2,7. Choosing 7. p₃ divides 8: 2. Forced to 2. Similarly, p₁=17. p₂ divides 18: 2,3. Choosing 3. p₃ divides 4: 2. Still.Wait, what if we have a prime p such that p +1 has an odd prime divisor q, and q +1 has an odd prime divisor r, etc. But as we saw, in practice, each step after an odd prime leads to a number which is even, so 2 is a divisor. If we choose an odd divisor, then the next term is an odd prime q, then q +1 is even, so again 2 is a divisor. So unless q +1 is a power of 2, we can choose another odd prime. However, if q +1 is a power of 2, then q is a Mersenne prime, which are rare. For example, 3, 7, 31, 127, etc. So if we have a Mersenne prime, then q +1 is 2^k, so the only prime divisor is 2. Therefore, in that case, the next term is 2. So unless we can have an infinite sequence of Mersenne primes where each q +1 is a power of 2, but Mersenne primes are primes of the form 2^k -1. However, it's not known whether there are infinitely many Mersenne primes. But even if there were, to have such a sequence, each subsequent prime would have to be a Mersenne prime as well. For example, starting from a Mersenne prime p₁=2^{k₁} -1, then p₂=2^{k₂} -1 divides p₁ +1 =2^{k₁}. But p₂ must divide 2^{k₁}, so p₂=2. But 2^{k₂} -1=2 implies k₂=2, but 2^2 -1=3, which is not 2. Contradiction. Therefore, it's impossible. Therefore, the only way p +1 is a power of 2 is if p is a Mersenne prime. But even so, the next term would be 2, leading to the cycle. Therefore, it's impossible to have an infinite sequence of odd primes in this way. Hence, any such sequence must eventually reach 2, and then cycle between 2 and 3.Therefore, regardless of the starting six-digit number x₁, the sequence x_n must eventually reach 2, and then enter the 2-3 cycle. Therefore, by the 19th and 20th terms, the sequence would certainly be in this cycle, so x₁₉ and x₂₀ are 2 and 3 in some order, hence their sum is 5.But to be thorough, let's check how many steps it could take at maximum. Suppose we start with the largest six-digit number, which is 999,999. Let's try to simulate the sequence, making choices that could potentially delay entering the cycle as much as possible.x₁ = 999,999x₁ +1 = 1,000,000. Factors: 2^6 *5^6. Prime divisors: 2 and 5. If we choose 5, then x₂=5.x₂=5, x₂+1=6. Factors: 2,3. Choose 3. x₃=3.x₃=3, x₃+1=4. Factors: 2. x₄=2.x₄=2, x₄+1=3. x₅=3.x₅=3, x₅+1=4. x₆=2.... and so on, cycling between 2 and 3.So here, starting from 999,999, even if we choose the non-2 prime divisor (5), we get to 5, then 3, then 2, entering the cycle at x₄. So only 4 steps.Another example: x₁=999,997 (prime? Let's assume it's prime for the sake of example).x₁=999,997 (prime)x₁ +1=999,998. Factors: 2 *499,999. Assume 499,999 is prime.x₂=499,999x₂ +1=500,000. Factors: 2^5 *5^6. Prime divisors: 2,5. Choose 5.x₃=5x₃ +1=6. Factors:2,3. Choose 3.x₄=3x₄ +1=4. x₅=2Cycle begins.So here, starting from a prime, x₂ is a large prime, but then x₃ is 5, leading to cycle in 5 steps.Another example, x₁=997, which is a prime three-digit, but similar idea.x₁=997x₂ is a prime divisor of 998. 998=2*499. 499 is prime.Choose x₂=499x₂ +1=500. Factors:2^2 *5^3. Choose 5.x₃=5x₃ +1=6. Choose 3.x₄=3x₄ +1=4. x₅=2Cycle.So again, within a few steps, we reach the cycle.Therefore, regardless of the starting six-digit number, even if we try to choose the largest possible prime divisors at each step, the number of steps needed to reach the 2-3 cycle is relatively small, definitely less than 19 steps. Therefore, by the time we reach x₁₉ and x₂₀, the sequence is already in the cycle. Hence, x₁₉ and x₂₀ are 2 and 3, so their sum is 5.To confirm, let's think about the cycle. Once the sequence reaches 2, the next term is 3, then back to 2, and so on. So if a term is 2, the next term is 3, and vice versa. Therefore, once in the cycle, the terms alternate between 2 and 3. Therefore, if the 19th term is 2, the 20th is 3, and their sum is 5. If the 19th term is 3, the 20th is 2, sum is still 5. Therefore, regardless of the position in the cycle, the sum x₁₉ +x₂₀ is always 5.Thus, the answer should be 5.Final Answerboxed{5}

✅:Given a sequence of positive integers ((x_n)) where (x_1) is a fixed six-digit number and for each (n geq 1), (x_{n+1}) is a prime divisor of (x_n + 1), we need to find (x_{19} + x_{20}).1. Initial Analysis: - The sequence starts with a six-digit number (x_1). - Each subsequent term (x_{n+1}) is a prime divisor of (x_n + 1). - The sequence will eventually enter a cycle due to the nature of prime divisors and the iterative process.2. Key Observations: - If (x_n) is even, (x_n + 1) is odd, and (x_{n+1}) could be an odd prime. - If (x_n) is odd, (x_n + 1) is even, so (x_{n+1}) could be 2. - Once the sequence reaches 2 or 3, it cycles between these values: (2 rightarrow 3 rightarrow 2 rightarrow 3 rightarrow ldots).3. Cycle Formation: - For any prime (p), (p + 1) is even, so 2 is always a divisor. - If the sequence reaches 2, the next term is 3, and vice versa, forming a cycle. - Regardless of the starting six-digit number, the sequence will eventually enter the 2-3 cycle.4. Steps to Cycle: - Even if starting from a large prime, the sequence reduces to smaller primes quickly. - Example: Starting with a large prime (x_1 = 999,983), the sequence reduces to 2 and 3 within a few steps. - The number of steps required to reach the cycle is logarithmic in the size of the starting number, ensuring that by the 19th and 20th terms, the sequence is in the cycle.5. Conclusion: - By the 19th and 20th terms, the sequence is in the cycle alternating between 2 and 3. - Therefore, (x_{19}) and (x_{20}) are 2 and 3 in some order, leading to their sum being (2 + 3 = 5).[boxed{5}]

🔑:1. Initial Setup and Sequence Definition: Let (x_n) be a sequence of positive integers where x_1 is a fixed six-digit number. For any n geq 1, x_{n+1} is a prime divisor of x_n + 1.2. Bounding the Sequence: Since x_1 is a six-digit number, we have x_1 leq 999999. Therefore, x_1 + 1 leq 1000000. [ x_2 text{ is a prime divisor of } x_1 + 1 implies x_2 leq sqrt{1000000} = 1000 ] Hence, x_2 < 1000.3. Continuing the Bounding Process: [ x_2 + 1 leq 1000 implies x_3 text{ is a prime divisor of } x_2 + 1 implies x_3 leq sqrt{1000} approx 31.62 ] Therefore, x_3 < 32.4. Further Bounding: [ x_3 + 1 leq 32 implies x_4 text{ is a prime divisor of } x_3 + 1 implies x_4 leq sqrt{32} approx 5.66 ] Hence, x_4 < 6.5. Possible Values for x_4: Since x_4 is a prime number less than 6, the possible values for x_4 are 2, 3, or 5.6. Analyzing Each Case: - Case 1: x_4 = 2 [ x_4 + 1 = 3 implies x_5 = 3 implies x_5 + 1 = 4 implies x_6 = 2 ] The sequence alternates between 2 and 3. - Case 2: x_4 = 3 [ x_4 + 1 = 4 implies x_5 = 2 implies x_5 + 1 = 3 implies x_6 = 3 ] The sequence alternates between 3 and 2. - Case 3: x_4 = 5 [ x_4 + 1 = 6 implies x_5 in {2, 3} ] If x_5 = 2, the sequence alternates between 2 and 3. If x_5 = 3, the sequence alternates between 3 and 2.7. Conclusion: In all cases, the sequence alternates between 2 and 3 or 3 and 2 starting from x_5. Therefore, (x_{19}, x_{20}) will be either (2, 3) or (3, 2).[x_{19} + x_{20} = 2 + 3 = boxed{5}]

🧠:Okay, let's try to figure out the smallest rational number r/s such that whenever 1/k + 1/m + 1/n is less than 1, it's also less than or equal to r/s. So, essentially, we need to find the supremum (the least upper bound) of all such sums that are still less than 1. Then, the smallest r/s would be this supremum. But since the problem mentions rational number, we have to make sure that this supremum is actually achievable by some k, m, n, and that it's a rational number. Hmm.First, maybe I should consider that if the sum is less than 1, then the supremum would be the maximum possible sum that is still less than 1. But since k, m, n are positive integers, the sum can approach 1 but never actually reach it. So, the supremum would be 1, but since we need the smallest rational number that is an upper bound for all such sums less than 1, maybe it's 1 itself. But wait, the question says "whenever 1/k + 1/m + 1/n < 1, then it's ≤ r/s". So, if we take r/s = 1, then certainly all such sums are less than 1 and hence ≤ 1. But is there a smaller rational number that still bounds all these sums?Wait, but if there are sums approaching 1 arbitrarily close, then any number less than 1 would not be an upper bound. For example, suppose we can make 1/k + 1/m + 1/n as close to 1 as we want. Then, the supremum would be 1, and since 1 is rational, the answer would be 1. But the problem states that the sum is strictly less than 1. So, the question is whether the supremum is actually 1, and if we can express 1 as r/s (which is 1/1), but maybe there is a smaller r/s such that all sums less than 1 are bounded above by r/s. Wait, but if the supremum is 1, then there's no smaller upper bound than 1. So perhaps 1 is the answer. But that seems too straightforward, and the problem is presented as a non-trivial problem. Maybe I need to check if there's a maximum sum less than 1, which would be the largest possible sum that is still less than 1, and that would be the r/s we need.But how can we have a maximum sum less than 1? If we can approach 1 as close as we like, then there's no maximum. But maybe in the case of Egyptian fractions (sums of distinct reciprocals), there are limitations, but here the problem allows any positive integers, not necessarily distinct. Wait, the problem doesn't specify that k, m, n have to be distinct. So maybe even repeating numbers. For example, taking k = m = n = 4. Then the sum is 3/4 = 0.75. If we take k = m = n = 3, the sum is 1, which is not allowed. So, if we take k = m = n = 4, sum is 3/4. If we take k = 2, m = 3, n = 7, the sum is 1/2 + 1/3 + 1/7 ≈ 0.5 + 0.333 + 0.142 ≈ 0.975, which is less than 1. So, that's a pretty high sum.Wait, but can we get even closer to 1? Let's see. For example, let's try k = 2, m = 3, n = 8. Then sum is 1/2 + 1/3 + 1/8 ≈ 0.5 + 0.333 + 0.125 ≈ 0.958. That's less than before. Hmm. What if k = 2, m = 3, n = 7, as above, gives ≈ 0.975. Is there a combination where the sum is even higher but still less than 1?Wait, let's check k=2, m=3, n=7: sum is 1/2 + 1/3 + 1/7. Let's compute exactly: (21 + 14 + 6)/42 = 41/42 ≈ 0.97619. So, 41/42 is about 0.976. If we take k=2, m=3, n=8, the sum is 1/2 + 1/3 + 1/8 = 12/24 + 8/24 + 3/24 = 23/24 ≈ 0.9583. That's actually lower. Wait, so maybe the 41/42 is higher. What if we take k=2, m=4, n=5: 1/2 + 1/4 + 1/5 = 10/20 + 5/20 + 4/20 = 19/20 = 0.95. Less than 41/42.Wait, maybe k=2, m=3, n=7 is the one that gives the highest sum less than 1. Let's check other possibilities. What about k=2, m=3, n=6: sum is 1/2 + 1/3 + 1/6 = 1, which is not allowed. So, that's exactly 1. So, n cannot be 6 here. So, if n=7, sum is 41/42. What if we try k=2, m=4, n=5: 19/20. No, lower. How about k=2, m=3, n=7: 41/42 ≈ 0.976. Let me see if there's a combination that gives a higher sum.Wait, let's try k=2, m=3, n=7. 1/2 + 1/3 + 1/7. Let's see, perhaps changing one of the denominators. For example, k=2, m=3, n=7: 41/42. If we decrease n from 7 to 6, we get 1/2 + 1/3 + 1/6 = 1, which is invalid. If we take n=8, as before, sum is 23/24 ≈ 0.958. So, n=7 is the closest. What if we fix k=2 and m=3, then n=7 is the smallest possible n that makes the sum less than 1. So, 41/42. What if we take k=2, m=4, n=5? 1/2 + 1/4 + 1/5 = 10/20 + 5/20 + 4/20 = 19/20 = 0.95. Lower.Alternatively, k=2, m=3, n=7: 41/42. Let me see another combination. How about k=2, m=3, n=43. Then the sum would be 1/2 + 1/3 + 1/43 ≈ 0.5 + 0.333 + 0.023 ≈ 0.856. That's way lower. So, no. So, it's better to have smaller denominators. But when we take k=2, m=3, n=7, that gives a high sum. Let's check k=3, m=3, n=4: 1/3 + 1/3 + 1/4 = 2/3 + 1/4 = 11/12 ≈ 0.916. Less than 41/42.Another combination: k=2, m=4, n=4: 1/2 + 1/4 + 1/4 = 1. Not allowed. So, if we take k=2, m=4, n=5: 19/20 as above. So, lower.Wait, maybe if we have two variables as 2 and 3, and then the third as 7. Let's check. Suppose we take k=2, m=3, n=8: sum is 1/2 + 1/3 + 1/8 = (12 + 8 + 3)/24 = 23/24 ≈ 0.958. Hmm. Less than 41/42.Alternatively, k=2, m=3, n=7: 41/42. Let's see if there's another triplet where the sum is higher than 41/42 but still less than 1. Let me think. Maybe k=2, m=3, n= something else. Let's try n=6, which gives exactly 1. So, invalid. n=7 gives 41/42. If we try k=2, m=4, n=5: 19/20. 19/20 is 0.95, which is less than 41/42 ≈ 0.976. So, 41/42 is higher.Wait, let's check other combinations. How about k=2, m=3, n=7. 41/42. If we take k=2, m=3, n=43, as before, sum is lower. So, perhaps 41/42 is the highest possible. Is there another combination?Wait, let's try k=3, m=3, n=4: 1/3 + 1/3 + 1/4 = 2/3 + 1/4 = 8/12 + 3/12 = 11/12 ≈ 0.916. Less than 41/42.How about k=2, m=5, n=5: 1/2 + 1/5 + 1/5 = 1/2 + 2/5 = 9/10 = 0.9. Still lower.Wait, maybe k=3, m=3, n=3: sum is 1, which is invalid. So, k=3, m=3, n=4 gives 11/12. Lower.Wait, maybe k=2, m=4, n=5: 19/20. 0.95. Still lower.Alternatively, k=2, m=3, n=7. Let me check another. For example, k=2, m=3, n=7: 41/42. Let's see if there's a way to get a sum higher than 41/42 without reaching 1. Let's suppose we take k=2, m=3, n=7. 41/42. What if we take k=2, m=3, n=7. If we try to make one denominator smaller, like n=6, but that gives 1. So, n has to be at least 7. What if we make k=2, m=4, n=5: 19/20. Hmm, lower.Alternatively, k=2, m=3, n=7 gives 41/42. Let's check if there's a different combination where two denominators are 2 and 4. For example, k=2, m=4, n= something. Let's see: 1/2 + 1/4 + 1/n < 1. So, 3/4 + 1/n < 1. So, 1/n < 1/4. So, n > 4. So, the minimal n is 5. Then, sum is 1/2 + 1/4 + 1/5 = 10/20 + 5/20 + 4/20 = 19/20. Which is 0.95. Less than 41/42.Alternatively, if we take k=3, m=3, n=4: 1/3 + 1/3 + 1/4 = 11/12 ≈ 0.916. So, lower.How about k=2, m=3, n=7. 41/42. Let me check another combination where one of the denominators is 2, another is 3, and the third is larger than 7. For example, n=8. Then sum is 1/2 + 1/3 + 1/8 ≈ 0.5 + 0.333 + 0.125 = 0.958. Which is 23/24. Less than 41/42. So, lower.Alternatively, what if we take k=2, m=5, n= something. Let's see: 1/2 + 1/5 + 1/n <1. So, 7/10 + 1/n <1. So, 1/n < 3/10. So, n>10/3≈3.333. So, n≥4. Then the minimal n is 4. Then sum is 1/2 +1/5 +1/4 = 10/20 +4/20 +5/20 =19/20=0.95. Again, lower.Alternatively, if we take k=2, m=3, n=7. Let's see if that's the maximum. Let me see another approach. Let's suppose that to maximize 1/k + 1/m + 1/n without reaching 1, we need to minimize the denominators as much as possible. So, the largest reciprocal is 1/2, then 1/3, etc. So, starting with 1/2, then 1/3. Then, adding another reciprocal such that the total is less than 1. So, 1/2 + 1/3 = 5/6. Then, 1 - 5/6 = 1/6. So, the next reciprocal has to be less than 1/6, which would mean denominator at least 7. But if we take 1/7, the total is 5/6 + 1/7 = 35/42 + 6/42 = 41/42. Which is exactly what we had before. If instead of 1/7, we take a smaller denominator, like 1/6, then the sum would be 5/6 + 1/6 = 1, which is not allowed. So, the next possible is 1/7, which gives the total sum 41/42. So, that seems like the maximum possible sum less than 1. So, that would be the supremum. But is this actually the case? Are there other triplets that can give a sum higher than 41/42 but still less than 1?Let me check. Suppose we take k=2, m=4, n=5. Then, sum is 1/2 +1/4 +1/5 = 10/20 +5/20 +4/20 =19/20=0.95. That's less than 41/42≈0.976.Alternatively, k=3, m=3, n=4: 1/3 +1/3 +1/4=2/3 +1/4=11/12≈0.916, which is less.What if we try k=2, m=3, n=7: 41/42≈0.976. Let me check another triplet where denominators are 2,3,7. Is there another combination with a larger sum? Let's see. For example, k=2, m=3, n=7. If we change k=2 to k=1, but k must be a positive integer. If k=1, then 1/k is 1, so even if m and n are infinite, the sum would be at least 1, which is not allowed. So, k=1 is invalid. Similarly, m=1 or n=1 would make the sum too big. So, denominators must be at least 2. So, the next smallest denominators would be 2,3,7.Alternatively, let's consider k=2, m=3, n=7. If we can find another triplet where two denominators are 2 and 3, and the third is adjusted to get a sum just below 1. Since 1/2 + 1/3 = 5/6, which leaves 1/6 for the third term. But 1/6 is exactly 1/6, which would make the total 1. So, the third term must be less than 1/6, which requires the denominator to be greater than 6. The smallest integer greater than 6 is 7, so 1/7. Thus, 1/2 +1/3 +1/7=41/42≈0.976. If we take the next denominator as 8, then 1/8=0.125, but 1/2 +1/3 +1/8=23/24≈0.958, which is lower.Therefore, 41/42 seems to be the maximum possible sum less than 1. Thus, the smallest rational number r/s that satisfies the condition is 41/42.But wait, let's verify if there exists another combination where the sum is higher than 41/42 but still less than 1. Suppose we take k=3, m=4, n=4. Then, sum is 1/3 +1/4 +1/4=1/3 +1/2=5/6≈0.833. Lower. Not helpful.Another possibility: k=2, m=3, n=7: 41/42. If we take k=2, m=3, n=6, sum is 1, which is invalid. So, n=7 is the next.Wait, let's check k=2, m=4, n=5: 19/20=0.95. Less than 41/42.How about k=2, m=3, n=7. Suppose we take k=2, m=5, n=5: sum is 1/2 +2/5=9/10=0.9. No.Alternatively, k=2, m=3, n=8: 23/24≈0.958. Less than 41/42.Alternatively, k=3, m=3, n=3: sum is 1, invalid. k=3, m=3, n=4: 11/12≈0.916.Alternatively, k=2, m=2, n= anything: 1/2 +1/2 +1/n=1 +1/n, which is greater than 1, so invalid. So, k and m cannot both be 2. Similarly, k=2, m=2, n=3: 1/2 +1/2 +1/3=1 +1/3>1. Invalid.So, the next best is k=2, m=3, n=7. Let me check another possibility. Suppose k=3, m=4, n=5. Sum is 1/3 +1/4 +1/5= (20 +15 +12)/60=47/60≈0.783. Lower.Wait, another approach: Maybe if we don't take the two smallest denominators as 2 and 3, but another combination. For example, taking k=2, m=4, n=5. But that gives 19/20=0.95. Still lower than 41/42. Or k=2, m=5, n=5: 9/10. No.Alternatively, take k=3, m=3, n=4: 11/12≈0.916. Less.So, seems like 41/42 is indeed the maximum possible sum less than 1. Therefore, the smallest rational number r/s such that all sums 1/k +1/m +1/n <1 are ≤ r/s is 41/42. Because if there was a smaller rational number, say 40/41≈0.9756, which is less than 41/42≈0.97619, then 41/42 would exceed that, so it cannot be an upper bound. Therefore, the smallest upper bound is 41/42.But to confirm, let's see if there exists a sum that is exactly 41/42, which is less than 1, so it is included in the condition. Yes, for k=2, m=3, n=7, the sum is 41/42, which is less than 1. Therefore, 41/42 is achievable. So, any upper bound must be at least 41/42, and since 41/42 is achievable, the smallest such upper bound is 41/42 itself. Therefore, the answer is boxed{dfrac{41}{42}}.Final Answerboxed{dfrac{41}{42}}

✅:To determine the smallest rational number (frac{r}{s}) such that (frac{1}{k} + frac{1}{m} + frac{1}{n} leq frac{r}{s}) whenever (k, m,) and (n) are positive integers satisfying (frac{1}{k} + frac{1}{m} + frac{1}{n} < 1), we need to find the supremum of all such sums that are still less than 1. We start by considering the largest possible sum of reciprocals of three positive integers that is still less than 1. The key is to minimize the denominators (k, m,) and (n) as much as possible while ensuring the sum does not reach 1. 1. Case (k = 2), (m = 3), (n = 7): - The sum is (frac{1}{2} + frac{1}{3} + frac{1}{7}). - Calculating this exactly: (frac{1}{2} + frac{1}{3} + frac{1}{7} = frac{21}{42} + frac{14}{42} + frac{6}{42} = frac{41}{42}).2. Verification with other combinations: - Other combinations such as (k = 2, m = 3, n = 8) yield (frac{23}{24} approx 0.958), which is less than (frac{41}{42}). - Combinations like (k = 2, m = 4, n = 5) yield (frac{19}{20} = 0.95), which is also less than (frac{41}{42}). - Testing various other combinations confirms that (frac{41}{42}) is indeed the highest sum less than 1.3. Conclusion: - The combination (k = 2, m = 3, n = 7) gives the sum (frac{41}{42}), which is the largest possible sum less than 1. - Since (frac{41}{42}) is achievable and any smaller rational number would not serve as an upper bound for all such sums, the smallest rational number (frac{r}{s}) is (frac{41}{42}).Thus, the final answer is (boxed{dfrac{41}{42}}).

🔑:To determine the smallest rational number frac{r}{s} such that frac{1}{k} + frac{1}{m} + frac{1}{n} leq frac{r}{s} whenever k, m, and n are positive integers that satisfy the inequality frac{1}{k} + frac{1}{m} + frac{1}{n} < 1, we need to maximize frac{1}{k} + frac{1}{m} + frac{1}{n} under the given constraint.1. Assume k geq m geq n: - This assumption helps us systematically explore the possible values of k, m, and n.2. Consider the maximum value of k: - If k geq 4, then frac{1}{k} leq frac{1}{4}. - Therefore, frac{1}{k} + frac{1}{m} + frac{1}{n} leq frac{1}{4} + frac{1}{4} + frac{1}{4} = frac{3}{4} < 1. - Hence, k must be less than 4. So, k = 2 or k = 3.3. Case 1: k = 3: - If k = 3, then frac{1}{k} = frac{1}{3}. - To maximize frac{1}{m} + frac{1}{n}, we need to consider the largest possible values for m and n such that frac{1}{3} + frac{1}{m} + frac{1}{n} < 1. - If m geq 4, then frac{1}{m} leq frac{1}{4}. - Therefore, frac{1}{3} + frac{1}{4} + frac{1}{n} leq frac{1}{3} + frac{1}{4} + frac{1}{4} = frac{11}{12} < 1. - Hence, m must be less than 4. So, m = 3.4. Subcase: k = 3, m = 3: - If k = 3 and m = 3, then frac{1}{k} + frac{1}{m} = frac{1}{3} + frac{1}{3} = frac{2}{3}. - To satisfy frac{1}{3} + frac{1}{3} + frac{1}{n} < 1, we need frac{1}{n} < frac{1}{3}, so n > 3. - The smallest integer n greater than 3 is n = 4. - Therefore, frac{1}{3} + frac{1}{3} + frac{1}{4} = frac{11}{12}.5. Case 2: k = 2: - If k = 2, then frac{1}{k} = frac{1}{2}. - To maximize frac{1}{m} + frac{1}{n}, we need to consider the largest possible values for m and n such that frac{1}{2} + frac{1}{m} + frac{1}{n} < 1. - If m geq 5, then frac{1}{m} leq frac{1}{5}. - Therefore, frac{1}{2} + frac{1}{5} + frac{1}{n} leq frac{1}{2} + frac{1}{5} + frac{1}{5} = frac{4}{5} < 1. - Hence, m must be less than 5. So, m = 3 or m = 4.6. Subcase: k = 2, m = 3: - If k = 2 and m = 3, then frac{1}{k} + frac{1}{m} = frac{1}{2} + frac{1}{3} = frac{5}{6}. - To satisfy frac{1}{2} + frac{1}{3} + frac{1}{n} < 1, we need frac{1}{n} < frac{1}{6}, so n > 6. - The smallest integer n greater than 6 is n = 7. - Therefore, frac{1}{2} + frac{1}{3} + frac{1}{7} = frac{41}{42}.7. Subcase: k = 2, m = 4: - If k = 2 and m = 4, then frac{1}{k} + frac{1}{m} = frac{1}{2} + frac{1}{4} = frac{3}{4}. - To satisfy frac{1}{2} + frac{1}{4} + frac{1}{n} < 1, we need frac{1}{n} < frac{1}{4}, so n > 4. - The smallest integer n greater than 4 is n = 5. - Therefore, frac{1}{2} + frac{1}{4} + frac{1}{5} = frac{19}{20}.8. Conclusion: - The maximum value of frac{1}{k} + frac{1}{m} + frac{1}{n} under the given constraint is frac{41}{42}, which occurs when k = 2, m = 3, n = 7.The final answer is boxed{frac{41}{42}}.

🧠:Okay, so I have this equation to solve: find all real numbers x, y, z greater than 1 such thatx + y + z + 3/(x - 1) + 3/(y - 1) + 3/(z - 1) = 2(sqrt(x + 2) + sqrt(y + 2) + sqrt(z + 2)).Hmm, that looks a bit complicated, but maybe I can simplify it by noticing some symmetry. Since the equation is symmetric in x, y, and z, maybe all the variables are equal? Let's assume that x = y = z. That might simplify things.Let me set x = y = z = a, where a > 1. Then the equation becomes:3a + 3*(3/(a - 1)) = 2*3*sqrt(a + 2).Simplify both sides:3a + 9/(a - 1) = 6*sqrt(a + 2).Divide both sides by 3:a + 3/(a - 1) = 2*sqrt(a + 2).Okay, so now I just need to solve this equation for a > 1. Let me denote the left side as L(a) = a + 3/(a - 1) and the right side as R(a) = 2*sqrt(a + 2).I need to find a such that L(a) = R(a). Let's try to manipulate this equation. Maybe square both sides to eliminate the square root. But before that, let's see if I can rearrange terms.First, write the equation as:a + 3/(a - 1) - 2*sqrt(a + 2) = 0.Let me denote this as f(a) = a + 3/(a - 1) - 2*sqrt(a + 2). I need to find the roots of f(a) for a > 1.Alternatively, maybe substitute t = sqrt(a + 2). Then a = t² - 2. Let's try that substitution.So substitute a = t² - 2 into the equation:(t² - 2) + 3/(t² - 2 - 1) - 2t = 0.Simplify the denominator: t² - 3.So equation becomes:t² - 2 + 3/(t² - 3) - 2t = 0.Let me combine terms:t² - 2t - 2 + 3/(t² - 3) = 0.Hmm, this might still be complicated. Maybe multiply both sides by (t² - 3) to eliminate the denominator:(t² - 2t - 2)(t² - 3) + 3 = 0.Expand the first product:First, multiply t² by (t² - 3): t^4 - 3t².Then, multiply -2t by (t² - 3): -2t^3 + 6t.Then, multiply -2 by (t² - 3): -2t² + 6.So putting all together:t^4 - 3t² - 2t^3 + 6t - 2t² + 6 + 3 = 0.Combine like terms:t^4 - 2t^3 - 5t² + 6t + 9 = 0.Wait, let me check the coefficients again.t^4 (from t²*t²)-2t^3 (from -2t*t²)-3t² - 2t² = -5t²6t (from -2t*(-3) + 6t?)Wait, no. Wait, expanding term by term:(t²)(t² - 3) = t^4 - 3t²(-2t)(t² - 3) = -2t^3 + 6t(-2)(t² - 3) = -2t² + 6So adding all these terms:t^4 - 3t² - 2t^3 + 6t - 2t² + 6.Then +3 from the original equation.So total is:t^4 - 2t^3 - 5t² + 6t + 6 + 3 = t^4 - 2t^3 - 5t² + 6t + 9 = 0.So the quartic equation is t^4 - 2t^3 -5t² +6t +9 =0.Hmm, quartic equations can be tough, but maybe factorable.Let me try to factor this quartic. Let's try rational roots. Possible rational roots are ±1, ±3, ±9.Test t=1: 1 -2 -5 +6 +9 = 1 -2= -1; -1 -5= -6; -6 +6=0; 0 +9=9 ≠0t= -1: 1 +2 -5 -6 +9=1+2=3; 3-5=-2; -2-6=-8; -8+9=1≠0t=3: 81 -54 -45 +18 +9=81-54=27; 27-45=-18; -18+18=0; 0+9=9≠0t= -3: 81 +54 -45 -18 +9=81+54=135; 135-45=90; 90-18=72;72+9=81≠0t=9: too big, not likely. So none of the rational roots work.Maybe factor as quadratics. Suppose the quartic factors into (t² + at + b)(t² + ct + d)= t^4 + (a + c)t^3 + (ac + b + d)t² + (ad + bc)t + bd.Set equal to t^4 -2t^3 -5t² +6t +9.So equate coefficients:1. a + c = -22. ac + b + d = -53. ad + bc =64. bd =9So from equation 4: bd=9. Possible integer pairs (b,d): (3,3), (1,9), (9,1), (-3,-3), (-1,-9), (-9,-1).Try positive first. Let's try b=3, d=3.Then equation 3: a*3 + c*3=6 → 3(a + c)=6. But from equation 1, a + c= -2, so 3*(-2)= -6 ≠6. Not possible.Next, try b=1, d=9. Then equation 3: a*9 + c*1=6. From equation 1, a + c= -2. So we have:9a + c=6a + c= -2Subtract the second equation from the first: 8a=8 → a=1. Then c= -3.Check equation 2: a*c + b + d =1*(-3) +1 +9= -3 +10=7≠-5. Not good.Next, b=9, d=1. Then equation3: a*1 + c*9=6. From equation1, a +c= -2. So:a +9c=6a +c= -2Subtract: 8c=8 → c=1, then a= -3.Check equation2: a*c + b + d= (-3)(1) +9 +1= -3 +10=7≠-5. Nope.Try negative pairs. Let's try b=-3, d=-3. Then equation3: a*(-3) + c*(-3)=6 → -3(a + c)=6. From equation1, a + c= -2, so -3*(-2)=6. Which matches! So equation3 is satisfied. Now check equation2: a*c + b + d= a*c + (-3) + (-3)=a*c -6. From equation1, a +c= -2. So we need a*c -6= -5 → a*c=1. So we have a + c= -2 and a*c=1. That's a system of equations.Solutions: a and c are roots of t² +2t +1=0 → (t +1)^2=0 → t=-1. So a=c=-1. Thus, the quartic factors as (t² - t -3)(t² - t -3)= (t² - t -3)^2. Wait, but let's check:Wait, original factors would be (t² + at + b)(t² + ct + d)= (t² - t -3)(t² - t -3). Wait, since a=-1, c=-1, b=-3, d=-3.Multiply out (t² - t -3)^2 = t^4 - 2t^3 + t² +6t +9. Wait, no:Wait, (t² - t -3)(t² - t -3):First: t²*t² = t^4t²*(-t) + (-t)*t² = -t^3 -t^3 = -2t^3t²*(-3) + (-t)*(-t) + (-3)*t² = -3t² + t² -3t² = (-3 +1 -3)t²= -5t²Then (-t)*(-3) + (-3)*(-t) = 3t +3t=6tFinally, (-3)*(-3)=9.So indeed, (t² - t -3)^2= t^4 -2t^3 -5t² +6t +9. Perfect! So the quartic factors as (t² - t -3)^2=0.Therefore, the solutions are t² - t -3=0 → t=[1 ±sqrt(1 +12)]/2= [1 ±sqrt(13)]/2.But since t=sqrt(a +2) and a>1, then t>sqrt(1 +2)=sqrt(3)≈1.732. So t must be positive and greater than sqrt(3). So t=[1 +sqrt(13)]/2 ≈(1 +3.605)/2≈4.605/2≈2.3025, which is greater than sqrt(3). The other root is [1 -sqrt(13)]/2≈negative, which we can disregard.Therefore, t=[1 +sqrt(13)]/2, so a= t² -2.Compute a:First, t=[1 +sqrt(13)]/2. Then t²= [1 + 2sqrt(13) +13]/4= [14 +2sqrt(13)]/4= [7 +sqrt(13)]/2.Therefore, a= [7 +sqrt(13)]/2 -2= [7 +sqrt(13) -4]/2= [3 +sqrt(13)]/2.So the solution is x=y=z=[3 +sqrt(13)]/2.But wait, I need to check if this actually satisfies the original equation, in case any step introduced extraneous solutions.Let me verify.Compute left side: 3a + 9/(a -1).Compute a= [3 +sqrt(13)]/2. Then a -1= [3 +sqrt(13)]/2 -2/2= [1 +sqrt(13)]/2.Thus, 3/(a -1)=3/([1 +sqrt(13)]/2)=6/(1 +sqrt(13)). Rationalize denominator: multiply numerator and denominator by (sqrt(13)-1):6*(sqrt(13)-1)/[(1 +sqrt(13))(sqrt(13)-1)]=6*(sqrt(13)-1)/(13 -1)=6*(sqrt(13)-1)/12=(sqrt(13)-1)/2.Therefore, 3/(a -1)= (sqrt(13)-1)/2. So sum of three such terms is 3*(sqrt(13)-1)/2.So left side: 3a + 3*(sqrt(13)-1)/2.Compute 3a: 3*[3 +sqrt(13)]/2= [9 +3sqrt(13)]/2.Adding 3*(sqrt(13)-1)/2: [9 +3sqrt(13) +3sqrt(13) -3]/2= [6 +6sqrt(13)]/2=3 +3sqrt(13).Right side: 2*3*sqrt(a +2).Compute sqrt(a +2). a +2= [3 +sqrt(13)]/2 +4/2= [7 +sqrt(13)]/2. So sqrt(a +2)=sqrt([7 +sqrt(13)]/2).Wait, but earlier when we substituted t=sqrt(a +2), we had t=[1 +sqrt(13)]/2. Wait, let's check:Wait, t was sqrt(a +2). So sqrt(a +2)=t=[1 +sqrt(13)]/2. Therefore, sqrt(a +2)=[1 +sqrt(13)]/2. Therefore, right side is 6*[1 +sqrt(13)]/2=3*(1 +sqrt(13))=3 +3sqrt(13).Which matches the left side, which was also 3 +3sqrt(13). So the solution checks out.Therefore, the symmetric solution x=y=z=[3 +sqrt(13)]/2 is a solution.But the problem says "determine all real numbers x>1, y>1, z>1". The equation is symmetric, so maybe there are non-symmetric solutions? Hmm. But usually, in symmetric equations, the only solutions are symmetric, unless there's some specific reason. But how can I be sure?Let me consider whether there could be non-symmetric solutions. Suppose, for example, that two variables are equal and the third is different. Let's say x = y ≠ z. Then we can set x = y = a and z = b, and see if such solutions exist.So substituting into the original equation:2a + b + 3/(a -1) + 3/(a -1) + 3/(b -1) = 2(2*sqrt(a +2) + sqrt(b +2)).Which simplifies to:2a + b + 6/(a -1) + 3/(b -1) = 4*sqrt(a +2) + 2*sqrt(b +2).This seems more complicated. Let me see if I can assume a particular relation between a and b. Alternatively, maybe try a = [3 +sqrt(13)]/2 and see if b can be different. But since the equation would require b to satisfy a similar equation, but since a is already fixed, perhaps not.Alternatively, suppose that two variables are equal to [3 +sqrt(13)]/2 and the third is different. Let's see.But this might get too messy. Alternatively, maybe the only solution is the symmetric one. Let's see.Suppose that the function f(a) = a + 3/(a -1) - 2*sqrt(a +2) has only one root. Earlier, we found that f(a)=0 at a=[3 +sqrt(13)]/2. Let's check if this is the only root.Compute f(a) as a approaches 1 from the right. As a→1+, 3/(a -1)→+infty, so f(a)→+infty. At a= [3 +sqrt(13)]/2≈ (3 +3.605)/2≈3.3025, which is the solution. Now as a approaches infinity, f(a)≈a -2*sqrt(a). Which tends to infinity as a increases. What about in between? Let's compute f(2):f(2)=2 + 3/(1) -2*sqrt(4)=2 +3 -4=1. So f(2)=1>0.f(1.5): a=1.5, then f(1.5)=1.5 + 3/(0.5) -2*sqrt(3.5)=1.5 +6 -2*1.8708≈7.5 -3.7416≈3.7584>0.f(3):3 +3/(2) -2*sqrt(5)=3 +1.5 -2*2.236≈4.5 -4.472≈0.028>0.Wait, so f(3)≈0.028>0, very close to zero. Wait, but according to our previous solution, the root is at approximately 3.3025. So at a=3.3025, f(a)=0.But f(3) is already positive. Wait, maybe I miscalculated. Wait, 3.3025 is approximately (3 +3.605)/2≈3.3025. Let's compute f(a) at a=3.3025.Compute sqrt(a +2)=sqrt(5.3025)≈2.3025. Then 2*sqrt(a +2)=4.605.a +3/(a -1)=3.3025 +3/(2.3025)≈3.3025 +1.302≈4.6045≈4.605. So approximately equal. So f(a)=0.At a=3, sqrt(5)=2.236, 2*sqrt(5)=4.472. Then a +3/(a -1)=3 +3/2=4.5. 4.5 -4.472≈0.028>0.At a=4, sqrt(6)=2.449, 2*sqrt(6)=4.898. Then a +3/(a-1)=4 +3/3=5. 5 -4.898≈0.102>0.So f(a) is positive at a=3 and increasing as a increases? Wait, but as a increases, a -2*sqrt(a +2) grows how? Let's see.For large a, f(a)≈a -2*sqrt(a). The leading term is a, which grows faster than sqrt(a), so f(a)→infty as a→infty.But between a= [solution]≈3.3025 and higher a, f(a) is positive. From a=1 to a=3.3025, f(a) decreases from +infty to 0, but wait, at a=2, f(a)=1, at a=1.5, f(a)=3.7584. So f(a) is decreasing from a=1 to a=3.3025? Wait, but when a increases from 1 to 3.3025, 3/(a -1) decreases, but a increases. Maybe the function first decreases and then increases?Wait, take the derivative of f(a):f(a)=a +3/(a -1) -2*sqrt(a +2).f’(a)=1 -3/(a -1)^2 -2*(1/(2*sqrt(a +2)))=1 -3/(a -1)^2 -1/sqrt(a +2).It's complicated, but maybe at the solution point, the derivative is zero or not. Not sure. But since f(a) approaches infinity as a approaches 1 from the right, and tends to infinity as a approaches infinity, and we have only one crossing point where f(a)=0, then likely there is only one solution. Hence, the equation f(a)=0 has only one real solution at a=[3 +sqrt(13)]/2.Therefore, the only solution is x=y=z=[3 +sqrt(13)]/2. Hence, the answer is x=y=z=(3 +√13)/2.But wait, before finalizing, let me check if there's another approach. Maybe using inequalities. The equation has terms like x + 3/(x -1) and 2*sqrt(x +2). Maybe relate these terms using AM-GM inequality or Cauchy-Schwarz.Consider the left side terms x + 3/(x -1). Let me set t = x -1, so t >0. Then x = t +1. Substitute into the expression:(t +1) + 3/t = t +1 +3/t.Similarly, the right side term 2*sqrt(x +2)=2*sqrt(t +3).So we have for each variable, the equation:t +1 +3/t = 2*sqrt(t +3).But since the variables are separated, each term for x, y, z must satisfy this equation individually. Therefore, each variable's term must satisfy t +1 +3/t =2*sqrt(t +3). So we can consider this equation in t>0.Hence, if we set u = sqrt(t +3), then u>=sqrt(3). Then t= u² -3. Substitute into the equation:(u² -3) +1 +3/(u² -3) =2u.Simplify:u² -2 +3/(u² -3) =2u.Multiply both sides by (u² -3):(u² -2)(u² -3) +3 =2u(u² -3).Expand left side:u^4 -3u² -2u² +6 +3 =u^4 -5u² +9.Right side:2u^3 -6u.Bring all terms to left:u^4 -5u² +9 -2u^3 +6u=0.Which is u^4 -2u^3 -5u² +6u +9=0. Which is the same quartic as before. Therefore, the same solution u=[1 +sqrt(13)]/2.Therefore, t= u² -3= ([1 +sqrt(13)]/2)^2 -3= [1 +2sqrt(13)+13]/4 -3= [14 +2sqrt(13)]/4 -12/4= [2 +2sqrt(13)]/4= [1 +sqrt(13)]/2.But wait, t= x -1= [1 +sqrt(13)]/2. Therefore, x= t +1= [1 +sqrt(13)]/2 +2/2= [3 +sqrt(13)]/2. Same as before.Therefore, each variable must satisfy this individually. Therefore, the only solution is when each variable is equal to [3 +sqrt(13)]/2.Hence, the only solution is x=y=z=(3 +√13)/2.Final AnswerThe real numbers are all equal to boxed{dfrac{3 + sqrt{13}}{2}}.

✅:Given the equation ( x + y + z + frac{3}{x-1} + frac{3}{y-1} + frac{3}{z-1} = 2(sqrt{x+2} + sqrt{y+2} + sqrt{z+2}) ) for ( x, y, z > 1 ), we start by considering the symmetry of the equation. Assuming ( x = y = z = a ), we simplify the equation to:[3a + frac{9}{a - 1} = 6sqrt{a + 2}]Dividing both sides by 3, we get:[a + frac{3}{a - 1} = 2sqrt{a + 2}]Let ( t = sqrt{a + 2} ). Then ( a = t^2 - 2 ), substituting this into the equation gives:[t^2 - 2 + frac{3}{t^2 - 3} = 2t]Multiplying through by ( t^2 - 3 ) and simplifying, we obtain the quartic equation:[t^4 - 2t^3 - 5t^2 + 6t + 9 = 0]Factoring this equation, we find it factors as ( (t^2 - t - 3)^2 = 0 ). Solving ( t^2 - t - 3 = 0 ), we get:[t = frac{1 pm sqrt{13}}{2}]Since ( t = sqrt{a + 2} ) must be positive, we take ( t = frac{1 + sqrt{13}}{2} ). Converting back to ( a ):[a = left( frac{1 + sqrt{13}}{2} right)^2 - 2 = frac{3 + sqrt{13}}{2}]Verifying this solution in the original equation confirms it satisfies the equality. Given the symmetry and the nature of the equation, the only solutions are when ( x = y = z ). Thus, the real numbers are:[boxed{dfrac{3 + sqrt{13}}{2}}]

🔑:1. Given the equation: [ x + y + z + frac{3}{x-1} + frac{3}{y-1} + frac{3}{z-1} = 2(sqrt{x+2} + sqrt{y+2} + sqrt{z+2}) ] we start by substituting ( x = p^2 - 2 ), ( y = q^2 - 2 ), and ( z = r^2 - 2 ) where ( p, q, r in mathbb{R} ) and ( p, q, r > sqrt{3} ).2. Substituting these into the equation, we get: [ (p^2 - 2) + (q^2 - 2) + (r^2 - 2) + frac{3}{p^2 - 3} + frac{3}{q^2 - 3} + frac{3}{r^2 - 3} = 2(p + q + r) ]3. Simplifying the left-hand side: [ p^2 + q^2 + r^2 - 6 + frac{3}{p^2 - 3} + frac{3}{q^2 - 3} + frac{3}{r^2 - 3} ]4. Equating both sides: [ p^2 + q^2 + r^2 - 6 + frac{3}{p^2 - 3} + frac{3}{q^2 - 3} + frac{3}{r^2 - 3} = 2(p + q + r) ]5. Rearranging terms: [ p^2 + q^2 + r^2 - 2(p + q + r) + frac{3}{p^2 - 3} + frac{3}{q^2 - 3} + frac{3}{r^2 - 3} = 6 ]6. We now consider the expression (frac{3}{p^2 - 3}). For ( p > sqrt{3} ), ( p^2 - 3 > 0 ), so (frac{3}{p^2 - 3} geq 0 ).7. We need to solve: [ p^2 - 2p + frac{3}{p^2 - 3} = 2 ]8. Multiplying through by ( p^2 - 3 ): [ p^4 - 2p^3 + 3 = 2(p^2 - 3) ]9. Simplifying: [ p^4 - 2p^3 - 2p^2 + 6 = 0 ]10. Factoring the polynomial: [ (p^2 - p - 3)^2 = 0 ]11. Solving for ( p ): [ p^2 - p - 3 = 0 ]12. Using the quadratic formula: [ p = frac{1 pm sqrt{13}}{2} ]13. Since ( p > sqrt{3} ), we take the positive root: [ p = frac{1 + sqrt{13}}{2} ]14. Similarly, for ( q ) and ( r ): [ q = r = frac{1 + sqrt{13}}{2} ]15. Substituting back, we get: [ x = y = z = left( frac{1 + sqrt{13}}{2} right)^2 - 2 = frac{3 + sqrt{13}}{2} ]The final answer is ( boxed{ x = y = z = frac{3 + sqrt{13}}{2} } ).