🧠:Okay, so I have this problem where I need to find the maximum value of the product of all roots of a specific 2023-degree real polynomial. The polynomial is given by P(x) = x^2023 + a1x^2022 + ... + a2023, and it has to satisfy two conditions: P(0) + P(1) = 0, and all 2023 roots are real and lie between 0 and 1 inclusive. The question is asking for the maximum possible product of these roots, r1 * r2 * ... * r2023. First, let me recall some fundamental concepts about polynomials and their roots. For a monic polynomial (which this is, since the leading coefficient is 1), the product of the roots is equal to (-1)^n times the constant term. Here, the polynomial is degree 2023, so the product of the roots r1 * r2 * ... * r2023 should be equal to (-1)^2023 * a2023. Since 2023 is odd, this simplifies to -a2023. Therefore, the product of the roots is -a2023. So, if I can figure out the maximum value of -a2023, that would give me the maximum product. But since we want the maximum value, and the roots are all between 0 and 1, their product will be non-negative because the product of numbers between 0 and 1 is non-negative. But since the product is -a2023, that would mean a2023 is negative, and thus -a2023 is positive. Therefore, maximizing the product is equivalent to minimizing a2023 (making it as negative as possible). But we also have the condition that P(0) + P(1) = 0. Let's compute P(0) and P(1). P(0) is the constant term a2023. P(1) would be 1 + a1 + a2 + ... + a2022 + a2023. Therefore, P(0) + P(1) = a2023 + 1 + a1 + a2 + ... + a2022 + a2023 = 1 + a1 + a2 + ... + a2022 + 2a2023 = 0. So, the equation we have is:1 + a1 + a2 + ... + a2022 + 2a2023 = 0. But the polynomial has all roots real and in [0,1]. So, perhaps we can express P(x) in terms of its roots. Since it's a monic polynomial with roots r1, r2, ..., r2023, we can write it as:P(x) = (x - r1)(x - r2)...(x - r2023).Expanding this, the constant term a2023 is (-1)^2023 * r1*r2*...*r2023 = -r1*r2*...*r2023. So, from earlier, the product of the roots is -a2023. Therefore, the product we're trying to maximize is equal to -a2023, which is the same as the absolute value of a2023 since a2023 is negative. But how does the condition P(0) + P(1) = 0 translate in terms of the roots? Let's compute P(0) and P(1) using the factored form. P(0) = (-1)^2023 * r1*r2*...*r2023 = - (r1*r2*...*r2023).P(1) = (1 - r1)(1 - r2)...(1 - r2023).Therefore, P(0) + P(1) = - (r1*r2*...*r2023) + (1 - r1)(1 - r2)...(1 - r2023) = 0.So, the key equation we have is:(1 - r1)(1 - r2)...(1 - r2023) = r1*r2*...*r2023.So, the product of (1 - ri) terms equals the product of ri terms. Let me denote the product of ri as Q, so Q = r1*r2*...*r2023. Then, the product of (1 - ri) is also Q. Therefore, Q = (1 - r1)(1 - r2)...(1 - r2023).So, we have Q = product(ri) = product(1 - ri). Our goal is to maximize Q, given that each ri is in [0,1], and Q = product(ri) = product(1 - ri). So, the problem reduces to finding the maximum Q such that Q = product(ri) = product(1 - ri), with each ri in [0,1]. This seems like an optimization problem where we need to maximize Q under the constraint that product(ri) = product(1 - ri). Since all the variables ri are symmetric in the equation, maybe the maximum occurs when all ri are equal. Let's test that hypothesis. Suppose all ri = r. Then, the equation becomes:r^2023 = (1 - r)^2023Taking both sides to the power of 1/2023, we get r = 1 - r, which implies 2r = 1, so r = 1/2. Therefore, if all roots are 1/2, then Q = (1/2)^2023, and indeed, product(ri) = product(1 - ri). But is this the maximum possible Q? Alternatively, perhaps having some roots at 0 or 1 could allow others to be larger? Wait, but if a root is 0, then the product Q becomes 0, which is not helpful. Similarly, if a root is 1, then (1 - ri) becomes 0, so the other product would also be 0. So, having any root at 0 or 1 would set Q to 0, which is the minimum possible. Therefore, to maximize Q, all roots should be strictly between 0 and 1. Therefore, the maximum Q might indeed occur when all roots are 1/2. But let's verify this. Suppose we have two variables, say in a simpler case with n=1. If n=1, then r1 = 1 - r1, so r1=1/2. For n=2, product(r1,r2) = product(1 - r1, 1 - r2). If r1=r2=1/2, then Q=(1/2)^2. Suppose instead we set r1= a and r2=1 - a. Then product(r1,r2) = a(1 - a), and product(1 - r1,1 - r2) = (1 - a)a. So, same product. Therefore, to maximize a(1 - a), which is maximized at a=1/2, so again, maximum at a=1/2. Therefore, for n=2, maximum Q is (1/2)^2. Similarly, for n=3, perhaps same. Therefore, in general, the maximum Q when all roots are 1/2. Therefore, the maximum product is (1/2)^2023. But wait, perhaps there is a better configuration where some roots are different but still the products are equal. Let's suppose that for some roots, we set some to a and others to 1 - a. Let's say for even n, but here n=2023 is odd. Suppose we split the roots into two groups: k roots at a and (2023 - k) roots at 1 - a. Then, product(ri) = a^k * (1 - a)^{2023 - k}, and product(1 - ri) = (1 - a)^k * a^{2023 - k}. For these products to be equal:a^k * (1 - a)^{2023 - k} = (1 - a)^k * a^{2023 - k}Divide both sides by a^k * (1 - a)^k:(1 - a)^{2023 - 2k} = a^{2023 - 2k}So, ( (1 - a)/a )^{2023 - 2k} = 1Therefore, either 2023 - 2k = 0, which is not possible since 2023 is odd, or (1 - a)/a = 1, which implies 1 - a = a, so a = 1/2. Therefore, the only possible solution in this case is a=1/2, regardless of k. Hence, even if we split the roots into two groups, they must still be at 1/2. Therefore, the only way for the products to be equal is if all roots are 1/2. Alternatively, perhaps if some roots are at different values. For example, let’s take three roots. Suppose two roots are a and one root is b. Then, product(ri) = a^2 * b, product(1 - ri) = (1 - a)^2 * (1 - b). Setting these equal:a^2 * b = (1 - a)^2 * (1 - b)This gives a relationship between a and b. Let's see if there's a solution where a ≠ 1/2. Let’s suppose a = 1/3, then:(1/3)^2 * b = (2/3)^2 * (1 - b)=> (1/9) b = (4/9)(1 - b)Multiply both sides by 9:b = 4(1 - b)=> b = 4 - 4b=> 5b = 4=> b = 4/5Therefore, product(ri) = (1/3)^2 * (4/5) ≈ 0.111 * 0.8 ≈ 0.0888. Compare this to if all three roots are 1/2: product is (1/2)^3 = 1/8 ≈ 0.125. So, in this case, splitting the roots into 1/3, 1/3, 4/5 gives a lower product. Hence, the maximum is still when all roots are 1/2.Alternatively, let's check if there's another combination where a and b could give a higher product. Let's suppose that for two variables, maybe a higher product. Let's take two roots: one at a and one at b. Then:a * b = (1 - a)(1 - b)So, ab = 1 - a - b + abCancelling ab:0 = 1 - a - bTherefore, a + b = 1So, in this case, if we have two roots, their product is a*b where a + b = 1. The maximum product of a and b under a + b = 1 is when a = b = 1/2, so product is 1/4. If a ≠ 1/2, then the product is less. Therefore, again, the maximum occurs at 1/2. Thus, even if we split roots into pairs where their sum is 1, the maximum product for each pair is 1/4, but if we have an odd number of roots, the extra root would have to be 1/2 to satisfy the equation. Wait, let's see. If we have an odd number, say 3 roots, two of them can be paired as a and 1 - a, but the third one must satisfy (for the product equality):a * (1 - a) * c = (1 - a) * a * (1 - c)So, same as a * (1 - a) * c = a * (1 - a) * (1 - c)Assuming a ≠ 0, 1, then cancelling a*(1 - a):c = 1 - c => c = 1/2.Therefore, the third root has to be 1/2. Then, the product is a*(1 - a)*(1/2). The maximum of a*(1 - a) is 1/4 at a=1/2, so the maximum product would be 1/4 * 1/2 = 1/8, which is the same as all three roots at 1/2: (1/2)^3 = 1/8. Therefore, even in this case, splitting into pairs and a single 1/2 gives the same product as all 1/2. So, in general, for an odd number of roots, if we pair them as a and 1 - a, the remaining root must be 1/2, but the product would still be equivalent to all roots at 1/2. Therefore, there's no gain in splitting them, but also no loss. However, if we don't split them symmetrically, we might end up with a lower product as shown in the earlier example. Therefore, the maximum product occurs when all roots are 1/2. But let's check another case. Suppose we have some roots set to a value higher than 1/2 and others lower. For example, let's take two roots, r1 and r2, with r1 > 1/2 and r2 < 1/2 such that r1 * r2 = (1 - r1)(1 - r2). Let’s set r1 = 1 - r2. Then, r1 = 1 - r2, so r1 + r2 = 1. Then, r1 * r2 = (1 - r1)(1 - r2) = r2 * r1, which is the same. Therefore, in this case, the product is r1 * r2, which is maximized when r1 = r2 = 1/2. Therefore, any deviation from 1/2 in a pair would decrease the product. Hence, the maximum product occurs when all roots are 1/2. Therefore, the maximum value of the product r1 * r2 * ... * r2023 is (1/2)^2023. But wait, let's confirm this with the original polynomial conditions. If all roots are 1/2, then the polynomial is (x - 1/2)^2023. Let's compute P(0) + P(1). P(0) = (-1/2)^2023 = - (1/2)^2023.P(1) = (1 - 1/2)^2023 = (1/2)^2023.Therefore, P(0) + P(1) = - (1/2)^2023 + (1/2)^2023 = 0. Which satisfies the condition. Therefore, this polynomial is valid and meets all the requirements, and gives the product of roots as (1/2)^2023. Hence, the maximum value is (1/2)^2023. But just to be thorough, let's consider if there could be a polynomial with some roots not equal to 1/2 that still satisfies P(0) + P(1) = 0 and has a higher product. Suppose, for example, we have one root at a and the rest at 1/2. Let's see if that's possible. Let’s suppose 2022 roots are at 1/2, and one root at a. Then, the polynomial is (x - a)(x - 1/2)^2022. Compute P(0) + P(1):P(0) = (-a)(-1/2)^2022 = -a*(1/2)^2022. Wait, hold on:Wait, expanding (x - a)(x - 1/2)^2022, the constant term is (-a)*(-1/2)^2022 = a*(1/2)^2022 * (-1)^2022. Since 2022 is even, (-1)^2022 = 1. So P(0) = a*(1/2)^2022.Similarly, P(1) = (1 - a)(1 - 1/2)^2022 = (1 - a)(1/2)^2022.Therefore, P(0) + P(1) = (a + (1 - a))(1/2)^2022 = 1*(1/2)^2022 ≠ 0. Therefore, this does not satisfy the condition P(0) + P(1) = 0 unless (1/2)^2022 = 0, which is not possible. Therefore, such a polynomial with one root different from 1/2 and the rest at 1/2 would not satisfy P(0) + P(1) = 0. Hence, this configuration is invalid. Alternatively, maybe having two roots different from 1/2. Let's suppose two roots at a and 1 - a, and the rest at 1/2. Then the polynomial is (x - a)(x - (1 - a))(x - 1/2)^2021. Let's compute P(0) + P(1):P(0) = (-a)(- (1 - a))(-1/2)^2021 = (-a)(-1 + a)(-1)^2021*(1/2)^2021. Since 2021 is odd, (-1)^2021 = -1. Therefore, P(0) = (-a)(-1 + a)(-1)*(1/2)^2021 = (-a)(a - 1)(1/2)^2021.Similarly, P(1) = (1 - a)(1 - (1 - a))(1 - 1/2)^2021 = (1 - a)(a)(1/2)^2021.Therefore, P(0) + P(1) = [(-a)(a - 1) + a(1 - a)]*(1/2)^2021.Let’s compute the bracket:First term: (-a)(a - 1) = -a^2 + a.Second term: a(1 - a) = a - a^2.Adding them: (-a^2 + a) + (a - a^2) = -2a^2 + 2a.Therefore, P(0) + P(1) = (-2a^2 + 2a)(1/2)^2021.Set this equal to zero:-2a^2 + 2a = 0 => -2a(a - 1) = 0 => a = 0 or a = 1.But if a is 0 or 1, then the roots at a or 1 - a would be 0 or 1, which makes the product zero. Therefore, this configuration also cannot yield a higher product than (1/2)^2023.Therefore, this approach doesn't work. Alternatively, maybe having three roots varied. But as we saw earlier, this leads to lower products. Another approach: suppose we use calculus to maximize Q = product(ri) under the constraint that product(ri) = product(1 - ri). Take the natural logarithm of both sides to turn products into sums:ln(Q) = sum(ln(ri)) = sum(ln(1 - ri)).Therefore, sum(ln(ri) - ln(1 - ri)) = 0.We need to maximize Q, which is equivalent to maximizing ln(Q) = sum(ln(ri)).Subject to the constraint sum(ln(ri) - ln(1 - ri)) = 0.So, this is an optimization problem with the objective function sum(ln(ri)) and constraint sum(ln(ri/(1 - ri))) = 0.We can use Lagrange multipliers here. Let’s set up the Lagrangian:L = sum_{i=1 to 2023} [ln(ri)] - λ sum_{i=1 to 2023} [ln(ri) - ln(1 - ri)].Taking derivative with respect to each ri:dL/dri = (1/ri) - λ (1/ri + 1/(1 - ri)) = 0.Wait, hold on. Let me check that again.Wait, the Lagrangian is L = sum(ln(ri)) - λ [sum(ln(ri) - ln(1 - ri))].So, expanding, L = sum(ln(ri)) - λ sum(ln(ri)) + λ sum(ln(1 - ri)).Therefore, L = (1 - λ) sum(ln(ri)) + λ sum(ln(1 - ri)).Then, derivative with respect to ri:dL/dri = (1 - λ)(1/ri) + λ (-1/(1 - ri)).Set this equal to zero:(1 - λ)/ri - λ/(1 - ri) = 0.Multiply through by ri(1 - ri):(1 - λ)(1 - ri) - λ ri = 0.Expand:(1 - λ)(1 - ri) - λ ri = 0(1 - λ) - (1 - λ)ri - λ ri = 0Combine terms with ri:(1 - λ) - ri [ (1 - λ) + λ ] = 0Since (1 - λ) + λ = 1:(1 - λ) - ri = 0Therefore, ri = 1 - λ.But this is true for all i. Therefore, all ri must be equal. Therefore, ri = 1 - λ for all i. Since all ri are equal, we can denote ri = r for all i. Then, from the constraint sum(ln(ri) - ln(1 - ri)) = 0:sum(ln(r) - ln(1 - r)) = 2023 [ln(r) - ln(1 - r)] = 0.Therefore, ln(r) - ln(1 - r) = 0 => ln(r/(1 - r)) = 0 => r/(1 - r) = e^0 = 1 => r = 1 - r => r = 1/2.Thus, the only critical point is when all ri = 1/2. Since this is the only critical point and we've checked other configurations which result in lower products, this must be the maximum. Therefore, the maximum value of the product r1*r2*...*r2023 is (1/2)^2023.Hence, the answer is boxed{left( dfrac{1}{2} right)^{2023}}.Final AnswerThe maximum value of ( r_1 cdot r_2 cdots r_{2023} ) is boxed{left( dfrac{1}{2} right)^{2023}}.

✅:Given a 2023-degree real polynomial ( P(x) = x^{2023} + a_1 x^{2022} + cdots + a_{2022} x + a_{2023} ) with the conditions ( P(0) + P(1) = 0 ) and all roots ( r_1, r_2, ldots, r_{2023} ) are real and lie in the interval ([0, 1]), we need to find the maximum value of the product ( r_1 cdot r_2 cdots r_{2023} ).1. Key Observations: - The product of the roots of the polynomial is given by ( (-1)^{2023} a_{2023} = -a_{2023} ). - The condition ( P(0) + P(1) = 0 ) translates to ( - (r_1 cdot r_2 cdots r_{2023}) + (1 - r_1)(1 - r_2) cdots (1 - r_{2023}) = 0 ).2. Simplification: - Let ( Q = r_1 cdot r_2 cdots r_{2023} ). The equation becomes ( Q = (1 - r_1)(1 - r_2) cdots (1 - r_{2023}) ).3. Optimization: - To maximize ( Q ), we need to find the configuration of roots ( r_i ) such that the product of the roots equals the product of ( 1 - r_i ). - By symmetry and the properties of products, the maximum product occurs when all roots are equal. Setting ( r_i = frac{1}{2} ) for all ( i ) satisfies the equation ( Q = (1 - frac{1}{2})^{2023} = (frac{1}{2})^{2023} ).4. Verification: - For the polynomial ( (x - frac{1}{2})^{2023} ), we check ( P(0) + P(1) ): - ( P(0) = -(frac{1}{2})^{2023} ) - ( P(1) = (frac{1}{2})^{2023} ) - Summing these gives ( 0 ), satisfying the condition.Thus, the maximum value of ( r_1 cdot r_2 cdots r_{2023} ) is (boxed{left( dfrac{1}{2} right)^{2023}}).

🔑:1. Achievability: - Consider the polynomial ( P(x) = prod_{i=1}^{2023} (x - r_i) ) where ( r_i = frac{1}{2} ) for all ( i ). - Then, ( P(x) = left( x - frac{1}{2} right)^{2023} ). - Evaluating ( P(0) ) and ( P(1) ): [ P(0) = left( 0 - frac{1}{2} right)^{2023} = left( -frac{1}{2} right)^{2023} = -frac{1}{2^{2023}} ] [ P(1) = left( 1 - frac{1}{2} right)^{2023} = left( frac{1}{2} right)^{2023} = frac{1}{2^{2023}} ] - Therefore, ( P(0) + P(1) = -frac{1}{2^{2023}} + frac{1}{2^{2023}} = 0 ), satisfying the given condition. - The product of the roots is: [ r_1 cdot r_2 cdots r_{2023} = left( frac{1}{2} right)^{2023} = 2^{-2023} ]2. Upper Bound: - Let ( P(x) = prod_{i=1}^{2023} (x - r_i) ) where ( 0 leq r_i leq 1 ) for all ( i ). - Given ( P(0) + P(1) = 0 ), we have: [ P(0) = (-1)^{2023} r_1 r_2 cdots r_{2023} = -r_1 r_2 cdots r_{2023} ] [ P(1) = (1 - r_1)(1 - r_2) cdots (1 - r_{2023}) ] [ P(0) + P(1) = -r_1 r_2 cdots r_{2023} + (1 - r_1)(1 - r_2) cdots (1 - r_{2023}) = 0 ] - Therefore: [ (1 - r_1)(1 - r_2) cdots (1 - r_{2023}) = r_1 r_2 cdots r_{2023} ] - Let ( q_i = frac{1}{r_i} - 1 ). Then ( 1 - r_i = frac{q_i}{1 + q_i} ) and: [ (1 - r_1)(1 - r_2) cdots (1 - r_{2023}) = frac{q_1}{1 + q_1} cdot frac{q_2}{1 + q_2} cdots frac{q_{2023}}{1 + q_{2023}} ] - Since ( q_1 q_2 cdots q_{2023} = 1 ), we have: [ r_1 r_2 cdots r_{2023} = left( frac{q_1}{1 + q_1} cdot frac{q_2}{1 + q_2} cdots frac{q_{2023}}{1 + q_{2023}} right)^{-1} ] - Using the AM-GM inequality: [ frac{q_1}{1 + q_1} cdot frac{q_2}{1 + q_2} cdots frac{q_{2023}}{1 + q_{2023}} leq left( frac{q_1 + q_2 + cdots + q_{2023}}{2023} right)^{2023} ] - Since ( q_1 q_2 cdots q_{2023} = 1 ), by AM-GM: [ frac{q_1 + q_2 + cdots + q_{2023}}{2023} geq 1 ] - Therefore: [ r_1 r_2 cdots r_{2023} leq left( frac{2023}{2023} right)^{2023} = 2^{-2023} ]Conclusion:[boxed{2^{-2023}}]

🧠:Okay, so I need to find all polynomials P(x) of odd degree d with integer coefficients such that for every positive integer n, there exist n positive integers x₁, x₂, ..., xₙ where for every pair i, j, the ratio P(x_i)/P(x_j) is between 1/2 and 2, and also that ratio is the d-th power of a rational number. Hmm, let me try to break this down.First, since P(x) is a polynomial of odd degree with integer coefficients, its leading term will dominate for large x. So as x gets large, P(x) will tend to positive or negative infinity depending on the leading coefficient. But since we're dealing with positive integers x_i and the ratio P(x_i)/P(x_j) has to be positive (because it's a d-th power, which for odd d can be positive or negative, but since we have positive integers x_i, maybe P(x_i) is positive?), so maybe the leading coefficient is positive. Otherwise, for large x, P(x) would be negative. So I can assume the leading coefficient is positive. That makes sense because otherwise, for sufficiently large x, P(x) would be negative, and taking ratios of negative numbers would complicate things, especially since we need the ratio to be a d-th power of a rational number, which would have to be positive if we are to avoid dealing with complex numbers. So yeah, leading coefficient positive. So P(x) is positive for all sufficiently large x, which is good because x_i are positive integers, so we can take x_i large enough.Next, the key condition is that for any n, there exist n positive integers x₁,...,xₙ such that 1/2 < P(x_i)/P(x_j) < 2, and P(x_i)/P(x_j) is a d-th power of a rational number. Let me parse this.First, for every pair i, j, the ratio is a d-th power of a rational number. So P(x_i)/P(x_j) = (a_{ij}/b_{ij})^d, where a_{ij} and b_{ij} are integers. Moreover, the ratio is between 1/2 and 2.So, if we can find x₁,...,xₙ such that each P(x_i) is a rational number raised to the d-th power times some common base, so that when you take the ratio, that base cancels out, leaving a ratio of d-th powers. But P(x_i) must be integers because the coefficients are integers and x_i are integers, so P(x_i) is an integer. Therefore, if P(x_i)/P(x_j) is a d-th power of a rational number, then it must be a d-th power of a rational number in reduced form, which implies that both numerator and denominator are d-th powers. Because if (a/b)^d = P(x_i)/P(x_j), then since P(x_i) and P(x_j) are integers, a^d divides P(x_i) and b^d divides P(x_j). But unless a and b are 1, this might be restrictive. Wait, but actually, if the ratio is (a/b)^d, then P(x_i) = k * a^d and P(x_j) = k * b^d for some integer k. But k must be common to all the P(x_i)'s? Because if you have multiple x_i, then P(x_i) must each be equal to k times some d-th power. Otherwise, the ratios would involve different k's, which might not cancel. Wait, but the ratio between any two P(x_i) and P(x_j) must be a d-th power. So if P(x_i) = k_i * c_i^d and P(x_j) = k_j * c_j^d, then the ratio is (k_i / k_j) * (c_i / c_j)^d. For this to be a d-th power of a rational number, the ratio k_i / k_j must itself be a d-th power. So unless all k_i are equal, this would require that k_i / k_j is a d-th power. So if all the k_i are equal, then the ratio is (c_i / c_j)^d, which is a d-th power. So maybe all the P(x_i) must be equal to a constant times a d-th power. Therefore, for some integer k, P(x) = k * (rational number)^d. But since P(x) has integer coefficients, maybe k must be a d-th power itself? Wait, maybe not. Let me think.Alternatively, perhaps P(x) itself is a d-th power of another polynomial. But if d is the degree, then that would require that P(x) = Q(x)^d where Q(x) is a linear polynomial (since d times the degree of Q would equal d, so Q is linear). So Q(x) = ax + b, then P(x) = (ax + b)^d. Then, for such a P(x), the ratio P(x_i)/P(x_j) = ((a x_i + b)/(a x_j + b))^d. So this ratio is indeed a d-th power of a rational number. Moreover, we need that for any n, there exist x₁,...,xₙ such that 1/2 < ((a x_i + b)/(a x_j + b))^d < 2. Since d is odd, we can take the d-th root without changing the inequality direction. Therefore, this reduces to 1/2^{1/d} < (a x_i + b)/(a x_j + b) < 2^{1/d}.But 2^{1/d} is a number slightly greater than 1, and 1/2^{1/d} is slightly less than 1. So the ratio (a x_i + b)/(a x_j + b) must be between these two numbers. Which suggests that all the terms a x_i + b are very close to each other. But how can we have n terms a x + b all within a multiplicative factor of 2^{1/d} of each other? For large n, this seems challenging because the numbers a x + b grow linearly with x, so their ratios approach 1 as x increases, but how can we have infinitely many such x?Wait, but the problem states for each positive integer n, there exists n positive integers x₁,...,xₙ such that all ratios are within that bound. So for each n, we need to find n numbers in the image of P that are within a factor of 2^{1/d} of each other. If P(x) is (a x + b)^d, then their ratios are ((a x_i + b)/(a x_j + b))^d, so the inner ratio must be between 1/2^{1/d} and 2^{1/d}. But 2^{1/d} is approximately 1 + (ln 2)/d for large d, but even for small d, say d=1, 2^{1/1}=2. Wait, but d is odd, so d could be 1,3,5,...Wait, if d=1, then the problem reduces to finding linear polynomials P(x) = a x + b with integer coefficients such that for any n, there are n positive integers x₁,...,xₙ with 1/2 < (a x_i + b)/(a x_j + b) < 2, and the ratio is a 1st power, i.e., just a rational number, which it is trivially. But in this case, for d=1, the ratio must be between 1/2 and 2. So for a linear polynomial, we need that for any n, there are n integers x such that a x + b are all within a factor of 2 of each other. So for example, if a=1, b=0, then P(x)=x. Then we need n integers x₁,...,xₙ such that each x_i / x_j is between 1/2 and 2. But this is impossible for n ≥ 3, because if you have three numbers x, y, z, each within a factor of 2 of each other, but then if x is the smallest, z has to be less than 2x, but y is between x and 2x, and z has to be between x and 2x, but if you have more numbers, you can't have them all within a factor of 2. Wait, actually, no. For example, if you take x, 2x, but then 2x / x = 2, but the ratio has to be less than 2. So actually, if you take numbers all in [k, 2k) for some k, then any two numbers in that interval have ratios between 1 and 2. But integers in [k, 2k) are k, k+1, ..., 2k-1. So if we take k=2^{m}, then the interval [2^m, 2^{m+1}) contains 2^m integers. So for any n, we can choose m such that 2^m ≥ n, and then take n integers from [2^m, 2^{m+1}). So actually, for P(x) = x, which is linear (d=1), we can satisfy the condition: for any n, pick numbers in a interval [k, 2k), which are all within a factor of 2 of each other. Hence, P(x) = x is a solution for d=1. Similarly, maybe other linear polynomials?Wait, suppose P(x) = a x + b. Then, the ratio (a x_i + b)/(a x_j + b). For this to be between 1/2 and 2, we need that a x_i + b and a x_j + b are within a factor of 2. To have infinitely many such x_i, we need that the polynomial can take on values in intervals where they are dense enough. But if a ≠ 1, say a=2, then P(x)=2x + b. Then, the values are even numbers plus b. If b is even, then P(x) is even, otherwise odd. But to have ratios between 1/2 and 2, similar to the linear case. But can we always find n numbers in an interval [k, 2k) for P(x)=2x + b?Wait, for example, let's take P(x)=2x + 1. Then, the values are 3,5,7,... To have ratios between 1/2 and 2, we need numbers such that the largest is less than twice the smallest. So if we take x=1,2: P(1)=3, P(2)=5. 5/3 ≈1.666 <2, so that's okay. For n=2, that works. For n=3, we need three numbers 3,5,7. But 7/3 ≈2.333 >2, so that doesn't work. So perhaps P(x)=2x +1 doesn't work. Wait, but maybe pick numbers starting from a higher x? For example, take x=4,5,6: P(4)=9, P(5)=11, P(6)=13. 13/9≈1.444<2. So that works. Then n=3 is okay. For n=4, take x=4,5,6,7: 9,11,13,15. 15/9≈1.666<2. So that works. Similarly, for larger n, as long as we pick x in a range where P(x) doesn't exceed twice the smallest P(x) in the range. For P(x)=2x +1, as x increases, the difference between consecutive terms is 2, so the ratio between consecutive terms is (2(x+1)+1)/(2x+1) = (2x+3)/(2x+1) = 1 + 2/(2x+1). As x increases, this ratio approaches 1. So for large x, we can have many terms in a row where each consecutive ratio is close to 1, hence all ratios are less than 2. For example, if we take x from N to N + n -1, then P(x) = 2x +1. The largest ratio is (2(N + n -1) +1)/(2N +1) = (2N + 2n -1)/(2N +1) = 1 + (2n -2)/(2N +1). We need this to be less than 2, so 1 + (2n -2)/(2N +1) <2 => (2n -2)/(2N +1) <1 => 2n -2 <2N +1 => 2N >2n -3 => N >n -1.5. So if we choose N =n, then 2N +1 =2n +1, and the ratio is (2(n +n -1) +1)/(2n +1) = (4n -2 +1)/(2n +1)= (4n -1)/(2n +1) ≈2 as n increases. Wait, that's approaching 2 from below. For example, n=1: N=1, x=1 to 1: ratio 1. For n=2: N=2, x=2 and 3: P(2)=5, P(3)=7, ratio 7/5=1.4 <2. For n=3: N=3, x=3,4,5: P(x)=7,9,11. 11/7≈1.57 <2. For n=4: N=4, x=4,5,6,7: P(x)=9,11,13,15. 15/9≈1.66 <2. For n=5: x=5,6,7,8,9: P=11,13,15,17,19. 19/11≈1.72 <2. So actually, if we choose N =n, the ratio is (2(N +n -1) +1)/(2N +1) = (2N +2n -2 +1)/(2N +1)= (2N +2n -1)/(2N +1). When N =n, this becomes (2n +2n -1)/(2n +1)= (4n -1)/(2n +1) ≈2 - 3/(2n +1). Which is less than 2. So as n increases, this approaches 2. So for any n, we can take N =n, and get n numbers with the largest ratio less than 2. Similarly, the smallest ratio is the reciprocal, which would be greater than 1/2. Wait, the smallest ratio is (2N +1)/(2(N +n -1) +1)= (2N +1)/(2N +2n -1). If N =n, this is (2n +1)/(4n -1) ≈ (2n)/(4n)=1/2. So as n increases, the minimal ratio approaches 1/2 from above. Therefore, for any n, choosing x from N to N +n -1 where N =n gives us n numbers where the ratios are between (2n +1)/(4n -1) and (4n -1)/(2n +1). Now, (2n +1)/(4n -1) = (2n +1)/(4n -1) = [2n +1]/[4n -1] = [2n +1]/[4n -1] = approx (2n)/(4n) =1/2 for large n. Similarly, (4n -1)/(2n +1)≈2. Therefore, for any n, these ratios are strictly between 1/2 and 2. Hence, for the polynomial P(x)=2x +1, we can satisfy the condition for any n. So P(x)=2x +1 is a solution for d=1. Similarly, any linear polynomial P(x)=ax +b with a and b integers, a>0, would work? Wait, let's check.Suppose P(x)=3x +1. Then, similar logic: For n numbers x starting at N, the ratio between the largest and smallest would be (3(N +n -1) +1)/(3N +1)= (3N +3n -3 +1)/(3N +1)= (3N +3n -2)/(3N +1). To have this ratio less than 2, we need 3N +3n -2 <2(3N +1) => 3N +3n -2 <6N +2 => 3n -4 <3N => N >(3n -4)/3. So choosing N =n would give (3n +3n -2)/(3n +1)= (6n -2)/(3n +1)=2 - 4/(3n +1) <2. Similarly, the minimal ratio is (3N +1)/(3(N +n -1) +1)= (3N +1)/(3N +3n -2). For N =n, this is (3n +1)/(6n -2)= (3n +1)/(2(3n -1))≈3n/(6n)=1/2. So similar to the previous case, the ratio is between 1/2 and 2. Hence, P(x)=3x +1 also works. Therefore, it seems that any linear polynomial P(x)=ax +b with a>0 and integer coefficients would work. Wait, but let's check for a=1, b=0. P(x)=x. Then, for any n, can we find n integers x₁,...,xₙ such that all ratios x_i/x_j are between 1/2 and 2. Yes, as I thought earlier, by taking x_i in [k, 2k). For example, take k=2^{m} and choose n numbers from [2^{m}, 2^{m+1}), which has 2^{m} numbers, so for any n, choose m such that 2^{m} ≥n. Therefore, P(x)=x works. Similarly, P(x)=ax +b with a≠1. So maybe all linear polynomials with positive leading coefficient and integer coefficients are solutions for d=1.But the problem says "polynomials of odd degree d". So d=1 is allowed, but also d=3,5,... So perhaps similar reasoning applies to higher odd degrees. But let's think.Suppose d=3. Then, we need a polynomial P(x) of degree 3 with integer coefficients, such that for any n, there exist n positive integers x₁,...,xₙ with 1/2 < P(x_i)/P(x_j) <2 and P(x_i)/P(x_j) is a cube of a rational number.If P(x) is a cubic polynomial, then for large x, P(x) ≈ a x^3. So the ratio P(x_i)/P(x_j) ≈ (x_i/x_j)^3. For this ratio to be between 1/2 and 2, we need (x_i/x_j)^3 between 1/2 and 2, so x_i/x_j between 1/2^{1/3} and 2^{1/3}. But 2^{1/3} is approximately 1.26, so the x_i must be within roughly 26% of each other. But how can we have n integers x₁,...,xₙ such that each x_i is within 26% of x_j for all i,j? For example, for n=3, we need three integers where each is between, say, k and 1.26k. But integers are discrete, so unless k is very large, there might not be enough integers in that interval. However, as k increases, the number of integers in [k, 2^{1/3}k] increases, but 2^{1/3} ~1.26, so the interval length is ~0.26k. For large k, this interval contains approximately 0.26k integers. So for any n, if we take k large enough such that 0.26k ≥n, then there will be at least n integers in [k, 2^{1/3}k]. Therefore, in principle, for any n, we can choose sufficiently large k and select n integers from [k, (1 + ε)k] where ε=2^{1/3} -1≈0.26. Then, the ratios x_i/x_j will be between 1 and 1.26, so their cubes will be between 1 and 2. Hence, the ratios P(x_i)/P(x_j) ≈(x_i/x_j)^3 will be between 1 and 2. But wait, but the problem requires the ratio to be exactly a cube of a rational number, not just approximately. So even if x_i/x_j is close to 1, unless x_i/x_j is exactly a rational number whose cube is between 1/2 and 2, it might not work.Wait, but if P(x) is a cubic polynomial, unless it's of the form (ax +b)^3, then the ratio P(x_i)/P(x_j) would be ((ax_i +b)/(ax_j +b))^3. Then, similar to the linear case, if we can choose x_i such that (ax_i +b)/(ax_j +b) is between 1/2^{1/3} and 2^{1/3}, then its cube would be between 1/2 and 2. But for such ratios to be rational numbers, we need (ax_i +b)/(ax_j +b) to be rational. However, (ax_i +b)/(ax_j +b) being rational doesn't necessarily mean it's a ratio of integers unless ax_i +b and ax_j +b are multiples of some common integer. Wait, but if ax +b is an integer for integer x, then yes, (ax_i +b)/(ax_j +b) is a rational number. Therefore, if P(x)=(ax +b)^3, then P(x_i)/P(x_j)=((ax_i +b)/(ax_j +b))^3, which is the cube of a rational number, as required. Moreover, if we can choose x_i such that the ratios (ax_i +b)/(ax_j +b) are between 1/2^{1/3} and 2^{1/3}, then their cubes will be between 1/2 and 2. So similar to the linear case, if we can find n integers x_i such that ax_i +b are all within a factor of 2^{1/3} of each other, then their cubes will satisfy the ratio condition.But how can we ensure that for any n, there are n integers x_i such that ax_i +b are within a multiplicative factor of 2^{1/3}? Let's consider the case when a=1, b=0, so P(x)=x^3. Then, we need n integers x_i such that x_i/x_j is between 1/2^{1/3} and 2^{1/3} for all i,j. As before, for large x, the number of integers in [x, 2^{1/3}x] is roughly (2^{1/3} -1)x. So for any n, choosing x large enough such that (2^{1/3} -1)x ≥n, then there will be at least n integers in that interval. Then, selecting those n integers would give us x_i such that their ratios are between 1 and 2^{1/3}, hence their cubes would have ratios between 1 and 2. Similarly, the reciprocal ratios would be between 1/2^{1/3} and 1, so all ratios between 1/2 and 2. Therefore, for P(x)=x^3, this condition is satisfied. Similarly, for P(x)=(ax +b)^3, we can choose x_i such that ax_i +b are in a sufficiently tight interval, leading to the ratios of their cubes being within 1/2 and 2. Therefore, polynomials of the form (ax +b)^d with a>0, integer coefficients, and odd d satisfy the conditions.But wait, the problem states "polynomials of odd degree d with integer coefficients". So if we can show that only polynomials of the form c(ax +b)^d with integers a,b,c, where a>0, c ≠0, and perhaps c being a perfect d-th power? Wait, but if P(x)=c(ax +b)^d, then P(x_i)/P(x_j)= (c(ax_i +b)^d)/(c(ax_j +b)^d)= ((ax_i +b)/(ax_j +b))^d. So the ratio is the d-th power of a rational number. The constant c cancels out. So even if c is not a d-th power, as long as P(x)=c Q(x)^d where Q(x) has integer coefficients, then the ratio P(x_i)/P(x_j)= (Q(x_i)/Q(x_j))^d. Wait, no, if P(x)=c Q(x)^d, then the ratio is (c Q(x_i)^d)/(c Q(x_j)^d)= (Q(x_i)/Q(x_j))^d. So c can be any integer, but Q(x) must have integer coefficients. Therefore, Q(x) must be a polynomial with integer coefficients. However, if d is the degree of P(x), then Q(x) must be linear, because if Q(x) is of degree k, then P(x)=c Q(x)^d would be degree k*d. Since P(x) is given to be of degree d, then k*d =d =>k=1. Therefore, Q(x) must be linear. Therefore, P(x) must be of the form c (ax +b)^d where a,b,c are integers, a≠0, and c≠0.But wait, if P(x) has integer coefficients, then c must be an integer, and (ax +b)^d must have integer coefficients. So (ax +b)^d will have integer coefficients if a and b are integers. Therefore, P(x)=c (ax +b)^d with integers a,b,c. However, c must also be a d-th power if we want P(x) to be a d-th power times something. Wait, but actually, no. Even if c is not a d-th power, P(x_i)/P(x_j)= (c (ax_i +b)^d)/(c (ax_j +b)^d)= ((ax_i +b)/(ax_j +b))^d. So the c cancels out. Therefore, even if c is not a d-th power, the ratio is still a d-th power of a rational number. Therefore, P(x) can be any scalar multiple of a d-th power of a linear polynomial with integer coefficients. However, we need P(x) to have integer coefficients. If P(x)=c(ax +b)^d, then as long as a,b,c are integers, P(x) will have integer coefficients. Therefore, all such polynomials are of the form c(ax +b)^d with a,b,c integers, a≠0, c≠0, and d odd.But wait, the problem specifies that the degree is d. So if we have P(x)=c(ax +b)^d, then the degree is d (since expanding (ax +b)^d gives a degree d polynomial). Therefore, the leading term is c a^d x^d. Therefore, to have P(x) of degree d, we need a≠0 and c≠0.But the problem says "polynomials of odd degree d". So d is odd, and we need to find all such polynomials with integer coefficients. So according to this reasoning, all polynomials of the form c(ax +b)^d with integer coefficients a,b,c, a>0, and d odd would satisfy the condition. However, we need to verify if there are other polynomials besides these that satisfy the condition.Suppose there is another polynomial not of this form. For example, suppose P(x) is a cubic polynomial not of the form c(ax +b)^3. Would such a polynomial satisfy the condition? Let's consider an example. Let P(x)=x^3 + x. This is a cubic polynomial. Let's see if it satisfies the condition. For any n, we need to find n positive integers x₁,...,xₙ such that 1/2 < (x_i^3 +x_i)/(x_j^3 +x_j) <2 and the ratio is a cube of a rational number.Take n=2. We need two integers x and y such that 1/2 < (x^3 +x)/(y^3 +y) <2 and the ratio is a cube of a rational number. Let's try x=1, y=1: ratio=1, which is 1^3. Okay. x=1, y=2: (1+1)/(8+2)=2/10=1/5 <1/2. Not good. x=2, y=1: (8+2)/(1+1)=10/2=5 >2. Not good. x=2, y=2: ratio=1. x=2, y=3: (8+2)/(27+3)=10/30=1/3 <1/2. x=3, y=2: (27+3)/(8+2)=30/10=3 >2. Hmm, so the ratios either blow up or shrink too much. What if we take larger x and y. Suppose x and y are large. Then P(x) ≈x^3, so the ratio ≈(x/y)^3. So similar to the pure x^3 case. But in that case, we could try to use the same approach as before: pick x and y close to each other so that (x/y)^3 is close to 1. However, the problem is that the actual ratio (x^3 +x)/(y^3 +y) might not be exactly a cube of a rational number. For example, take x= k and y=k+1. Then, the ratio is (k^3 +k)/( (k+1)^3 + (k+1)) ≈(k^3)/(k^3 +3k^2 +3k +1 +k +1) ≈1/(1 +3/k + ...). So it's approximately 1 - 3/k. But this is not a cube of a rational number unless specific conditions are met. It's unclear how to ensure that this ratio is exactly a cube. Therefore, unless P(x) is of the form c(ax +b)^d, where the ratio simplifies to ((ax_i +b)/(ax_j +b))^d, which is a d-th power, it's hard to see how the ratio could be a d-th power for arbitrary n.Therefore, it's plausible that only polynomials of the form c(ax +b)^d satisfy the condition. Moreover, since the problem requires the ratio P(x_i)/P(x_j) to be exactly a d-th power of a rational number, the structure of P(x) must be such that this ratio simplifies in this way. The most straightforward way for this to happen is if P(x) itself is a d-th power of a linear polynomial, scaled by a constant. Because then the ratio becomes ((ax_i +b)/(ax_j +b))^d, which is a d-th power. Additionally, the constant c cancels out in the ratio, so it doesn't affect the condition.Now, we need to check if constants are allowed. Suppose P(x)=c(ax +b)^d. Then, even if c is not 1, the ratio is still a d-th power, as c cancels out. Therefore, any scalar multiple of a d-th power of a linear polynomial would work. However, we must ensure that P(x) has integer coefficients. If a, b, c are integers, then P(x)=c(ax +b)^d certainly has integer coefficients. Conversely, if P(x) is of the form c(ax +b)^d with integer coefficients, then expanding it would give integer coefficients. So that's fine.But could there be other polynomials? Suppose P(x) is a product of multiple linear factors raised to powers. For example, P(x)=(ax +b)^d(cx +e)^k. But unless k=0, the degree would be d +k* (degree of cx +e). But since we need the overall degree to be d (which is odd), and if we have multiple factors, unless they are all raised to the same power d, but then the degree would be higher. Wait, no. For example, if P(x)=(ax +b)(cx +d)^(d-1). Then the degree is 1 + (d-1)*1 =d. However, such a polynomial would have ratios that are products of terms, which might not be d-th powers. For example, take d=3: P(x)=(ax +b)(cx +d)^2. Then P(x_i)/P(x_j)= [(ax_i +b)(cx_i +d)^2]/[(ax_j +b)(cx_j +d)^2]. This is not necessarily a cube of a rational number unless the terms can be expressed as a cube. It's not clear how to ensure that this ratio is a cube for arbitrary n. Therefore, such polynomials likely do not satisfy the condition.Therefore, the only viable candidates are polynomials of the form c(ax +b)^d with integer coefficients. Now, we need to check if these are the only solutions.Another way to approach this is to consider the multiplicative condition. For the ratio P(x_i)/P(x_j) to be a d-th power of a rational number, it must be that the prime factorization of P(x_i) and P(x_j) have exponents that are multiples of d. However, since P(x) has integer coefficients, P(x_i) and P(x_j) are integers. Therefore, if P(x_i) is divisible by a prime p, then the exponent of p in P(x_i) must be a multiple of d. This suggests that P(x_i) must be a perfect d-th power times a unit (which is ±1, but since we need positive ratios, they must be positive). Therefore, P(x_i) must be a perfect d-th power. Wait, but if P(x_i) is a d-th power for each x_i, then their ratios would automatically be d-th powers. But the problem doesn't require P(x_i) to be a d-th power, only that their ratios are d-th powers. So it's possible that P(x_i) =k_i^d * m where m is a fixed integer, so that P(x_i)/P(x_j)= (k_i/k_j)^d. This would require that P(x) =m * k(x)^d where k(x) is an integer-valued polynomial. However, since P(x) has integer coefficients, k(x) must be such that k(x)^d has integer coefficients. The only linear polynomials k(x) with integer coefficients raised to the d-th power would give P(x)=m(ax +b)^d. Therefore, this again leads us to P(x)=c(ax +b)^d with integers a,b,c.Therefore, combining all these observations, it seems that the only polynomials satisfying the given conditions are those of the form c(ax +b)^d where a,b,c are integers with a>0 and d odd. However, we need to confirm the constant c. Since the ratio P(x_i)/P(x_j)= ( (ax_i +b)/(ax_j +b) )^d, the constant c cancels out. Therefore, c can be any integer. However, we must ensure that P(x)=c(ax +b)^d has integer coefficients. Expanding (ax +b)^d gives a polynomial with integer coefficients if a and b are integers, and multiplying by integer c preserves integer coefficients. Therefore, c can indeed be any integer.However, the problem states that the polynomial has odd degree d. So the leading term is c*(a^d)x^d. For the degree to be exactly d, we need a≠0 and c≠0. Since the problem allows any odd degree d, and we need a>0 to ensure positivity for large x (as discussed earlier), we can conclude that the solutions are all polynomials of the form P(x)=c(ax +b)^d where a,b,c are integers with a>0, c≠0, and d odd.But wait, the problem says "polynomials of odd degree d". So the degree is fixed as d, so when we write P(x)=c(ax +b)^d, that's degree d as required. Additionally, since the problem requires integer coefficients, which is satisfied by a,b,c integers. The leading coefficient is c*a^d, which must be an integer, which it is since a and c are integers.Therefore, the conclusion is that all such polynomials are the ones of the form c(ax +b)^d with integers a>0, b, c≠0, and d odd. However, we might need to check if a=0 is allowed, but a=0 would make P(x)=c*b^d, a constant polynomial, which has degree 0, not odd. Hence, a must be non-zero, and since we need a>0 for positivity.Wait, but the problem states "polynomials of odd degree d". So d must be at least 1 and odd. Therefore, the answer should be all polynomials of the form c(ax +b)^d where a,b,c are integers with a>0, c≠0, and d is an odd positive integer.But to confirm, let's check if constants are allowed. If d=0, but d must be odd, but degree 0 is even (since 0 is even). So no, d must be at least 1 and odd. Hence, the answer is as above.But let me check if there are any other possibilities. For example, suppose P(x) is a monic polynomial of degree d, say P(x)=x^d + lower degree terms. Would such a polynomial satisfy the condition? For example, P(x)=x^3 + x^2 +x +1. For large x, P(x)≈x^3, so ratios would approximate (x_i/x_j)^3. However, the lower-degree terms might perturb the ratio. For example, take x=1 and y=2. P(1)=1+1+1+1=4, P(2)=8+4+2+1=15. Ratio 15/4=3.75 >2, which doesn't satisfy the condition. So even though for large x, the ratios approach (x_i/x_j)^3, for small x, the ratios can exceed 2. Therefore, such a polynomial would not satisfy the condition for n=2 with x=1 and x=2. Therefore, non-monic polynomials or polynomials with additional terms besides the d-th power do not satisfy the condition. Hence, only polynomials that are exact d-th powers of linear polynomials (scaled by a constant) work.Therefore, the final answer is that all such polynomials are the ones of the form P(x)=c(ax +b)^d where a,b,c are integers with a>0, c≠0, and d is an odd positive integer.But the problem says "polynomials of odd degree d". So the degree is exactly d, which for P(x)=c(ax +b)^d is indeed d, as long as a≠0. Since we have a>0, this holds. Therefore, these are precisely the polynomials that satisfy the given conditions.Final AnswerThe polynomials are all integer multiples of the d-th power of a linear polynomial with integer coefficients. Thus, the solutions are boxed{P(x) = c(ax + b)^d} where a, b, and c are integers with a > 0, c neq 0, and d is an odd positive integer.

✅:To find all polynomials ( P(x) ) of odd degree ( d ) with integer coefficients satisfying the given condition, we analyze the requirements step by step:1. Condition Analysis: - For each positive integer ( n ), there exist ( n ) positive integers ( x_1, x_2, ldots, x_n ) such that ( frac{1}{2} < frac{P(x_i)}{P(x_j)} < 2 ) and ( frac{P(x_i)}{P(x_j)} ) is the ( d )-th power of a rational number for all pairs ( i, j ).2. Key Observations: - The ratio ( frac{P(x_i)}{P(x_j)} ) must be a ( d )-th power of a rational number, implying ( P(x_i) ) and ( P(x_j) ) must be related through a ( d )-th power. - For large ( x ), the polynomial ( P(x) ) is dominated by its leading term. This suggests ( P(x) ) could be a ( d )-th power of a linear polynomial.3. Polynomial Structure: - If ( P(x) = c(ax + b)^d ) where ( a, b, c ) are integers, ( a > 0 ), and ( d ) is odd, then the ratio ( frac{P(x_i)}{P(x_j)} = left( frac{ax_i + b}{ax_j + b} right)^d ), which is a ( d )-th power of a rational number. - For any ( n ), choosing ( x_i ) such that ( ax_i + b ) are within a factor of ( 2^{1/d} ) ensures the ratio ( left( frac{ax_i + b}{ax_j + b} right)^d ) is between ( frac{1}{2} ) and ( 2 ).4. Verification: - Polynomials of the form ( c(ax + b)^d ) with integer coefficients satisfy the conditions. Any other polynomial structure (e.g., with additional terms) would not guarantee the ratio to be a ( d )-th power of a rational number for all pairs ( i, j ).5. Conclusion: - The only polynomials meeting all criteria are those that are integer multiples of the ( d )-th power of a linear polynomial with integer coefficients. Final AnswerThe polynomials are all integer multiples of the ( d )-th power of a linear polynomial with integer coefficients. Thus, the solutions are:[boxed{P(x) = c(ax + b)^d}]where ( a ), ( b ), and ( c ) are integers with ( a > 0 ), ( c neq 0 ), and ( d ) is an odd positive integer.

🔑:1. Problem Restatement and Initial Assumptions: We need to find all polynomials ( P(x) ) of odd degree ( d ) with integer coefficients such that for each positive integer ( n ), there exist ( n ) positive integers ( x_1, x_2, ldots, x_n ) satisfying: [ frac{1}{2} < frac{P(x_i)}{P(x_j)} < 2 ] and (frac{P(x_i)}{P(x_j)}) is the ( d )-th power of a rational number for every pair of indices ( i ) and ( j ) with ( 1 leq i, j leq n ).2. Polynomial Form and Rational Root Theorem: Let ( P(x) = a_d x^d + a_{d-1} x^{d-1} + cdots + a_1 x + a_0 ) where ( a_i in mathbb{Z} ) and ( a_d neq 0 ). Assume ( a_d > 0 ) without loss of generality. We aim to show that ( P(x) ) must be of the form ( P(x) = c(rx + s)^d ) for some integers ( c, r, s ) with ( c neq 0 ).3. Reduction to Leading Term: Write ( P(x) = a_d (x + q)^d + Q(x) ) for some ( Q(x) in mathbb{Q}[x] ) with ( deg(Q) leq d-2 ). This can be done by choosing ( q = frac{d a_{d-1}}{a_d} ) so that the coefficient of ( x^{d-1} ) in ( a_d (x + q)^d ) matches ( a_{d-1} ).4. Main Lemma and Rational Approximation: We need to show that for sufficiently large ( x_1, ldots, x_n in mathbb{N} ), we have: [ y_i = leftlceil sqrt[d]{frac{a_d}{T}} cdot (x_i + q) rightrceil ] for all ( i in {1, cdots, n} ), where ( T ) is a ( d )-th power free integer.5. Contradiction Argument: Assume for contradiction that: [ y_i geq leftlceil sqrt[d]{frac{a_d}{T}} cdot (x_i + q) rightrceil + 1 geq sqrt[d]{frac{a_d}{T}} cdot (x_i + q) + 1 ] Then: [ a_d (x + q)^d + Q(x) = T cdot y_i^d > T cdot left( sqrt[d]{frac{a_d}{T}} (x_i + q) + 1 right)^d ] Using the Binomial Theorem: [ Q(x_i) > T cdot sum_{i=1}^{d-1} binom{d}{i} left( frac{a_d}{T} right)^{frac{i}{d}} (x_i + q)^i = sum_{i=0}^{d-1} binom{d}{i} a_d^{frac{i}{d}} T^{frac{d-i}{d}} (x_i + q)^i ] This implies: [ Q(x_i) > d cdot a_d^{frac{d-1}{d}} T^{frac{1}{d}} (x_i + q)^{d-1} ] Contradicting the fact that ( Q(x) ) has degree at most ( d-2 ).6. Conclusion: Since ( Q(x) ) must be identically zero, we have: [ P(x) = a_d (x + q)^d ] By the Rational Root Theorem, ( q ) must be rational, and thus ( P(x) ) must be of the form ( P(x) = c(rx + s)^d ) for integers ( c, r, s ) with ( c neq 0 ).The final answer is ( boxed{ P(x) = c(rx + s)^d } ) for integers ( c, r, s ) with ( c neq 0 ).

🧠:Okay, so I need to find the product of the real roots of the equation ( x^4 + 3x^3 + 5x^2 + 21x -14 = 0 ). The options are given as A) -2, B) 7, C) -14, D) 21, E) None of the preceding. Let me think step by step.First, I remember that for a polynomial equation, the product of all roots (real and complex) can be found using Vieta's formula. For a quartic equation ( x^4 + a x^3 + b x^2 + c x + d = 0 ), the product of all roots is ( d ) if the leading coefficient is 1. Here, the constant term is -14, so the product of all four roots is -14. But the question asks specifically for the product of the real roots, not all roots. That complicates things because we don't know how many real roots there are. So I can't just directly apply Vieta's formula for all roots.Next, I need to figure out how many real roots this equation has. Maybe using Descartes' Rule of Signs or analyzing the graph of the polynomial.Let's check Descartes' Rule of Signs for the number of positive real roots. The polynomial is ( x^4 + 3x^3 + 5x^2 + 21x -14 ). The coefficients are +, +, +, +, -, so the number of sign changes is 1. Therefore, there is exactly 1 positive real root.For negative real roots, substitute ( x = -y ) where ( y > 0 ), so the polynomial becomes ( (-y)^4 + 3(-y)^3 + 5(-y)^2 + 21(-y) -14 = y^4 - 3y^3 + 5y^2 -21y -14 ). The coefficients are +, -, +, -, -, so the sign changes are: + to -, -, to +, + to -, - to -. That's 3 sign changes. Therefore, there could be 3 or 1 negative real roots.So total possible real roots are 1 (positive) + 3 or 1 (negative) = 4 or 2 real roots. But since it's a quartic, the number of real roots must be even or odd? Wait, complex roots come in pairs, so if there's 1 positive real root, then the remaining roots must include an odd number of real roots. Wait, but complex roots come in pairs, so total non-real roots are even. So total real roots must be even (since 4 is even). Hmm, Descartes says 1 positive real root, and 3 or 1 negative real roots. But total real roots would be 4 or 2. Wait, if it's 1 positive and 3 negative, that's 4 real roots. If it's 1 positive and 1 negative, that's 2 real roots. Either way, the number of real roots is even. So possible 2 or 4 real roots. But since the leading coefficient is positive and the degree is even, as ( x ) approaches infinity, the polynomial goes to infinity, and as ( x ) approaches negative infinity, it also goes to infinity. The constant term is -14, so the y-intercept is at (0, -14). Let me try to sketch the graph roughly. It starts from positive infinity on the left, goes down to -14 at x=0, then goes back to positive infinity on the right. So there must be at least two real roots: one on the left side (negative) and one on the right side (positive). But according to Descartes, exactly 1 positive real root. So perhaps 1 positive, 1 negative, and two complex roots? Or 1 positive, 3 negative. Hmm. Wait, if there are 1 positive and 3 negative roots, that would make 4 real roots, but Descartes allows for 3 or 1 negative roots, so 3 negative roots would mean 3 sign changes, but after substitution, we had 3 sign changes, which allows for 3 or 1. So possible.But how do I figure out the exact number? Maybe test some possible roots to factor the polynomial.Let me try rational roots. The possible rational roots are factors of the constant term divided by factors of the leading coefficient. Since the leading coefficient is 1, possible rational roots are ±1, ±2, ±7, ±14.Let me test x=1: 1 + 3 + 5 + 21 -14 = 1 + 3 is 4, +5 is 9, +21 is 30, -14 is 16. Not zero.x=-1: 1 -3 + 5 -21 -14 = 1 -3 = -2, +5 = 3, -21 = -18, -14 = -32. Not zero.x=2: 16 + 24 + 20 + 42 -14 = 16+24=40, +20=60, +42=102, -14=88. Not zero.x=-2: 16 -24 + 20 -42 -14 = 16-24=-8, +20=12, -42=-30, -14=-44. Not zero.x=7: 2401 + 1029 + 245 + 147 -14. That's way too big. Definitely not zero.x=-7: Similarly, way too big. x=14 and x=-14 are even bigger. So none of the rational roots work. Therefore, the polynomial doesn't have rational roots. So maybe it factors into quadratics or something else.Alternatively, try factoring the polynomial as a product of quadratics. Let's assume it factors into ( (x^2 + a x + b)(x^2 + c x + d) ). Then expanding gives:( x^4 + (a + c)x^3 + (ac + b + d)x^2 + (ad + bc)x + b d ).Comparing coefficients with the original polynomial:1. Coefficient of ( x^3 ): ( a + c = 3 )2. Coefficient of ( x^2 ): ( a c + b + d = 5 )3. Coefficient of ( x ): ( a d + b c = 21 )4. Constant term: ( b d = -14 )We need to find integers a, c, b, d that satisfy these equations.Starting with the constant term: ( b d = -14 ). The possible integer pairs for (b, d) are (1, -14), (-1, 14), (2, -7), (-2, 7), (7, -2), (-7, 2), (14, -1), (-14, 1).Let me try different pairs.First pair: (b, d) = (2, -7)Then, from equation 1: a + c = 3.From equation 2: a c + 2 + (-7) = 5 ⇒ a c -5 =5 ⇒ a c =10.From equation 3: a*(-7) + 2*c =21 ⇒ -7a + 2c =21.We have a + c =3 and a c=10. Let's solve for a and c.From a + c =3 ⇒ c=3 -a. Substitute into a c=10:a(3 -a)=10 ⇒ 3a -a²=10 ⇒ a² -3a +10=0. Discriminant: 9 -40 = -31 <0. No real solutions. So this pair doesn't work.Next pair: (b, d)= (-2,7)Equation 2: a c + (-2) +7=5 ⇒ a c +5=5 ⇒ a c=0.Equation 3: a*7 + (-2)c=21 ⇒7a -2c=21.From a +c=3 and a c=0. If a c=0, then either a=0 or c=0.If a=0, then c=3. Then equation3: 0 -2*3= -6 ≠21. Not working.If c=0, then a=3. Then equation3:7*3 -0=21=21. That works. So a=3, c=0, b=-2, d=7.Therefore, the polynomial factors as ( (x^2 +3x -2)(x^2 +0x +7)= (x^2 +3x -2)(x^2 +7) ).Let me check if this is correct.Multiply them out:First, ( x^2 times x^2 = x^4 ).Then, ( x^2 times 7 =7x^2 ).Then, ( 3x times x^2 =3x^3 ), ( 3x times7=21x ).Then, -2 ×x^2= -2x^2, -2×7= -14.So combining all terms:x^4 +3x^3 + (7x^2 -2x^2) +21x -14 = x^4 +3x^3 +5x^2 +21x -14. Yes, that's correct!So the polynomial factors into ( (x^2 +3x -2)(x^2 +7) =0 ).Therefore, the roots are the roots of ( x^2 +3x -2 =0 ) and ( x^2 +7=0 ).Solving ( x^2 +7=0 ): ( x = pm isqrt{7} ). So those are complex roots.Solving ( x^2 +3x -2=0 ): Using quadratic formula, ( x = [-3 pm sqrt{9 +8}]/2 = [-3 pm sqrt{17}]/2 ). These are real roots, two of them.So the real roots are ( [-3 + sqrt{17}]/2 ) and ( [-3 - sqrt{17}]/2 ). Their product can be found by multiplying them.Alternatively, for a quadratic equation ( x^2 + px + q =0 ), the product of roots is q. Here, the quadratic is ( x^2 +3x -2 =0 ), so the product of its roots is -2. Therefore, the product of the real roots is -2.Looking back at the options, A is -2. So the answer should be A.Wait, but let me verify once again. The original polynomial is quartic, and we factored it into two quadratics. One quadratic gives two real roots, the other gives two complex roots. The product of the real roots is the product of the roots of ( x^2 +3x -2 ), which is indeed -2, as per Vieta. So yes, that's correct.Therefore, the correct answer is A) -2.But wait, just to be thorough, let me compute the product manually. Let the roots be ( r_1 = [-3 + sqrt{17}]/2 ) and ( r_2 = [-3 - sqrt{17}]/2 ). Then,( r_1 times r_2 = [(-3 + sqrt{17})(-3 - sqrt{17})]/4 = [9 - ( sqrt{17} )^2 ] /4 = [9 -17]/4 = (-8)/4 = -2 ). Yep, that's correct. So product is -2.Therefore, the answer is A) -2. So I think that's the correct choice.Final Answerboxed{A}

✅:To find the product of the real roots of the equation ( x^4 + 3x^3 + 5x^2 + 21x - 14 = 0 ), we start by considering the polynomial factorization. Using Descartes' Rule of Signs, we determine there is exactly 1 positive real root and 3 or 1 negative real roots. Testing possible rational roots (factors of -14) shows none are roots of the polynomial. Next, we factor the polynomial into quadratics. Assuming it factors as ( (x^2 + ax + b)(x^2 + cx + d) ), we solve the system of equations derived from the coefficients. We find the polynomial factors into ( (x^2 + 3x - 2)(x^2 + 7) ).The roots of ( x^2 + 7 = 0 ) are complex, while the roots of ( x^2 + 3x - 2 = 0 ) are real. Using Vieta's formulas, the product of the roots of ( x^2 + 3x - 2 = 0 ) is -2. Thus, the product of the real roots is (-2).[boxed{A}]

🔑:1. Given the polynomial equation: [ x^4 + 3x^3 + 5x^2 + 21x - 14 = 0 ] We need to find the product of the real roots of this equation.2. We start by attempting to factorize the polynomial. Notice that we can group terms to facilitate factorization: [ x^4 + 3x^3 + 5x^2 + 21x - 14 = (x^4 + 5x^2 - 14) + (3x^3 + 21x) ]3. Factor out common terms from each group: [ (x^2 + 7)(x^2 - 2) + 3x(x^2 + 7) ]4. Notice that both groups contain the factor ((x^2 + 7)): [ (x^2 + 7)(x^2 - 2) + 3x(x^2 + 7) = (x^2 + 7)(x^2 - 2 + 3x) ]5. Simplify the expression inside the parentheses: [ (x^2 + 7)(x^2 + 3x - 2) ]6. Now, we need to find the real roots of the quadratic equation (x^2 + 3x - 2 = 0). We use the quadratic formula: [ x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ] where (a = 1), (b = 3), and (c = -2).7. Calculate the discriminant: [ Delta = b^2 - 4ac = 3^2 - 4 cdot 1 cdot (-2) = 9 + 8 = 17 ]8. Since the discriminant is positive, the quadratic equation has two distinct real roots: [ x = frac{-3 pm sqrt{17}}{2} ]9. The product of the roots of a quadratic equation (ax^2 + bx + c = 0) is given by (frac{c}{a}). For the equation (x^2 + 3x - 2 = 0), the product of the roots is: [ frac{c}{a} = frac{-2}{1} = -2 ]10. Therefore, the product of the real roots of the original polynomial equation is (boxed{-2}).

🧠:Alright, so the baker has 188 cups of flour and 113 cups of sugar. She wants to use all of both by making loaves of bread and cakes. Each loaf needs 3 cups of flour and 0.5 cups of sugar, and each cake needs 2 cups of flour and 2 cups of sugar. We need to figure out how many cakes she should make. Let me think through this step by step.First, let me define the variables. Let’s let B be the number of loaves of bread and C be the number of cakes. Then, the total flour used will be 3B (from the bread) plus 2C (from the cakes), and that has to equal 188 cups. Similarly, the total sugar used will be 0.5B (from the bread) plus 2C (from the cakes), which has to equal 113 cups. So, I can set up two equations here:1. Flour: 3B + 2C = 1882. Sugar: 0.5B + 2C = 113Now, I need to solve this system of equations to find the values of B and C. Since we’re interested in the number of cakes, which is C, maybe I can solve for C.Let me write the equations again for clarity:1. 3B + 2C = 1882. 0.5B + 2C = 113Hmm, both equations have a 2C term. Maybe I can subtract the second equation from the first to eliminate C. Let me try that.Subtracting equation 2 from equation 1:(3B + 2C) - (0.5B + 2C) = 188 - 113Calculating the left side: 3B - 0.5B + 2C - 2C = 2.5BRight side: 75So, 2.5B = 75To solve for B, divide both sides by 2.5:B = 75 / 2.5Let me compute that. 2.5 goes into 75 how many times? 2.5 * 30 = 75, right? Because 2.5 * 10 = 25, so 25 * 3 = 75, so 10 * 3 = 30. Yeah, so B = 30.So, the baker makes 30 loaves of bread. Now, plug this back into one of the equations to find C. Let me use equation 2 because the numbers might be smaller.Equation 2: 0.5B + 2C = 113Plugging B = 30:0.5*30 + 2C = 1130.5*30 is 15, so:15 + 2C = 113Subtract 15 from both sides:2C = 113 - 15 = 98Divide both sides by 2:C = 49So, she should make 49 cakes. Let me check if this works with the flour as well.Using equation 1: 3B + 2C = 1883*30 + 2*49 = 90 + 98 = 188Yes, that adds up. And for sugar:0.5*30 + 2*49 = 15 + 98 = 113That also checks out. So, both equations are satisfied. Therefore, the number of cakes she should make is 49.Wait a second, let me just make sure I didn't make any calculation errors. Let me go through the steps again.Starting with the two equations:3B + 2C = 1880.5B + 2C = 113Subtracting the second equation from the first gives:(3B - 0.5B) + (2C - 2C) = 188 - 113Which simplifies to 2.5B = 75Then B = 75 / 2.5. Let me convert 2.5 to a fraction to check. 2.5 is 5/2, so dividing by 5/2 is the same as multiplying by 2/5. So 75 * 2/5 = (75/5)*2 = 15*2 = 30. Yep, that's correct.Then substituting B = 30 into the second equation:0.5*30 + 2C = 1130.5*30 is 15, so 15 + 2C = 113. Subtract 15: 2C = 98. So C = 49. Correct.Check in the first equation:3*30 + 2*49 = 90 + 98 = 188. Correct.So all steps seem right. Maybe another way to solve it is by using substitution or matrices, but elimination worked here.Alternatively, let's try expressing one variable in terms of another. Let's solve the first equation for B.From 3B + 2C = 188,3B = 188 - 2CB = (188 - 2C)/3Then substitute into the second equation:0.5*((188 - 2C)/3) + 2C = 113Multiply 0.5 by the fraction:(0.5*(188 - 2C))/3 + 2C = 113Simplify 0.5*(188 - 2C):= 94 - CSo, (94 - C)/3 + 2C = 113Multiply both sides by 3 to eliminate the denominator:94 - C + 6C = 3*113Which is 94 + 5C = 339Subtract 94:5C = 339 - 94 = 245Then C = 245 / 5 = 49Same result. So that confirms it. Therefore, regardless of the method, the number of cakes is 49.I think that's solid. Both methods lead to C = 49. So, the answer must be 49 cakes.Final AnswerThe baker should make boxed{49} cakes.

✅:The baker has 188 cups of flour and 113 cups of sugar. Each loaf of bread requires 3 cups of flour and 0.5 cups of sugar, while each cake requires 2 cups of flour and 2 cups of sugar. We need to determine the number of cakes she should make to use up all the flour and sugar.Define variables:- ( B ) = number of loaves of bread- ( C ) = number of cakesSet up the system of equations based on the given constraints:1. Flour: ( 3B + 2C = 188 )2. Sugar: ( 0.5B + 2C = 113 )Subtract the second equation from the first to eliminate ( C ):[(3B + 2C) - (0.5B + 2C) = 188 - 113]Simplify:[2.5B = 75]Solve for ( B ):[B = frac{75}{2.5} = 30]Substitute ( B = 30 ) into the second equation:[0.5(30) + 2C = 113]Simplify:[15 + 2C = 113]Solve for ( C ):[2C = 98 implies C = 49]Verify the solution by substituting ( B = 30 ) and ( C = 49 ) back into the original equations:- Flour: ( 3(30) + 2(49) = 90 + 98 = 188 )- Sugar: ( 0.5(30) + 2(49) = 15 + 98 = 113 )Both equations are satisfied, confirming the solution is correct.[boxed{49}]

🔑:1. Let ( c ) be the number of cakes and ( b ) be the number of loaves of bread.2. Each loaf of bread requires 3 cups of flour and 0.5 cups of sugar.3. Each cake requires 2 cups of flour and 2 cups of sugar.4. We are given that the baker has 188 cups of flour and 113 cups of sugar.We can set up the following system of linear equations based on the given information:[3b + 2c = 188 quad text{(Equation 1: Flour)}][frac{1}{2}b + 2c = 113 quad text{(Equation 2: Sugar)}]5. To eliminate ( c ), we can subtract Equation 2 from Equation 1. First, we need to clear the fraction in Equation 2 by multiplying through by 2:[b + 4c = 226 quad text{(Equation 3: Sugar, cleared of fractions)}]6. Now, subtract Equation 1 from Equation 3:[(b + 4c) - (3b + 2c) = 226 - 188][b + 4c - 3b - 2c = 38][-2b + 2c = 38][-2b + 2c = 38 implies 2c - 2b = 38 implies c - b = 19 quad text{(Equation 4)}]7. From Equation 4, we can express ( c ) in terms of ( b ):[c = b + 19]8. Substitute ( c = b + 19 ) back into Equation 1:[3b + 2(b + 19) = 188][3b + 2b + 38 = 188][5b + 38 = 188][5b = 150][b = 30]9. Substitute ( b = 30 ) back into the expression for ( c ):[c = b + 19][c = 30 + 19][c = 49]The final answer is ( boxed{49} ).

🧠:Okay, let's try to tackle this problem. So we have a cubic monic polynomial P with roots a, b, and c. That means the polynomial can be written as P(x) = (x - a)(x - b)(x - c). Since it's monic, the leading coefficient is 1. We are given that P(1) = 91 and P(-1) = -121. We need to compute the maximum possible value of (ab + bc + ca)/(abc + a + b + c). Hmm, let's see.First, let's recall that for a cubic polynomial with roots a, b, c, the expanded form is x³ - (a + b + c)x² + (ab + bc + ca)x - abc. So the coefficients are related to the sums and products of the roots. Specifically, if we denote:- S = a + b + c- Q = ab + bc + ca- R = abcThen the polynomial is x³ - Sx² + Qx - R.Given that P is monic, these are indeed the coefficients. Now, let's compute P(1) and P(-1).Calculating P(1):P(1) = (1 - a)(1 - b)(1 - c) = 1 - S + Q - R = 91.Similarly, P(-1) = (-1 - a)(-1 - b)(-1 - c). Let's compute that:First, (-1 - a) = -(1 + a), similarly for the other terms. So:P(-1) = (-1 - a)(-1 - b)(-1 - c) = (-1)³(1 + a)(1 + b)(1 + c) = - (1 + a)(1 + b)(1 + c).Expanding (1 + a)(1 + b)(1 + c):1 + (a + b + c) + (ab + bc + ca) + abc = 1 + S + Q + R.Therefore, P(-1) = - (1 + S + Q + R) = -121.So now we have two equations:1. 1 - S + Q - R = 912. - (1 + S + Q + R) = -121Let me write these equations more clearly:Equation 1: 1 - S + Q - R = 91 ⇒ -S + Q - R = 90Equation 2: -1 - S - Q - R = -121 ⇒ -S - Q - R = -120 ⇒ S + Q + R = 120Wait, let me check that again. The second equation:P(-1) = - (1 + S + Q + R) = -121 ⇒ 1 + S + Q + R = 121 ⇒ S + Q + R = 120Yes, that's correct. So equation 2 is S + Q + R = 120Equation 1: 1 - S + Q - R = 91 ⇒ -S + Q - R = 90So now we have two equations:1. -S + Q - R = 902. S + Q + R = 120We can try to solve these two equations for variables S, Q, R. But there are three variables and only two equations, so we need another relation. However, since we are dealing with roots of a cubic, there might be some constraints on S, Q, R. Alternatively, maybe we can express some variables in terms of others.Let me denote equations again:Equation 1: -S + Q - R = 90Equation 2: S + Q + R = 120If we add Equation 1 and Equation 2, we get:(-S + Q - R) + (S + Q + R) = 90 + 120Simplifying:0 + 2Q + 0 = 210 ⇒ 2Q = 210 ⇒ Q = 105So Q = 105. That's the sum of the products of the roots two at a time.Now, substitute Q = 105 into the equations.Equation 1: -S + 105 - R = 90 ⇒ -S - R = 90 - 105 ⇒ -S - R = -15 ⇒ S + R = 15Equation 2: S + 105 + R = 120 ⇒ S + R = 15So both equations reduce to S + R = 15. Therefore, we have S + R = 15 and Q = 105.So the variables S and R are related by S + R = 15. So R = 15 - S.Our target expression is (ab + bc + ca)/(abc + a + b + c) = Q/(R + S). Since Q is 105, and R + S is 15. So Q/(R + S) = 105/15 = 7.Wait, that seems straightforward. But the question says "compute the maximum possible value of...". So if Q/(R + S) is always 105/15 = 7, then the value is fixed. But the problem states "maximum possible value", implying that there could be different possible values depending on S and R, but according to our equations, Q is fixed at 105, and R + S is fixed at 15. So Q/(R + S) is always 105/15 = 7. Therefore, the value is fixed, so the maximum possible value is 7.But that seems too straightforward. Maybe I made a mistake here. Let me check again.Wait, so the problem is asking for (ab + bc + ca)/(abc + a + b + c) which is Q/(R + S). From the equations, Q is 105, and R + S is 15, so indeed, 105/15 = 7. So regardless of the values of S and R (as long as they satisfy S + R = 15), this ratio is always 7. Therefore, the maximum possible value is 7.But why does the problem ask for the maximum possible value? If it's fixed, then the maximum is 7. But maybe there is a mistake in the reasoning.Wait, let's check all steps again.We have a cubic monic polynomial P(x) with roots a, b, c. Therefore, P(x) = (x - a)(x - b)(x - c) = x³ - Sx² + Qx - R, where S = a + b + c, Q = ab + bc + ca, R = abc.Compute P(1) = (1 - a)(1 - b)(1 - c) = 1 - S + Q - R. Given that this equals 91, so:1 - S + Q - R = 91 ⇒ -S + Q - R = 90. (Equation 1)Compute P(-1) = (-1 - a)(-1 - b)(-1 - c) = - (1 + a)(1 + b)(1 + c). Expanding (1 + a)(1 + b)(1 + c) = 1 + S + Q + R, so P(-1) = - (1 + S + Q + R) = -121. Therefore:- (1 + S + Q + R) = -121 ⇒ 1 + S + Q + R = 121 ⇒ S + Q + R = 120. (Equation 2)Adding Equation 1 and Equation 2:(-S + Q - R) + (S + Q + R) = 90 + 120 ⇒ 2Q = 210 ⇒ Q = 105.Then substituting Q = 105 into both equations:From Equation 1: -S - R = 90 - Q = 90 - 105 = -15 ⇒ S + R = 15From Equation 2: S + R = 120 - Q = 120 - 105 = 15. So same result.Therefore, regardless of the specific values of S and R, as long as they satisfy S + R = 15, the ratio (ab + bc + ca)/(abc + a + b + c) = Q/(R + S) = 105/15 = 7.Hence, this ratio is fixed, so the maximum possible value is 7.But the problem says "compute the maximum possible value". If the value is fixed, then the maximum is just 7. However, maybe there's something wrong here. Let me think again.Wait, perhaps there are constraints on S, Q, R beyond just S + R = 15 and Q = 105. Because S, Q, R must correspond to real numbers a, b, c such that the polynomial has real roots? Wait, the problem doesn't specify that the roots are real. It just says a cubic monic polynomial. So the roots could be complex. But the question is about the maximum possible value of the given expression. If complex roots are allowed, then perhaps S, Q, R can be complex, but in the problem statement, we are dealing with P(1) and P(-1) which are real numbers (91 and -121). So S, Q, R must be real numbers because the coefficients of the polynomial are real (since P(x) is a polynomial with real coefficients if the roots are real or complex conjugates). Wait, but even if roots are complex, the coefficients would still be real. Therefore, S, Q, R are real numbers.Therefore, even if the roots are complex, S, Q, R are real. So the equations we have are:Q = 105S + R = 15So regardless of whether the roots are real or complex, Q is fixed at 105, and S + R is fixed at 15. Therefore, the ratio Q/(R + S) is always 105/15 = 7, so the maximum possible value is 7.Therefore, the answer should be 7. But I need to confirm again. The problem is from David Altizio, so maybe there's a trick here. Let me check once more.Wait, let's compute (ab + bc + ca)/(abc + a + b + c). As we have Q = 105, and R + S = 15, so 105/15 = 7. So the value is 7. Therefore, the maximum possible value is 7. There is no variation here because Q is fixed, and R + S is fixed. Therefore, regardless of the values of S and R, the ratio is fixed. Therefore, the answer is 7. So maybe the problem is designed to look like an optimization problem, but in reality, the ratio is fixed, so the maximum is 7.But let me check with actual numbers. Suppose we choose specific values for S and R. For example, if we take S = 0, then R = 15. Then Q = 105. Then the ratio is 105/(15) = 7. If we take S = 15, then R = 0. Then the ratio is still 105/(15) = 7. If we take S = 10, R = 5. Then ratio is 105/15 = 7. So regardless of S and R, the ratio is fixed. Therefore, the maximum possible value is 7.Therefore, the answer is 7. So boxed answer is boxed{7}Final Answerboxed{7}

✅:Given a cubic monic polynomial ( P ) with roots ( a ), ( b ), and ( c ), we know that ( P(1) = 91 ) and ( P(-1) = -121 ). We need to compute the maximum possible value of (dfrac{ab + bc + ca}{abc + a + b + c}).The polynomial ( P(x) ) can be written as ( (x - a)(x - b)(x - c) ), which expands to ( x^3 - Sx^2 + Qx - R ) where ( S = a + b + c ), ( Q = ab + bc + ca ), and ( R = abc ).First, we calculate ( P(1) ):[P(1) = (1 - a)(1 - b)(1 - c) = 1 - S + Q - R = 91]This gives the equation:[1 - S + Q - R = 91 implies -S + Q - R = 90 quad text{(Equation 1)}]Next, we calculate ( P(-1) ):[P(-1) = (-1 - a)(-1 - b)(-1 - c) = - (1 + a)(1 + b)(1 + c) = - (1 + S + Q + R) = -121]This gives the equation:[- (1 + S + Q + R) = -121 implies 1 + S + Q + R = 121 implies S + Q + R = 120 quad text{(Equation 2)}]Adding Equation 1 and Equation 2:[(-S + Q - R) + (S + Q + R) = 90 + 120 implies 2Q = 210 implies Q = 105]Substituting ( Q = 105 ) into both equations, we get:[-S - R = 90 - 105 implies -S - R = -15 implies S + R = 15][S + 105 + R = 120 implies S + R = 15]Thus, ( Q = 105 ) and ( S + R = 15 ). The expression we need to evaluate is:[dfrac{ab + bc + ca}{abc + a + b + c} = dfrac{Q}{R + S} = dfrac{105}{15} = 7]Therefore, the maximum possible value is (boxed{7}).

🔑:1. Let ( P(x) ) be a monic cubic polynomial with roots ( alpha, beta, gamma ). Therefore, we can write: [ P(x) = x^3 + ax^2 + bx + c ] Given that ( P(1) = 91 ) and ( P(-1) = -121 ), we need to find the maximum value of: [ frac{alpha beta + beta gamma + gamma alpha}{alpha beta gamma + alpha + beta + gamma} ]2. By Vieta's formulas, we know: [ alpha + beta + gamma = -a ] [ alpha beta + beta gamma + gamma alpha = b ] [ alpha beta gamma = -c ]3. Using the given conditions: [ P(1) = 1^3 + a cdot 1^2 + b cdot 1 + c = 1 + a + b + c = 91 ] Therefore: [ a + b + c = 90 ]4. Similarly: [ P(-1) = (-1)^3 + a cdot (-1)^2 + b cdot (-1) + c = -1 + a - b + c = -121 ] Therefore: [ a - b + c = -120 ]5. We now have the system of equations: [ a + b + c = 90 ] [ a - b + c = -120 ]6. Subtract the second equation from the first: [ (a + b + c) - (a - b + c) = 90 - (-120) ] [ 2b = 210 ] [ b = 105 ]7. Add the two equations: [ (a + b + c) + (a - b + c) = 90 + (-120) ] [ 2a + 2c = -30 ] [ a + c = -15 ]8. We need to find the maximum value of: [ frac{alpha beta + beta gamma + gamma alpha}{alpha beta gamma + alpha + beta + gamma} = frac{b}{-(a + c)} ] Substituting the values we found: [ frac{b}{-(a + c)} = frac{105}{-(-15)} = frac{105}{15} = 7 ]The final answer is (boxed{7})

🧠:Okay, so I need to solve part (a) and part (b) of this problem. Let me start with part (a). The problem says that Lisa writes a positive whole number on the blackboard and in each turn, she deletes the last digit, then takes the remaining number (or 0 if there was only one digit) and adds four times the deleted digit. The new number is the result of this calculation. She repeats this until she gets a number that was already on the board before. We need to show that this sequence always ends.First, for part (a), I need to show that regardless of the starting number, the process will eventually reach a number that has already appeared, thus terminating the sequence. Since the numbers are generated in each turn, the key here is to argue that the sequence must eventually repeat a number because there are only finitely many possibilities. However, since the numbers can get larger or smaller, I need to check if the sequence is guaranteed to enter a cycle regardless of the starting number.Let me think. Let's denote the number on the board as N. When Lisa performs a move, she removes the last digit, say d, and then replaces the remaining number (which is floor(N/10)) with floor(N/10) + 4*d. Wait, no, actually, according to the problem statement: "the last digit is deleted from the number on the board and then the remaining shorter number (or 0 if the number was one digit) becomes four times the deleted number added." Wait, maybe I misread. Let me parse that again.Original number: Let's call it N. She deletes the last digit. Let's say the last digit is d, so the remaining number is M = floor(N / 10). Then, the new number is M + 4*d. So, in each step, N becomes floor(N / 10) + 4*(last digit of N). So, the transformation is T(N) = floor(N / 10) + 4*(N mod 10).For example, starting with 2022: floor(2022 / 10) = 202, last digit is 2. So, 202 + 4*2 = 210. Then 210 becomes 21 + 4*0 = 21, then 2 + 4*1 = 6, then 0 + 4*6 = 24, then 2 + 4*4 = 18, then 1 + 4*8 = 33, then 3 + 4*3 = 15, then 1 + 4*5 = 21. Since 21 was already there, the process stops.So, the transformation is T(N) = floor(N / 10) + 4*(N mod 10). Now, we need to show that starting from any N, the sequence N, T(N), T(T(N)), ... must eventually repeat a number.To show that the sequence always ends, we can use the pigeonhole principle. Since there are infinitely many numbers, but each number is transformed into another number. However, the problem is that numbers can potentially grow. For instance, if you have a number that ends with a high digit, adding 4*d could make it larger. For example, starting with 9: T(9) = 0 + 4*9 = 36. Then T(36) = 3 + 4*6 = 27, then T(27) = 2 + 4*7 = 30, T(30) = 3 + 4*0 = 3, T(3) = 0 + 4*3 = 12, T(12) = 1 + 4*2 = 9. So here, it cycles back to 9 after a few steps. But in this case, numbers can both increase and decrease.Wait, but even if numbers can sometimes increase, since the process is deterministic, each number leads to exactly one next number. Therefore, starting from any N, the sequence must eventually repeat a number because there are only finitely many numbers modulo some value. Wait, but actually, there are infinitely many numbers. However, if we can show that the numbers are bounded in some way, so that after some steps, the numbers can't exceed a certain value, then by the pigeonhole principle, the sequence must eventually repeat.Alternatively, perhaps there is a maximum number that can be achieved through this transformation, so that once numbers exceed that maximum, they start decreasing. Let's try to analyze the transformation.Let N be a k-digit number with digits d_{k-1}d_{k-2}...d_1d_0. Then, T(N) = floor(N / 10) + 4*d_0. Let's compare T(N) to N.N = 10*floor(N / 10) + d_0.So, T(N) = floor(N / 10) + 4*d_0 = (N - d_0)/10 + 4*d_0 = N/10 - d_0/10 + 4*d_0 = N/10 + (4 - 1/10)*d_0.Therefore, T(N) = N/10 + (39/10)*d_0. Since d_0 is between 0 and 9, (39/10)*d_0 ≤ 39/10*9 = 35.1. So, T(N) ≤ N/10 + 35.1.If N is large enough, say N > 351, then N/10 + 35.1 < N/10 + N/10 = N/5. So, if N > 351, then T(N) < N/5. Therefore, for N > 351, the sequence is decreasing. Thus, starting from any N, once the numbers get above 351, they will start decreasing by a factor of 1/5 each step. Hence, eventually, the sequence will enter a number ≤ 351, and from there, since there are only finitely many numbers below 351, the sequence must eventually repeat a number by the pigeonhole principle.Therefore, the sequence must always terminate. That's part (a). So, part (a) can be argued by showing that after some steps, the numbers decrease until they are below a certain bound, after which they can't decrease indefinitely and must repeat.Now, moving to part (b). Lisa starts with the number 53^{2022} - 1. We need to find the last number on the board before a repeat occurs. So, similar to the example, starting with 2022, the process ends when 21 is reached again. The last number would be the one before the repeat. Wait, but in the example, the process ends when 21 is reached for the second time, so the last number on the board would be 21. But in the problem statement, they say "the last number on the board" when the process ends. Wait, the problem says "until she gets a number for the first time was on the board." Wait, maybe mistranslation. The original says: "until she gets a number for the first time was on the board." Probably, it should be "until she gets a number that was already on the board before." So, the process stops when a number is repeated. Therefore, the sequence is the numbers from the start up to, but not including, the repeated number. So, the last number is the one before the repeat. Wait, but in the example, starting with 2022, the sequence is 2022 → 210 → 21 → 6 → 24 → 18 → 33 → 15 → 21. Then, since 21 was already there, the process ends. Therefore, the last number on the board is 21, which is the repeated one. Wait, but in the example description, they say "the result" is 21. So, maybe the last number is the first repeated one, i.e., the cycle starts at 21. Therefore, the last number is 21. So, in that case, the problem might be asking for the first repeated number, which is the start of the cycle. Therefore, perhaps part (b) is asking for the cycle entry point, i.e., the number that first gets repeated when starting from 53^{2022} - 1.Alternatively, the problem says "what is the last number on the board?" when the process ends. Since the process ends when a number is repeated, the last number written on the board is the one that causes the termination, which is the repeated number. So, in the example, 21 is written again, so the last number is 21. Therefore, in part (b), we need to find the number that is first repeated when starting from 53^{2022} - 1. Hence, that would be the entry point of the cycle.So, perhaps part (b) requires us to find the number that is the first repeated number in the sequence starting from 53^{2022} - 1. Therefore, we need to analyze the transformation T(N) = floor(N / 10) + 4*(N mod 10) starting from 53^{2022} - 1 and find the first repeated number in the sequence.Given that 53^{2022} - 1 is a very large number, we cannot compute it directly. Therefore, we need to find some patterns or cycles in the transformation that would allow us to compute the result modulo some number, perhaps.Alternatively, since the process reduces the number each time unless the last digit is large, but given that 53^{2022} - 1 is a number which is congruent to -1 mod 5, so 53 ≡ 3 mod 5, so 53^{2022} ≡ 3^{2022} mod 5. Since 3^4 ≡ 1 mod 5, 2022 divided by 4 is 505*4 + 2, so 3^{2022} ≡ 3^2 ≡ 9 ≡ 4 mod 5. Therefore, 53^{2022} ≡ 4 mod 5, so 53^{2022} - 1 ≡ 3 mod 5. Therefore, 53^{2022} - 1 ≡ 3 mod 5. However, I'm not sure how this helps.Alternatively, note that the transformation T(N) can be considered modulo some number. Let me see. Let's consider the transformation T(N) = floor(N / 10) + 4*(N mod 10). Let me write this as T(N) = (N - (N mod 10)) / 10 + 4*(N mod 10) = N / 10 - (N mod 10)/10 + 4*(N mod 10) = N / 10 + (39/10)*(N mod 10). But since we are dealing with integers, T(N) must be an integer. Therefore, T(N) = (N - d)/10 + 4d where d = N mod 10.So, T(N) = (N - d)/10 + 4d = (N + 39d)/10. Therefore, for T(N) to be integer, (N + 39d) must be divisible by 10. Since d = N mod 10, N ≡ d mod 10, so N ≡ d mod 10. Then, N + 39d ≡ d + 39d = 40d ≡ 0 mod 10. Therefore, yes, T(N) is always an integer.Now, let's analyze the transformation T(N) modulo some small numbers. Perhaps modulo 13 or 7? Let's see. Let me see if the transformation T is linear modulo some modulus.Alternatively, note that in the example, the cycle starts at 21. Let's see the cycle in the example:21 → 6 → 24 → 18 → 33 → 15 → 21. So, the cycle length is 6. So, the cycle is of length 6 starting from 21.If we can find that the transformation T has cycles, then starting numbers eventually fall into a cycle. For part (a), we have to show that regardless of starting number, you eventually reach a cycle, which we did by bounding the number. For part (b), starting from 53^{2022} - 1, we need to see where it ends up.Alternatively, perhaps the transformation T is related to multiplication by 4 modulo some number. Let me see. If we consider the transformation T(N) = floor(N / 10) + 4*(N mod 10). Let's see, if N is written as 10a + b, where a is the number formed by all digits except the last, and b is the last digit (0 ≤ b ≤9), then T(N) = a + 4b.Therefore, T(N) = a + 4b where N = 10a + b. This resembles a linear transformation. If we think of N as the pair (a, b), then T(N) is a new number formed by a + 4b. However, a is itself a number, which can be represented as 10c + d, and so on.Alternatively, perhaps this transformation is similar to multiplying by 4 in some modulus. Let me check. Suppose we have N = 10a + b, then T(N) = a + 4b. Let's compare this to 4*N mod something. 4*N = 40a + 4b. If we take 4*N mod 39, we get 40a + 4b mod 39. Since 40 ≡ 1 mod 39, so 40a + 4b ≡ a + 4b mod 39. Therefore, 4*N ≡ T(N) mod 39. So, T(N) ≡ 4*N mod 39.Wait, that's interesting. Let me confirm that. Let N = 10a + b. Then 4*N = 40a + 4b. Since 40 ≡ 1 mod 39, then 40a ≡ a mod 39. Therefore, 40a + 4b ≡ a + 4b mod 39. But T(N) = a + 4b. Therefore, T(N) ≡ 4*N mod 39. Therefore, T(N) ≡ 4N mod 39.Therefore, the transformation T is congruent to multiplication by 4 modulo 39. This is a key insight. Therefore, if we consider the sequence N_0, N_1, N_2, ..., where N_{k+1} = T(N_k), then each term satisfies N_{k+1} ≡ 4*N_k mod 39. Therefore, the sequence modulo 39 is N_0, 4*N_0, 4^2*N_0, ..., 4^k*N_0 mod 39.Since 4 and 39 are coprime (gcd(4,39)=1), the multiplicative order of 4 modulo 39 can be calculated. The order of 4 modulo 13 is 6 because 4^6 ≡ 1 mod 13 (since 4^3=64≡12≡-1 mod 13, so 4^6≡1 mod 13). Modulo 3, 4≡1 mod 3, so the order of 4 modulo 3 is 1. Therefore, by the Chinese Remainder Theorem, the order of 4 modulo 39 is lcm(6,1)=6. Therefore, 4^6 ≡1 mod 39. Therefore, 4^6 ≡1 mod 39. Therefore, the sequence modulo 39 cycles every 6 steps.Therefore, if we can show that the transformation T is equivalent to multiplying by 4 modulo 39, then the sequence modulo 39 cycles every 6 steps. However, this is only modulo 39. The actual numbers may not cycle unless they reduce to the same number, but the congruence is helpful.Given that, since T(N) ≡ 4*N mod 39, then after k transformations, N_k ≡ 4^k*N_0 mod 39. Therefore, if we start with N_0 = 53^{2022} -1, then N_k ≡ 4^k*(53^{2022} -1) mod 39.But 53 mod 39 is 53 - 39 = 14. So, 53 ≡14 mod 39. Therefore, 53^{2022} ≡14^{2022} mod 39. Let's compute 14 mod 39. Let me compute 14^{2022} mod 39.Note that 39 = 3*13. We can compute 14^{2022} mod 3 and mod 13 separately.First, mod 3: 14 ≡ 2 mod 3, so 14^{2022} ≡2^{2022} mod 3. Since 2^2 ≡1 mod 3, so 2^{2022} = (2^2)^1011 ≡1^1011 ≡1 mod 3.Mod 13: 14 ≡1 mod 13. Therefore, 14^{2022} ≡1^{2022} ≡1 mod13.Therefore, by Chinese Remainder Theorem, 14^{2022} ≡1 mod 3 and ≡1 mod 13, so ≡1 mod 39. Therefore, 53^{2022} ≡14^{2022} ≡1 mod39. Therefore, 53^{2022} -1 ≡0 mod39. Therefore, N_0 = 53^{2022} -1 ≡0 mod39.Therefore, N_0 ≡0 mod39, so N_1 = T(N_0) ≡4*N_0 ≡0 mod39, and similarly, all subsequent terms N_k ≡0 mod39. Therefore, the entire sequence starting from N_0 is congruent to 0 modulo39. Therefore, all numbers in the sequence are divisible by 39.However, in the example, the cycle includes numbers like 21, which is 21 ≡21 mod39. But 21 is not 0 mod39. So, perhaps my conclusion is not directly helpful. Wait, but in the example, the starting number was 2022, which is 2022 mod39. Let's compute 2022 /39: 39*51=1989, 2022-1989=33. So 2022≡33 mod39. Then, T(2022)=210, which is 210 mod39: 39*5=195, 210-195=15, so 210≡15 mod39. Then T(210)=21≡21 mod39. Then T(21)=6≡6 mod39. Then T(6)=24≡24 mod39. Then T(24)=18≡18 mod39. Then T(18)=33≡33 mod39. Then T(33)=15≡15 mod39. Then T(15)=21≡21 mod39. So the cycle here is 21→6→24→18→33→15→21. All these numbers are different mod39 except when they repeat.But in our problem, starting with N_0 ≡0 mod39, so all numbers will be ≡0 mod39. Therefore, the sequence will consist of numbers divisible by39. So, the cycle that the sequence enters must consist of numbers divisible by39. However, in the example, the cycle contains numbers not divisible by39. So, perhaps starting from a multiple of39, the sequence will enter a different cycle, perhaps of numbers divisible by39. Let's check.Take N=39. Let's compute T(39): floor(39/10)=3, last digit=9. So, 3 + 4*9=3+36=39. So, T(39)=39. Therefore, 39 maps to itself. So, if we start with 39, the sequence is just 39→39, so it's a fixed point. Similarly, let's check another multiple of39, say 78. T(78)=7 + 4*8=7+32=39. Then T(39)=39. So, 78→39→39. So, 78 leads to 39.Another multiple: 117. T(117)=11 +4*7=11+28=39. So, 117→39. 156: 15 +4*6=15+24=39. Similarly, 195: 19 +4*5=19+20=39. 234:23 +4*4=23+16=39. 273:27 +4*3=27+12=39. 312:31 +4*2=31+8=39. 351:35 +4*1=35+4=39. 390:39 +4*0=39. So, all multiples of39 (from 39 upwards) map to39 in one step. 39 maps to itself.Therefore, starting from any multiple of39, the sequence will reach39 in one step and then stay there. Therefore, the cycle for multiples of39 is just the fixed point39.But in our problem, the starting number is53^{2022} -1, which is≡0 mod39. Therefore, the sequence starting from53^{2022} -1 will first go to39, then stay at39 indefinitely. Therefore, the process will end when39 is reached again. However, since the starting number is≡0 mod39, which is not necessarily39 itself. For example, if the starting number is39, then the next number is39, so it terminates immediately. If the starting number is78, then it goes to39, then next step is39 again, so the sequence ends when39 is repeated. So, the last number on the board is39.Wait, but if starting from N_0=78, which is≡0 mod39, then the sequence is78→39→39. Since39 was already on the board, the process stops. Therefore, the last number on the board is39. Similarly, starting from any multiple of39 (greater than39), the sequence reduces it to39 in one step, then39 repeats, so the last number is39. Therefore, for part (b), if the starting number is a multiple of39, then the sequence will end at39. Since53^{2022} -1≡0 mod39, then the sequence starting from53^{2022} -1 will eventually reach39, and then terminate when39 is written again. Therefore, the last number on the board is39.But wait, need to verify this. Let's take a larger multiple of39, say39*10=390. Then T(390)=39 +4*0=39. So, 390→39→39. So, the last number is39. Similarly, 39*100=3900. T(3900)=390 +4*0=390. Then T(390)=39 +0=39. Then T(39)=39. So, 3900→390→39→39. So, the last number is39. Therefore, regardless of how large the starting multiple of39 is, the process reduces it step by step to39, which then repeats. Therefore, the last number is39.Therefore, for part (b), the answer is39. So, boxed, boxed{39}.But let me check for another number. Take N=13. 13 is not a multiple of39. Let's compute T(13)=1 +4*3=1+12=13. So, T(13)=13. So,13 is a fixed point. Similarly, N=26: T(26)=2 +4*6=2+24=26. So,26 is a fixed point. Wait, so there are multiple cycles. For example, numbers like13,26, etc., map to themselves. So, the transformation T has multiple cycles: fixed points like13,26,39, etc.But in our case, since we start with a multiple of39, we enter the cycle at39. But if we start with other numbers, they might enter different cycles. For example, starting with2022, which is33 mod39, we entered the cycle21→6→24→18→33→15→21. But that cycle doesn't include multiples of39.But in our problem, starting with a number ≡0 mod39, which is53^{2022} -1, so the sequence will stay within multiples of39, and since the transformation reduces the number each time until it reaches39, then the process stops. Hence, the last number is39.Therefore, part (b) answer is39. Let me confirm with another example. Let's take N=39 itself. Starting with39, next step is39, so the process stops immediately. The last number is39. If N=78, then78→39→39, so last number is39. If N=117, then117→39→39, last number39. Therefore, regardless of the starting multiple of39, the last number is39.Therefore, the answer to part (b) is boxed{39}.Final Answer(a) The sequence always terminates because numbers eventually decrease until entering a cycle. (b) The last number on the board is boxed{39}.

✅:(a) To show that the sequence of moves always ends, we consider the transformation ( T(N) = leftlfloor frac{N}{10} rightrfloor + 4 cdot (N mod 10) ). For sufficiently large ( N ), this transformation reduces the number. Specifically, if ( N > 351 ), then ( T(N) < frac{N}{5} ). Once the number becomes small enough (i.e., ( N leq 351 )), the pigeonhole principle ensures that the sequence must eventually repeat a number, as there are only finitely many numbers below 351. Thus, the sequence will always terminate.(b) Starting with ( 53^{2022} - 1 ), we need to determine the last number on the board. We observe that the transformation ( T(N) ) is equivalent to multiplying by 4 modulo 39. Since ( 53 equiv 14 mod 39 ), we have ( 53^{2022} equiv 14^{2022} mod 39 ). Calculating ( 14^{2022} mod 39 ), we find it is congruent to 1 modulo 39. Thus, ( 53^{2022} - 1 equiv 0 mod 39 ). Any number congruent to 0 modulo 39 will transform to another multiple of 39. The process reduces such numbers until reaching 39, which is a fixed point. Therefore, the last number on the board is (boxed{39}). Final Answer(a) The sequence always terminates because numbers eventually decrease until entering a cycle. (b) The last number on the board is (boxed{39}).

🔑:1. Show that the sequence of moves always ends. Let's denote the number on the board as ( X ). In each turn, Lisa performs the following operation: [ X = 10a + b quad text{(where ( a ) is the number formed by all digits except the last one, and ( b ) is the last digit)} ] The new number ( Y ) is given by: [ Y = a + 4b ] We need to show that this sequence always ends. First, observe that if ( X ) is a single-digit number, the sequence will eventually repeat because there are only 10 possible single-digit numbers (0 through 9). Next, consider the case when ( X ) has more than one digit. If ( X geq 40 ), then ( a geq 4 ). We can show that ( X ) strictly decreases: [ X = 10a + b quad text{and} quad Y = a + 4b ] Since ( a geq 4 ), we have: [ 3a geq 12 > b implies 10a + b > a + 4b implies X > Y ] Therefore, the number on the board always decreases strictly until it falls below 40. Once it falls below 40, it will eventually repeat because all numbers from 0 to 39 are represented in the sequences provided. Hence, the sequence of moves always ends.2. If Lisa begins with the number ( 53^{2022} - 1 ), what is the last number on the board? We need to determine the final number on the board when Lisa starts with ( 53^{2022} - 1 ). First, note that: [ X to Y implies 4X equiv Y pmod{13} ] because: [ 40a + 4b equiv a + 4b pmod{13} ] Each operation multiplies the number on the board by 4 modulo 13. Since ( 53 equiv 1 pmod{13} ), we have: [ 53^{2022} equiv 1^{2022} equiv 1 pmod{13} ] Therefore: [ 53^{2022} - 1 equiv 1 - 1 equiv 0 pmod{13} ] The initial number is divisible by 13, so the final number will also be divisible by 13. The possible numbers are 13, 26, or 39. Next, we consider the modulo 3 condition. The number on the board does not change modulo 3: [ 53 equiv 2 pmod{3} implies 53^{2022} equiv 2^{2022} pmod{3} ] Since ( 2^2 equiv 1 pmod{3} ), we have: [ 2^{2022} equiv (2^2)^{1011} equiv 1^{1011} equiv 1 pmod{3} ] Therefore: [ 53^{2022} - 1 equiv 1 - 1 equiv 0 pmod{3} ] The final number must also be divisible by 3. The only number that is divisible by both 13 and 3 is 39. Therefore, the last number on the board is 39.The final answer is (boxed{39})