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| problem_data_list = [ | |
| """```json | |
| [ | |
| { | |
| "problem": "Consider the multiset of letters {A, A, B, B, C, C, D, D}. Find the number of distinct arrangements of these eight letters such that no two identical letters are adjacent.", | |
| "topic": "Combinatorics", | |
| "answer": "864", | |
| "reasoning": "1. Calculate the total number of distinct arrangements of the 8 letters: 8! / (2! * 2! * 2! * 2!) = 40320 / 16 = 2520. | |
| 2. Use the principle of inclusion-exclusion. Let S be the set of all arrangements. Let P_X be the property that the two letters X are adjacent (XX). We want to find |S| - |P_A U P_B U P_C U P_D|. | |
| 3. Calculate the sum of the sizes of single properties (e.g., |P_A|): Treat AA as a single block. We arrange {AA, B, B, C, C, D, D}. Number of arrangements = 7! / (2! * 2! * 2!) = 5040 / 8 = 630. There are 4 such properties (P_A, P_B, P_C, P_D). Sum = 4 * 630 = 2520. | |
| 4. Calculate the sum of the sizes of pairwise intersections (e.g., |P_A inter P_B|): Treat AA and BB as blocks. We arrange {AA, BB, C, C, D, D}. Number of arrangements = 6! / (2! * 2!) = 720 / 4 = 180. There are C(4, 2) = 6 pairs of properties. Sum = 6 * 180 = 1080. | |
| 5. Calculate the sum of the sizes of three-way intersections (e.g., |P_A inter P_B inter P_C|): Treat AA, BB, CC as blocks. We arrange {AA, BB, CC, D, D}. Number of arrangements = 5! / 2! = 120 / 2 = 60. There are C(4, 3) = 4 triplets of properties. Sum = 4 * 60 = 240. | |
| 6. Calculate the size of the four-way intersection (|P_A inter P_B inter P_C inter P_D|): Treat AA, BB, CC, DD as blocks. We arrange {AA, BB, CC, DD}. Number of arrangements = 4! = 24. There is C(4, 4) = 1 such case. Sum = 1 * 24 = 24. | |
| 7. Apply the inclusion-exclusion principle: Number of arrangements with no adjacent identical letters = |S| - (Sum(|P_i|) - Sum(|P_i inter P_j|) + Sum(|P_i inter P_j inter P_k|) - |P_A inter P_B inter P_C inter P_D|) = 2520 - (2520 - 1080 + 240 - 24) = 2520 - 2520 + 1080 - 240 + 24 = 1080 - 240 + 24 = 840 + 24 = 864." | |
| } | |
| ] | |
| ```""", | |
| """```json | |
| [ | |
| { | |
| "problem": "Consider the multiset of letters $\{A, A, B, B, C, C, D, D\}$. Find the number of distinct arrangements of these eight letters such that no two identical letters are adjacent.", | |
| "topic": "Combinatorics", | |
| "answer": "864", | |
| "reasoning": "1. Calculate the total number of distinct arrangements of the multiset $\{A, A, B, B, C, C, D, D\}$. This is given by the multinomial coefficient $8! / (2!2!2!2!) = 40320 / 16 = 2520$. | |
| 2. Use the Principle of Inclusion-Exclusion to find the number of arrangements where at least one pair of identical letters is adjacent. Let $P_A$ be the property that AA are adjacent, $P_B$ that BB are adjacent, $P_C$ that CC are adjacent, and $P_D$ that DD are adjacent. We want to find $N(\overline{P_A} \cap \overline{P_B} \cap \overline{P_C} \cap \overline{P_D}) = N(\text{total}) - N(P_A \cup P_B \cup P_C \cup P_D)$. | |
| 3. Calculate the terms for Inclusion-Exclusion: | |
| - Sum of single properties: $N(P_i)$. Treat a pair (e.g., AA) as a single unit. We arrange $\{AA, B, B, C, C, D, D\}$. Number of arrangements is $7! / (2!2!2!) = 5040 / 8 = 630$. There are $\binom{4}{1}=4$ such terms, so the sum is $4 \times 630 = 2520$. | |
| - Sum of pairs of properties: $N(P_i \cap P_j)$. Treat two pairs (e.g., AA, BB) as single units. We arrange $\{AA, BB, C, C, D, D\}$. Number of arrangements is $6! / (2!2!) = 720 / 4 = 180$. There are $\binom{4}{2}=6$ such terms, so the sum is $6 \times 180 = 1080$. | |
| - Sum of triplets of properties: $N(P_i \cap P_j \cap P_k)$. Treat three pairs (e.g., AA, BB, CC) as single units. We arrange $\{AA, BB, CC, D, D\}$. Number of arrangements is $5! / 2! = 120 / 2 = 60$. There are $\binom{4}{3}=4$ such terms, so the sum is $4 \times 60 = 240$. | |
| - Sum of quadruplets of properties: $N(P_A \cap P_B \cap P_C \cap P_D)$. Treat all four pairs as single units. We arrange $\{AA, BB, CC, DD\}$. Number of arrangements is $4! = 24$. There is $\binom{4}{4}=1$ such term, so the sum is $1 \times 24 = 24$. | |
| 4. Apply the Inclusion-Exclusion Principle: $N(P_A \cup P_B \cup P_C \cup P_D) = \sum N(P_i) - \sum N(P_i \cap P_j) + \sum N(P_i \cap P_j \cap P_k) - N(P_A \cap P_B \cap P_C \cap P_D) = 2520 - 1080 + 240 - 24 = 1656$. | |
| 5. The number of arrangements with no two identical letters adjacent is the total number of arrangements minus those where at least one pair is adjacent: $2520 - 1656 = 864$." | |
| } | |
| ] | |
| ```""", | |
| """```json | |
| [ | |
| { | |
| "problem": "Find the number of integer solutions to the equation $x_1 + x_2 + x_3 = 10$, subject to the following constraints:\n1. $1 \le x_1 \le 5$\n2. $x_2$ is a non-negative multiple of 3\n3. $2 \le x_3 \le 7$", | |
| "topic": "Combinatorics", | |
| "answer": "10", | |
| "reasoning": "The equation is $x_1 + x_2 + x_3 = 10$. Constraints are $1 \le x_1 \le 5$, $x_2 \in \{0, 3, 6, 9, ...\}$, and $2 \le x_3 \le 7$. Rewrite the equation as $x_2 = 10 - x_1 - x_3$. For $x_2$ to be a non-negative multiple of 3, we need $10 - x_1 - x_3 \ge 0$ and $10 - x_1 - x_3 \equiv 0 \pmod 3$. Let $S = x_1 + x_3$. The conditions are $S \le 10$ and $S \equiv 10 \pmod 3 \equiv 1 \pmod 3$. We examine the possible integer values for $x_1$ from 1 to 5 and for $x_3$ from 2 to 7. We need to find pairs $(x_1, x_3)$ within these ranges such that their sum $S = x_1 + x_3$ satisfies $S \le 10$ and $S \equiv 1 \pmod 3$. | |
| Possible values for $S$ are integers in the range $[1+2, 5+7] = [3, 12]$ that are $\le 10$ and $\equiv 1 \pmod 3$. These are $S \in \{4, 7, 10\}$. | |
| Case 1: $x_1 + x_3 = 4$. Pairs $(x_1, x_3)$ satisfying $1 \le x_1 \le 5$ and $2 \le x_3 \le 7$: $(1, 3), (2, 2)$. (2 solutions: $(1, 6, 3), (2, 6, 2)$) | |
| Case 2: $x_1 + x_3 = 7$. Pairs $(x_1, x_3)$ satisfying $1 \le x_1 \le 5$ and $2 \le x_3 \le 7$: $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2)$. (5 solutions: $(1, 3, 6), (2, 3, 5), (3, 3, 4), (4, 3, 3), (5, 3, 2)$) | |
| Case 3: $x_1 + x_3 = 10$. Pairs $(x_1, x_3)$ satisfying $1 \le x_1 \le 5$ and $2 \le x_3 \le 7$: $(3, 7), (4, 6), (5, 5)$. (3 solutions: $(3, 0, 7), (4, 0, 6), (5, 0, 5)$) | |
| The total number of integer solutions is the sum of solutions from each case: $2 + 5 + 3 = 10$." | |
| } | |
| ] | |
| ```""", | |
| """```json | |
| [ | |
| { | |
| "problem": "A total of 30 identical candies are to be distributed among four distinct children, labeled Child 1, Child 2, Child 3, and Child 4. Let $x_i$ be the number of candies received by Child $i$, where $x_i \ge 0$ is an integer for $i=1, 2, 3, 4$. The distribution must satisfy the following conditions:\n\n1. The total number of candies given to Child 1 and Child 2 is exactly 15.\n2. Child 1 receives at least 5 candies.\n3. Child 2 receives no more than 8 candies.\n4. Child 3 receives an even number of candies, and receives at least 2 candies.\n\nFind the number of distinct ways to distribute the candies according to these conditions.", | |
| "topic": "Combinatorics, Integer Solutions", | |
| "answer": "63", | |
| "reasoning": "The problem asks for the number of integer solutions to $x_1 + x_2 + x_3 + x_4 = 30$ with $x_i \ge 0$ subject to additional constraints. | |
| Condition 1 states $x_1 + x_2 = 15$. Since the total number of candies is 30, this implies $x_3 + x_4 = 30 - (x_1 + x_2) = 30 - 15 = 15$. The problem can be solved by finding the number of valid pairs $(x_1, x_2)$ and the number of valid pairs $(x_3, x_4)$ independently and multiplying the results. | |
| Part 1: Find the number of integer solutions for $(x_1, x_2)$ such that $x_1 + x_2 = 15$, $x_1 \ge 0$, $x_2 \ge 0$, $x_1 \ge 5$, and $x_2 \le 8$. | |
| From $x_1 + x_2 = 15$, we have $x_1 = 15 - x_2$. | |
| The constraint $x_1 \ge 5$ becomes $15 - x_2 \ge 5$, which simplifies to $x_2 \le 10$. | |
| The constraint $x_2 \le 8$ is given. | |
| Combining $x_2 \le 10$ and $x_2 \le 8$, we get $x_2 \le 8$. | |
| Also, $x_1 \ge 0$ means $15 - x_2 \ge 0$, so $x_2 \le 15$. This is satisfied by $x_2 \le 8$. | |
| And $x_2 \ge 0$. | |
| So we need integer solutions to $x_1 + x_2 = 15$ with $0 \le x_2 \le 8$. | |
| The possible values for $x_2$ are $0, 1, 2, 3, 4, 5, 6, 7, 8$. | |
| For each value of $x_2$, $x_1 = 15 - x_2$ is uniquely determined. Let's check if the $x_1$ constraint ($x_1 \ge 5$) is met. | |
| If $x_2=8$, $x_1=7$. $7 \ge 5$ (satisfied). | |
| If $x_2=0$, $x_1=15$. $15 \ge 5$ (satisfied). | |
| For $0 \le x_2 \le 8$, $x_1 = 15 - x_2$ will range from $15 - 8 = 7$ to $15 - 0 = 15$. All these values are $\ge 7$, hence $\ge 5$. | |
| So, the possible integer values for $x_2$ are $0, 1, \ldots, 8$. The number of such values is $8 - 0 + 1 = 9$. | |
| There are 9 distinct pairs $(x_1, x_2)$ satisfying the conditions. | |
| Part 2: Find the number of integer solutions for $(x_3, x_4)$ such that $x_3 + x_4 = 15$, $x_3 \ge 0$, $x_4 \ge 0$, $x_3$ is even, and $x_3 \ge 2$. | |
| $x_3$ must be an even integer and $x_3 \ge 2$. | |
| Also, since $x_4 \ge 0$ and $x_3 + x_4 = 15$, we must have $x_3 \le 15$. | |
| So, $x_3$ must be an even integer in the range $[2, 15]$. | |
| The possible values for $x_3$ are $2, 4, 6, 8, 10, 12, 14$. | |
| There are 7 such integer values for $x_3$. | |
| For each value of $x_3$, $x_4 = 15 - x_3$ is uniquely determined. Let's check if the $x_4$ constraint ($x_4 \ge 0$) is met. | |
| If $x_3=14$, $x_4=1$. $1 \ge 0$ (satisfied). | |
| If $x_3=2$, $x_4=13$. $13 \ge 0$ (satisfied). | |
| For all values of $x_3$ in $\{2, 4, \ldots, 14\}$, $x_4 = 15 - x_3$ will range from $15 - 14 = 1$ to $15 - 2 = 13$. All these values are $\ge 1$, hence $\ge 0$. | |
| So, there are 7 distinct pairs $(x_3, x_4)$ satisfying the conditions. | |
| The total number of distinct ways to distribute the candies is the product of the number of possibilities for $(x_1, x_2)$ and $(x_3, x_4)$. | |
| Total ways = (Number of valid $(x_1, x_2)$ pairs) $\times$ (Number of valid $(x_3, x_4)$ pairs) | |
| Total ways = $9 \times 7 = 63$. | |
| " | |
| } | |
| ] | |
| ```""", | |
| """```json | |
| [ | |
| { | |
| "problem": "Let $A(x) = 1+x^2+x^4+\dots+x^{20}$ and $B(x) = 1+x^3+x^6+\dots+x^{30}$. Find the coefficient of $x^{25}$ in the product $A(x)B(x)$.", | |
| "topic": "Coefficient finding in polynomial products", | |
| "answer": "3", | |
| "reasoning": "The polynomial $A(x)$ is a sum of terms of the form $x^{2i}$ where $i \in \{0, 1, \dots, 10\}$. The polynomial $B(x)$ is a sum of terms of the form $x^{3j}$ where $j \in \{0, 1, \dots, 10\}$. The product $A(x)B(x)$ contains terms obtained by multiplying a term from $A(x)$ and a term from $B(x)$. To find the coefficient of $x^{25}$ in $A(x)B(x)$, we need to find pairs of indices $(i, j)$ such that $x^{2i} \cdot x^{3j} = x^{2i+3j} = x^{25}$, where $0 \le i \le 10$ and $0 \le j \le 10$. We need to find the number of integer solutions $(i, j)$ to the equation $2i + 3j = 25$ subject to the constraints $0 \le i \le 10$ and $0 \le j \le 10$. | |
| We can iterate through possible values of $j$ (or $i$). Since $2i = 25 - 3j$, $25 - 3j$ must be a non-negative even number. | |
| Also, $0 \le j \le 10$. | |
| If $j=0$, $2i = 25$, $i = 12.5$, not an integer. | |
| If $j=1$, $2i = 25 - 3 = 22$, $i = 11$. $i=11$ is outside the range $0 \le i \le 10$. | |
| If $j=2$, $2i = 25 - 6 = 19$, not even. | |
| If $j=3$, $2i = 25 - 9 = 16$, $i = 8$. $i=8$ is in range $0 \le i \le 10$. $j=3$ is in range $0 \le j \le 10$. Solution: $(i, j) = (8, 3)$. | |
| If $j=4$, $2i = 25 - 12 = 13$, not even. | |
| If $j=5$, $2i = 25 - 15 = 10$, $i = 5$. $i=5$ is in range $0 \le i \le 10$. $j=5$ is in range $0 \le j \le 10$. Solution: $(i, j) = (5, 5)$. | |
| If $j=6$, $2i = 25 - 18 = 7$, not even. | |
| If $j=7$, $2i = 25 - 21 = 4$, $i = 2$. $i=2$ is in range $0 \le i \le 10$. $j=7$ is in range $0 \le j \le 10$. Solution: $(i, j) = (2, 7)$. | |
| If $j=8$, $2i = 25 - 24 = 1$, not even. | |
| If $j=9$, $2i = 25 - 27 = -2$, $i = -1$. $i=-1$ is outside the range $0 \le i \le 10$. | |
| If $j=10$, $2i = 25 - 30 = -5$, not even. | |
| The valid pairs $(i, j)$ are $(8, 3)$, $(5, 5)$, and $(2, 7)$. | |
| For each valid pair $(i, j)$, the product of the corresponding terms from $A(x)$ and $B(x)$ is $x^{2i} \cdot x^{3j} = x^{2i+3j} = x^{25}$. Since all coefficients in $A(x)$ and $B(x)$ are 1, the coefficient of $x^{2i+3j}$ in the product $A(x)B(x)$ is $1 \cdot 1 = 1$. | |
| The coefficient of $x^{25}$ is the sum of the coefficients for each valid pair $(i, j)$. | |
| The coefficient is $1$ (for $(8, 3)$) + $1$ (for $(5, 5)$) + $1$ (for $(2, 7)$) = 3. | |
| Alternatively, solve $2i + 3j = 25$ for $i \in \{0, 1, \dots, 10\}$. | |
| $3j = 25 - 2i$. We need $25 - 2i$ to be divisible by 3 and $0 \le (25 - 2i)/3 \le 10$. | |
| $25 - 2i \equiv 0 \pmod 3$. Since $25 \equiv 1 \pmod 3$ and $2i \equiv -i \cdot 2 \pmod 3 \equiv -i \cdot (-1) \pmod 3 \equiv i \pmod 3$, we have $1 + i \equiv 0 \pmod 3$, so $i \equiv -1 \equiv 2 \pmod 3$. | |
| Possible values for $i$ in the range $0 \le i \le 10$ are $2, 5, 8$. | |
| If $i=2$, $3j = 25 - 4 = 21$, $j = 7$. $0 \le 7 \le 10$, valid. $(i, j) = (2, 7)$. | |
| If $i=5$, $3j = 25 - 10 = 15$, $j = 5$. $0 \le 5 \le 10$, valid. $(i, j) = (5, 5)$. | |
| If $i=8$, $3j = 25 - 16 = 9$, $j = 3$. $0 \le 3 \le 10$, valid. $(i, j) = (8, 3)$. | |
| If $i=11$, $11 > 10$, not in range. | |
| The valid pairs are $(2, 7), (5, 5), (8, 3)$. There are 3 such pairs. | |
| The coefficient of $x^{25}$ is 3. | |
| " | |
| } | |
| ] | |
| ```""", | |
| """```json | |
| [ | |
| { | |
| "problem": "Let $A(x) = 1+x^2+x^4+\dots+x^{20}$ and $B(x) = 1+x^3+x^6+\dots+x^{30}$. Find the coefficient of $x^{25}$ in the product $A(x)B(x)$.", | |
| "topic": "Coefficient of a polynomial product", | |
| "answer": "3", | |
| "reasoning": "The polynomial $A(x)$ is a sum of terms $x^{2k}$ for $k \in \{0, 1, \dots, 10\}$. The polynomial $B(x)$ is a sum of terms $x^{3j}$ for $j \in \{0, 1, \dots, 10\}$. The product $A(x)B(x)$ consists of terms of the form $x^{2k} \cdot x^{3j} = x^{2k+3j}$. To find the coefficient of $x^{25}$, we need to find the number of pairs of integers $(k, j)$ such that $2k+3j = 25$, with the constraints $0 \le k \le 10$ and $0 \le j \le 10$. We solve the equation $2k+3j=25$ subject to these constraints. | |
| From $2k = 25 - 3j$, we require $25 - 3j$ to be a non-negative even number. | |
| $25 - 3j \ge 0 \implies 3j \le 25 \implies j \le 8$. | |
| $25 - 3j$ is even implies $3j$ is odd, which means $j$ must be odd. | |
| Considering the constraint $0 \le j \le 10$, the possible odd values for $j$ are 1, 3, 5, 7, 9. | |
| Combined with $j \le 8$, the possible values for $j$ are 1, 3, 5, 7. | |
| For each possible value of $j$, we find the corresponding value of $k$ and check if it satisfies $0 \le k \le 10$: | |
| If $j=1$, $2k = 25 - 3(1) = 22 \implies k=11$. This violates $k \le 10$. | |
| If $j=3$, $2k = 25 - 3(3) = 16 \implies k=8$. This satisfies $0 \le 8 \le 10$. Valid pair: $(k,j)=(8,3)$. | |
| If $j=5$, $2k = 25 - 3(5) = 10 \implies k=5$. This satisfies $0 \le 5 \le 10$. Valid pair: $(k,j)=(5,5)$. | |
| If $j=7$, $2k = 25 - 3(7) = 4 \implies k=2$. This satisfies $0 \le 2 \le 10$. Valid pair: $(k,j)=(2,7)$. | |
| If $j=9$, $2k = 25 - 3(9) = -2 \implies k=-1$. This violates $k \ge 0$. | |
| The valid pairs $(k, j)$ are $(8, 3)$, $(5, 5)$, and $(2, 7)$. | |
| Each term in $A(x)$ and $B(x)$ has a coefficient of 1. Thus, the coefficient of $x^{25}$ in $A(x)B(x)$ is the number of such valid pairs $(k, j)$. | |
| There are 3 valid pairs. | |
| The coefficient of $x^{25}$ is 3." | |
| } | |
| ] | |
| ```""", | |
| """ | |
| ```json | |
| [ | |
| { | |
| "problem": "Solve the following problem using a **concise chain of thought**.\n Clearly and directly walk through the key reasoning steps needed to arrive at the answer.\n Avoid unnecessary elaboration or backtracking—keep it efficient, but logically sound.\n\n Problem: Let $S(n, m) = \sum_{k=0}^n k^2 \binom{n}{k} \binom{m}{k}$ for positive integers $n$ and $m$. Find a closed-form expression for $S(n,m)$ in terms of $n$ and $m$.\n\n Final Answer:\n", | |
| "topic": "Combinatorial Identities", | |
| "answer": "$nm \\binom{n+m-2}{n-1}$", | |
| "reasoning": "1. Rewrite the term $k^2 \\binom{n}{k} \\binom{m}{k}$ using the identity $k \\binom{N}{k} = N \\binom{N-1}{k-1}$ twice. The summation runs from $k=0$, but the $k=0$ term is zero. The identity $k \\binom{n}{k} = n \\binom{n-1}{k-1}$ holds for $k \ge 1$, and $k \\binom{m}{k} = m \\binom{m-1}{k-1}$ holds for $k \ge 1$. | |
| 2. $k^2 \\binom{n}{k} \\binom{m}{k} = (k \\binom{n}{k}) (k \\binom{m}{k}) = \\left( n \\binom{n-1}{k-1} \\right) \\left( m \\binom{m-1}{k-1} \\right)$ for $k \\ge 1$. | |
| 3. Substitute this back into the sum $S(n,m)$. The $k=0$ term is 0, so the sum can start from $k=1$. | |
| $S(n,m) = \\sum_{k=1}^n nm \\binom{n-1}{k-1} \\binom{m-1}{k-1}$. | |
| 4. Factor out the constants $nm$: $S(n,m) = nm \\sum_{k=1}^n \\binom{n-1}{k-1} \\binom{m-1}{k-1}$. | |
| 5. Change the index of summation by setting $j=k-1$. When $k=1$, $j=0$. When $k=n$, $j=n-1$. The sum becomes: | |
| $S(n,m) = nm \\sum_{j=0}^{n-1} \\binom{n-1}{j} \\binom{m-1}{j}$. | |
| 6. Apply the identity $\\sum_{j=0}^r \\binom{A}{j} \\binom{B}{j} = \\binom{A+B}{A} = \\binom{A+B}{B}$. This identity is derived from Vandermonde's Identity by using $\\binom{B}{j} = \\binom{B}{B-j}$, giving $\\sum_{j=0}^r \\binom{A}{j} \\binom{B}{B-j} = \\binom{A+B}{B}$. The sum effectively runs up to $\min(n-1, m-1)$. With $A=n-1$ and $B=m-1$, the sum is equal to $\\binom{(n-1)+(m-1)}{n-1} = \\binom{n+m-2}{n-1}$ or $\\binom{(n-1)+(m-1)}{m-1} = \\binom{n+m-2}{m-1}$. | |
| 7. Substitute the result of the sum back into the expression for $S(n,m)$: | |
| $S(n,m) = nm \\binom{n+m-2}{n-1}$. | |
| 8. Using the symmetry property $\\binom{N}{K} = \\binom{N}{N-K}$, we can also write $\\binom{n+m-2}{n-1} = \\binom{n+m-2}{(n+m-2)-(n-1)} = \\binom{n+m-2}{m-1}$. Both forms are valid closed-form expressions. Let's choose one, e.g., $nm \\binom{n+m-2}{n-1}$. This holds for positive integers $n,m$. For $n=1$, $m \binom{m-1}{0} = m$. For $m=1$, $n \binom{n-1}{n-1} = n$. This matches the values computed directly from the sum definition for these cases." | |
| } | |
| ] | |
| ``` | |
| """, | |
| """```json | |
| [ | |
| { | |
| "problem": "Let $p$ be a prime number such that $M_p = 2^p-1$ is a Mersenne prime. Let $E_p = 2^{p-1}M_p$ be the associated even perfect number. Find the remainder when $\sigma(E_p^2)$, the sum of the positive divisors of $E_p^2$, is divided by $M_p$.", | |
| "topic": "Number Theory", | |
| "answer": "$2^{p-1}-1$", | |
| "reasoning": "The given perfect number is $E_p = 2^{p-1}M_p$, where $M_p = 2^p-1$ is a prime number. We want to find the sum of divisors of $E_p^2$. First, find the prime factorization of $E_p^2$: $E_p^2 = (2^{p-1}M_p)^2 = 2^{2(p-1)}M_p^2$. Since 2 and $M_p$ are distinct primes, the sum of divisors function $\sigma$ is multiplicative. Thus, $\sigma(E_p^2) = \sigma(2^{2(p-1)}) \sigma(M_p^2)$. | |
| The sum of divisors of a prime power $q^k$ is $\sigma(q^k) = \frac{q^{k+1}-1}{q-1}$. | |
| For the factor $2^{2(p-1)}$, the sum of divisors is $\sigma(2^{2(p-1)}) = \frac{2^{2(p-1)+1}-1}{2-1} = 2^{2p-2+1}-1 = 2^{2p-1}-1$. | |
| For the factor $M_p^2$, the sum of divisors is $\sigma(M_p^2) = \frac{M_p^{2+1}-1}{M_p-1} = \frac{M_p^3-1}{M_p-1} = M_p^2 + M_p + 1$. | |
| So, $\sigma(E_p^2) = (2^{2p-1}-1)(M_p^2 + M_p + 1)$. | |
| We need to find the remainder when $\sigma(E_p^2)$ is divided by $M_p$. This is $\sigma(E_p^2) \pmod{M_p}$. | |
| $\sigma(E_p^2) \equiv (2^{2p-1}-1)(M_p^2 + M_p + 1) \pmod{M_p}$. | |
| Since $M_p \equiv 0 \pmod{M_p}$, we have $M_p^2 + M_p + 1 \equiv 0^2 + 0 + 1 \equiv 1 \pmod{M_p}$. | |
| So, $\sigma(E_p^2) \equiv (2^{2p-1}-1)(1) \pmod{M_p} \equiv 2^{2p-1}-1 \pmod{M_p}$. | |
| We know that $M_p = 2^p-1$, which implies $2^p \equiv 1 \pmod{M_p}$. | |
| Then $2^{2p-1} = 2^{2p} \cdot 2^{-1} = (2^p)^2 \cdot 2^{-1}$. | |
| $2^{2p-1} \equiv (1)^2 \cdot 2^{-1} \pmod{M_p} \equiv 2^{-1} \pmod{M_p}$. | |
| To find the modular inverse of 2 modulo $M_p$, we look for a number $x$ such that $2x \equiv 1 \pmod{M_p}$. Since $2^p = M_p + 1$, we have $2 \cdot 2^{p-1} = M_p + 1$. | |
| This means $2 \cdot 2^{p-1} \equiv 1 \pmod{M_p}$. | |
| So, $2^{-1} \equiv 2^{p-1} \pmod{M_p}$. | |
| Substituting this back, we get $2^{2p-1}-1 \equiv 2^{-1}-1 \pmod{M_p} \equiv 2^{p-1}-1 \pmod{M_p}$. | |
| The remainder when $\sigma(E_p^2)$ is divided by $M_p$ is $2^{p-1}-1$. Since $p$ is prime, $p \ge 2$, so $2^{p-1}-1 < 2^p-1 = M_p$, confirming it is a valid remainder." | |
| } | |
| ] | |
| ``` | |
| """, | |
| """```json | |
| [ | |
| { | |
| "problem": "Let $p$ be a prime number such that $M_p = 2^p-1$ is a Mersenne prime. Let $E_p = 2^{p-1}M_p$ be the associated even perfect number. Find the remainder when $\sigma(E_p^2)$, the sum of the positive divisors of $E_p^2$, is divided by $M_p$.", | |
| "topic": "Number Theory", | |
| "answer": "$2^{p-1}-1$", | |
| "reasoning": "Let $E_p = 2^{p-1}M_p$. We are given that $p$ is a prime and $M_p = 2^p-1$ is a Mersenne prime, which means $M_p$ is also a prime number. | |
| We need to find the sum of the divisors of $E_p^2$. | |
| First, find the prime factorization of $E_p^2$: | |
| $E_p^2 = (2^{p-1}M_p)^2 = (2^{p-1})^2 (M_p)^2 = 2^{2(p-1)} M_p^2 = 2^{2p-2} M_p^2$. | |
| Since 2 and $M_p$ are distinct primes, the sum of divisors function $\sigma(n)$ can be applied directly to the prime factorization. | |
| The formula for the sum of divisors of $n = q_1^{a_1} q_2^{a_2} \cdots q_k^{a_k}$ is $\sigma(n) = \sigma(q_1^{a_1}) \sigma(q_2^{a_2}) \cdots \sigma(q_k^{a_k})$. | |
| Also, $\sigma(q^a) = 1 + q + \cdots + q^a = \frac{q^{a+1}-1}{q-1}$. | |
| Applying this to $E_p^2$: | |
| $\sigma(E_p^2) = \sigma(2^{2p-2}) \sigma(M_p^2)$. | |
| Calculate each part: | |
| $\sigma(2^{2p-2}) = \frac{2^{(2p-2)+1}-1}{2-1} = \frac{2^{2p-1}-1}{1} = 2^{2p-1}-1$. | |
| $\sigma(M_p^2) = \frac{M_p^{2+1}-1}{M_p-1} = \frac{M_p^3-1}{M_p-1} = \frac{(M_p-1)(M_p^2+M_p+1)}{M_p-1} = M_p^2+M_p+1$. | |
| So, $\sigma(E_p^2) = (2^{2p-1}-1)(M_p^2+M_p+1)$. | |
| We need to find the remainder when $\sigma(E_p^2)$ is divided by $M_p$. This is $\sigma(E_p^2) \pmod{M_p}$. | |
| $\sigma(E_p^2) \equiv (2^{2p-1}-1)(M_p^2+M_p+1) \pmod{M_p}$. | |
| Consider the second factor modulo $M_p$: | |
| $M_p^2+M_p+1 \equiv 0^2 + 0 + 1 \equiv 1 \pmod{M_p}$ since $M_p \equiv 0 \pmod{M_p}$. | |
| So, $\sigma(E_p^2) \equiv (2^{2p-1}-1) \cdot 1 \pmod{M_p}$. | |
| $\sigma(E_p^2) \equiv 2^{2p-1}-1 \pmod{M_p}$. | |
| Now, we use the property $M_p = 2^p-1$, which implies $2^p \equiv 1 \pmod{M_p}$. | |
| We want to simplify $2^{2p-1} \pmod{M_p}$. | |
| $2^{2p-1} = 2^{2p} \cdot 2^{-1} = (2^p)^2 \cdot 2^{-1}$. | |
| Using $2^p \equiv 1 \pmod{M_p}$, we get: | |
| $2^{2p-1} \equiv (1)^2 \cdot 2^{-1} \equiv 2^{-1} \pmod{M_p}$. | |
| To find $2^{-1} \pmod{M_p}$, we can use $2^p \equiv 1 \pmod{M_p}$. Multiplying by $2^{-1}$ (which exists since $M_p=2^p-1$ is odd), we get $2^{p-1} \equiv 2^{-1} \pmod{M_p}$. | |
| Alternatively, write $2^{2p-1} = 2^p \cdot 2^{p-1}$. | |
| $2^{2p-1} \equiv 1 \cdot 2^{p-1} \equiv 2^{p-1} \pmod{M_p}$. | |
| Substituting this back into the expression for the remainder: | |
| $\sigma(E_p^2) \equiv 2^{p-1}-1 \pmod{M_p}$. | |
| The expression $2^{p-1}-1$ is the remainder if it is between $0$ and $M_p-1$. | |
| Since $p$ is a prime, $p \ge 2$. | |
| If $p=2$, $2^{p-1}-1 = 2^1-1 = 1$. $M_2 = 2^2-1 = 3$. $1$ is the remainder modulo 3. | |
| If $p \ge 3$, $p-1 \ge 2$, so $2^{p-1} \ge 4$. | |
| $2^{p-1}-1 \ge 3$. | |
| Also, $2^{p-1}-1 < 2^p-1 = M_p$. | |
| So, $0 \le 2^{p-1}-1 < M_p$ for any prime $p$. | |
| Therefore, $2^{p-1}-1$ is the correct remainder. | |
| The final answer is $2^{p-1}-1$. | |
| " | |
| } | |
| ] | |
| ``` | |
| """, | |
| """```json | |
| [ | |
| { | |
| "problem": "Let $x$ be a real number such that $\lfloor x^2 \\rfloor = 1000000$. Find the number of possible integer values for $\lfloor (x+2)^2 \\rfloor - \lfloor x^2 \\rfloor$.", | |
| "topic": "Number Theory", | |
| "answer": "3", | |
| "reasoning": "Let $N = 1000000$. The condition $\lfloor x^2 \rfloor = N$ implies $N \le x^2 < N+1$. We are asked to find the number of possible integer values for $V = \lfloor (x+2)^2 \rfloor - \lfloor x^2 \rfloor$. Expanding the expression, we get $V = \lfloor x^2 + 4x + 4 \rfloor - \lfloor x^2 \rfloor$. Using the property $\lfloor y+k \rfloor = \lfloor y \rfloor + k$ for an integer $k$, we have $\lfloor x^2 + 4x + 4 \rfloor = \lfloor x^2 + 4x \rfloor + 4$. Thus, $V = \lfloor x^2 + 4x \rfloor + 4 - \lfloor x^2 \rfloor$. Let $f = x^2 - \lfloor x^2 \rfloor = x^2 - N$. Since $N \le x^2 < N+1$, we have $0 \le f < 1$. The expression becomes $V = \lfloor (N+f) + 4x \rfloor + 4 - N = \lfloor N + f + 4x \rfloor + 4 - N = N + \lfloor f + 4x \rfloor + 4 - N = \lfloor f + 4x \rfloor + 4$. We need to find the possible integer values of $\lfloor f + 4x \rfloor = \lfloor x^2 - N + 4x \rfloor$. | |
| The condition $N \le x^2 < N+1$ implies $\sqrt{N} \le |x| < \sqrt{N+1}$. Since $N=1000000$, $\sqrt{N}=1000$. So $1000 \le |x| < \sqrt{1000001}$. | |
| Case 1: $x \ge 0$. | |
| In this case, $1000 \le x < \sqrt{1000001}$. | |
| Multiplying by 4, we get $4000 \le 4x < 4\sqrt{1000001}$. | |
| Let $g(x) = x^2 - N + 4x$. We need to find the range of $\lfloor g(x) \rfloor$ for $x \in [1000, \sqrt{1000001})$. | |
| The function $g(x) = x^2 + 4x - 1000000$ is increasing for $x > -2$, so it is increasing on $[1000, \sqrt{1000001})$. | |
| The minimum value of $g(x)$ is $g(1000) = 1000^2 - 1000000 + 4(1000) = 1000000 - 1000000 + 4000 = 4000$. | |
| The limit as $x$ approaches $\sqrt{1000001}$ from below is $g(\sqrt{1000001}^-) = (\sqrt{1000001})^2 - 1000000 + 4\sqrt{1000001} = 1000001 - 1000000 + 4\sqrt{1000001} = 1 + 4\sqrt{1000001}$. | |
| Let's estimate $4\sqrt{1000001} = \sqrt{16 \times 1000001} = \sqrt{16000016}$. | |
| $4000^2 = 16000000$, $4001^2 = 16008001$, $4002^2 = 16016004$. | |
| So $4000 < \sqrt{16000016} < 4001$ is incorrect. $4000^2=16000000$, $4001^2 = 16008001$, $4002^2=16016004$, $4003^2 = 16024009$. | |
| $\sqrt{16000016}$ is between $4000$ and $4001$. | |
| $4000^2 = 16000000$. $4000.001^2 = (4000+0.001)^2 = 16000000 + 8 + 0.000001 = 16000008.000001$. | |
| $4000.002^2 = (4000+0.002)^2 = 16000000 + 16 + 0.000004 = 16000016.000004$. | |
| So $\sqrt{16000016} \approx 4000.001999...$ | |
| $4\sqrt{1000001} \approx 4000.001999...$ | |
| The range of $g(x)$ for $x \ge 0$ is $[4000, 1 + 4\sqrt{1000001}) = [4000, 1 + 4000.001999...) = [4000, 4001.001999...)$. | |
| The possible integer values for $\lfloor g(x) \rfloor$ are $\lfloor 4000 \rfloor = 4000$ and $\lfloor 4001.something \rfloor = 4001$. Both values are achievable because the interval $[4000, 4001.001999...)$ contains numbers in $[4000, 4001)$ and numbers in $[4001, 4001.001999...)$. | |
| Case 2: $x < 0$. | |
| In this case, $-\sqrt{1000001} < x \le -1000$. | |
| Multiplying by 4, we get $-4\sqrt{1000001} < 4x \le -4000$. | |
| $-4000.001999... < 4x \le -4000$. | |
| Let $h(x) = x^2 - N + 4x$. We need the range of $\lfloor h(x) \rfloor$ for $x \in (-\sqrt{1000001}, -1000]$. | |
| The function $h(x) = x^2 + 4x - 1000000$ is decreasing for $x < -2$, so it is decreasing on $(-\sqrt{1000001}, -1000]$. | |
| The maximum value of $h(x)$ is $h(-1000) = (-1000)^2 - 1000000 + 4(-1000) = 1000000 - 1000000 - 4000 = -4000$. | |
| The limit as $x$ approaches $-\sqrt{1000001}$ from above is $h(-\sqrt{1000001}^+) = (\sqrt{1000001})^2 - 1000000 + 4(-\sqrt{1000001}) = 1000001 - 1000000 - 4\sqrt{1000001} = 1 - 4\sqrt{1000001}$. | |
| $1 - 4\sqrt{1000001} \approx 1 - 4000.001999... = -3999.001999...$ | |
| The range of $h(x)$ for $x < 0$ is $(1 - 4\sqrt{1000001}, -4000] = (-3999.001999..., -4000]$. | |
| The only integer in this interval is $-4000$. So the only possible value for $\lfloor h(x) \rfloor$ is $-4000$. | |
| Combining both cases, the possible integer values for $\lfloor x^2 - 1000000 + 4x \rfloor$ are $\{4000, 4001\}$ from Case 1 and $\{-4000\}$ from Case 2. | |
| The set of possible integer values for $\lfloor x^2 - 1000000 + 4x \rfloor$ is $\{-4000, 4000, 4001\}$. | |
| The possible integer values for $\lfloor (x+2)^2 \rfloor - \lfloor x^2 \rfloor = 4 + \lfloor x^2 - 1000000 + 4x \rfloor$ are: | |
| $4 + (-4000) = -3996$ | |
| $4 + 4000 = 4004$ | |
| $4 + 4001 = 4005$ | |
| These three values $-3996, 4004, 4005$ are distinct integer values. All three values are achievable as shown in the analysis of the ranges of $\lfloor g(x) \rfloor$ and $\lfloor h(x) \rfloor$. | |
| The number of possible integer values is 3. | |
| The final answer is $\boxed{3}$." | |
| } | |
| ] | |
| ``` | |
| """, | |
| """2025-05-03 22:12:54 | INFO | [JOB START] Master Job ID: 84eaf825-e75f-4af3-9b8f-0a5291a448fd | Logging progress every 3 seconds | |
| 2025-05-03 22:12:54 | INFO | [JOB PROGRESS] Completed: 0/2 | Running: 1 | Attempted: 0 (Completed: 0, Failed: 0, Filtered: 0, Duplicate: 0) | |
| START A fair coin is flipped repeatedly. What is the expected number of flips until the sequence HHH appears for the first time OR the sequence TTT appears for the first time? | |
| 2025-05-03 22:12:57 | INFO | [JOB PROGRESS] Completed: 0/2 | Running: 1 | Attempted: 0 (Completed: 0, Failed: 0, Filtered: 0, Duplicate: 0) | |
| 2025-05-03 22:13:00 | INFO | [JOB PROGRESS] Completed: 0/2 | Running: 1 | Attempted: 0 (Completed: 0, Failed: 0, Filtered: 0, Duplicate: 0) | |
| 2025-05-03 22:13:03 | INFO | [JOB PROGRESS] Completed: 0/2 | Running: 1 | Attempted: 0 (Completed: 0, Failed: 0, Filtered: 0, Duplicate: 0) | |
| 2025-05-03 22:13:06 | INFO | [JOB PROGRESS] Completed: 0/2 | Running: 1 | Attempted: 0 (Completed: 0, Failed: 0, Filtered: 0, Duplicate: 0) | |
| 2025-05-03 22:13:09 | INFO | [JOB PROGRESS] Completed: 0/2 | Running: 1 | Attempted: 0 (Completed: 0, Failed: 0, Filtered: 0, Duplicate: 0) | |
| 2025-05-03 22:13:12 | INFO | [JOB PROGRESS] Completed: 0/2 | Running: 1 | Attempted: 0 (Completed: 0, Failed: 0, Filtered: 0, Duplicate: 0) | |
| RAW DATA: ```json | |
| [ | |
| { | |
| "problem": "A fair coin is flipped repeatedly. What is the expected number of flips until the sequence HHH appears for the first time OR the sequence TTT appears for the first time?", | |
| "topic": "Expected Value", | |
| "answer": "7", | |
| "reasoning": "Let E be the expected number of flips. Define states by the longest suffix that is a prefix of HHH or TTT: S0 (empty string), SH (H), ST (T), SHH (HH), STT (TT), SHT (HT), STH (TH). Let E_s be the expected additional flips from state s. We want E = E_S0.\nE_S0 = 1 + 0.5 E_SH + 0.5 E_ST\nE_SH = 1 + 0.5 E_SHH + 0.5 E_SHT\nE_ST = 1 + 0.5 E_STH + 0.5 E_STT\nE_SHH = 1 + 0.5 * 0 (HHH achieved) + 0.5 E_ST (suffix T)\nE_STT = 1 + 0.5 E_SH (suffix H) + 0.5 * 0 (TTT achieved)\nE_SHT = 1 + 0.5 E_SH (suffix THT->H) + 0.5 E_STT (suffix HTT->TT)\nE_STH = 1 + 0.5 E_SHH (suffix THH->HH) + 0.5 E_ST (suffix THT->T)\nBy symmetry, E_SH = E_ST. This implies E_SHH = E_STT and E_SHT = E_STH.\nFrom E_SHH = 1 + 0.5 E_ST and E_STT = 1 + 0.5 E_SH, with E_SH = E_ST, we get E_SHH = E_STT = 1 + 0.5 E_SH.\nFrom E_SHT = 1 + 0.5 E_SH + 0.5 E_STT = 1 + 0.5 E_SH + 0.5(1 + 0.5 E_SH) = 1.5 + 0.75 E_SH.\nFrom E_STH = 1 + 0.5 E_SHH + 0.5 E_ST = 1 + 0.5(1 + 0.5 E_ST) + 0.5 E_ST = 1.5 + 0.75 E_ST = 1.5 + 0.75 E_SH.\nSubstitute into E_SH equation: E_SH = 1 + 0.5 E_SHH + 0.5 E_SHT = 1 + 0.5(1 + 0.5 E_SH) + 0.5(1.5 + 0.75 E_SH).\nE_SH = 1 + 0.5 + 0.25 E_SH + 0.75 + 0.375 E_SH\nE_SH = 2.25 + 0.625 E_SH\nE_SH - 0.625 E_SH = 2.25\n0.375 E_SH = 2.25\n(3/8) E_SH = 9/4\nE_SH = (9/4) * (8/3) = 6.\nSo, E_ST = E_SH = 6.\nFinally, E = E_S0 = 1 + 0.5 E_SH + 0.5 E_ST = 1 + 0.5 * 6 + 0.5 * 6 = 1 + 3 + 3 = 7." | |
| } | |
| ] | |
| ``` | |
| PARSE DONE {'problem': 'A fair coin is flipped repeatedly. What is the expected number of flips until the sequence HHH appears for the first time OR the sequence TTT appears for the first time?', 'topic': 'Expected Value', 'answer': '7', 'reasoning': 'Let E be the expected number of flips. Define states by the longest suffix that is a prefix of HHH or TTT: S0 (empty string), SH (H), ST (T), SHH (HH), STT (TT), SHT (HT), STH (TH). Let E_s be the expected additional flips from state s. We want E = E_S0.\nE_S0 = 1 + 0.5 E_SH + 0.5 E_ST\nE_SH = 1 + 0.5 E_SHH + 0.5 E_SHT\nE_ST = 1 + 0.5 E_STH + 0.5 E_STT\nE_SHH = 1 + 0.5 * 0 (HHH achieved) + 0.5 E_ST (suffix T)\nE_STT = 1 + 0.5 E_SH (suffix H) + 0.5 * 0 (TTT achieved)\nE_SHT = 1 + 0.5 E_SH (suffix THT->H) + 0.5 E_STT (suffix HTT->TT)\nE_STH = 1 + 0.5 E_SHH (suffix THH->HH) + 0.5 E_ST (suffix THT->T)\nBy symmetry, E_SH = E_ST. This implies E_SHH = E_STT and E_SHT = E_STH.\nFrom E_SHH = 1 + 0.5 E_ST and E_STT = 1 + 0.5 E_SH, with E_SH = E_ST, we get E_SHH = E_STT = 1 + 0.5 E_SH.\nFrom E_SHT = 1 + 0.5 E_SH + 0.5 E_STT = 1 + 0.5 E_SH + 0.5(1 + 0.5 E_SH) = 1.5 + 0.75 E_SH.\nFrom E_STH = 1 + 0.5 E_SHH + 0.5 E_ST = 1 + 0.5(1 + 0.5 E_ST) + 0.5 E_ST = 1.5 + 0.75 E_ST = 1.5 + 0.75 E_SH.\nSubstitute into E_SH equation: E_SH = 1 + 0.5 E_SHH + 0.5 E_SHT = 1 + 0.5(1 + 0.5 E_SH) + 0.5(1.5 + 0.75 E_SH).\nE_SH = 1 + 0.5 + 0.25 E_SH + 0.75 + 0.375 E_SH\nE_SH = 2.25 + 0.625 E_SH\nE_SH - 0.625 E_SH = 2.25\n0.375 E_SH = 2.25\n(3/8) E_SH = 9/4\nE_SH = (9/4) * (8/3) = 6.\nSo, E_ST = E_SH = 6.\nFinally, E = E_S0 = 1 + 0.5 E_SH + 0.5 E_ST = 1 + 0.5 * 6 + 0.5 * 6 = 1 + 3 + 3 = 7.'} | |
| START Let $x$ and $y$ be positive integers such that $x^y$ is a perfect square and $y^x$ is a perfect cube. Find the minimum possible value of $x+y$. | |
| 2025-05-03 22:13:15 | INFO | [JOB PROGRESS] Completed: 1/2 | Running: 1 | Attempted: 1 (Completed: 1, Failed: 0, Filtered: 0, Duplicate: 0) | |
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| 2025-05-03 22:13:54 | INFO | [JOB PROGRESS] Completed: 1/2 | Running: 1 | Attempted: 1 (Completed: 1, Failed: 0, Filtered: 0, Duplicate: 0) | |
| 2025-05-03 22:13:57 | INFO | [JOB PROGRESS] Completed: 1/2 | Running: 1 | Attempted: 1 (Completed: 1, Failed: 0, Filtered: 0, Duplicate: 0) | |
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| 2025-05-03 22:14:11 | WARNING | Invalid JSON syntax in LLM response: Invalid control character at: line 6 column 694 (char 924) | |
| RAW DATA: ```json | |
| [ | |
| { | |
| "problem": "Let $x$ and $y$ be positive integers such that $x^y$ is a perfect square and $y^x$ is a perfect cube. Find the minimum possible value of $x+y$.", | |
| "topic": "Number Theory", | |
| "answer": "2", | |
| "reasoning": "Let the prime factorization of $x$ be $\prod p_i^{\alpha_i}$ and $y$ be $\prod p_i^{\beta_i}$.\nFor $x^y = \prod p_i^{\alpha_i y}$ to be a perfect square, $\alpha_i y$ must be even for all $i$. This means for every $p_i$ with $\alpha_i > 0$, either $\alpha_i$ is even or $y$ is even. Equivalently, $v_{p_i}(x)$ must be a multiple of $2/\gcd(y, 2)$ for all $i$.\nFor $y^x = \prod p_i^{\beta_i x}$ to be a perfect cube, $\beta_i x$ must be a multiple of 3 for all $i$. This means for every $p_i$ with $\beta_i > 0$, either $\beta_i$ is a multiple of 3 or $x$ is a multiple of 3. Equivalently, $v_{p_i}(y)$ must be a multiple of $3/\gcd(x, 3)$ for all $i$. | |
| Let $g_2 = \gcd(y, 2)$ and $g_3 = \gcd(x, 3)$. The conditions are that $v_p(x)$ is a multiple of $2/g_2$ for all primes $p$, and $v_p(y)$ is a multiple of $3/g_3$ for all primes $p$. We consider the four cases for the possible values of $(g_2, g_3)$: | |
| Case 1: $g_2=1$ ($y$ is odd) and $g_3=1$ ($x$ is not a multiple of 3). | |
| $v_p(x)$ must be a multiple of $2/1=2$ for all $p$, so $x$ is a perfect square ($x=k^2$). | |
| $v_p(y)$ must be a multiple of $3/1=3$ for all $p$, so $y$ is a perfect cube ($y=m^3$). | |
| $y=m^3$ is odd $\implies m$ is odd. | |
| $x=k^2$ is not a multiple of 3 $\implies k$ is not a multiple of 3. | |
| We seek to minimize $x+y = k^2 + m^3$ with $k \not\equiv 0 \pmod 3$ and $m$ odd. The smallest possible values for $k$ are $1, 2, 4, 5, \ldots$ and for $m$ are $1, 3, 5, \ldots$. | |
| Trying small values: $(k, m) = (1, 1) \implies (x, y) = (1^2, 1^3) = (1, 1)$. $x+y = 1+1=2$. $(1,1)$ satisfies the case conditions ($1$ is odd, $1$ is not a multiple of 3). $1^1=1$ is a square, $1^1=1$ is a cube. Valid. | |
| $(k, m) = (2, 1) \implies (x, y) = (2^2, 1^3) = (4, 1)$. $x+y = 4+1=5$. Valid. | |
| $(k, m) = (1, 3) \implies (x, y) = (1^2, 3^3) = (1, 27)$. $x+y = 1+27=28$. Valid. | |
| The minimum in this case is 2. | |
| Case 2: $g_2=1$ ($y$ is odd) and $g_3=3$ ($x$ is a multiple of 3). | |
| $v_p(x)$ must be a multiple of $2/1=2$ for all $p$, so $x$ is a perfect square ($x=k^2$). | |
| $v_p(y)$ must be a multiple of $3/3=1$ for all $p$, so $y$ is any integer. | |
| $y$ is odd. | |
| $x=k^2$ is a multiple of 3 $\implies k$ is a multiple of 3. | |
| We seek to minimize $x+y = k^2 + y$ with $k \equiv 0 \pmod 3$ and $y$ odd. | |
| Smallest $k$ is 3, so smallest $x=3^2=9$. Smallest odd $y$ is 1. | |
| $(x, y) = (9, 1)$. $x+y=9+1=10$. $(9,1)$ satisfies the case conditions ($1$ is odd, $9$ is a multiple of 3). $9^1=9$ is a square, $1^9=1$ is a cube. Valid. | |
| Case 3: $g_2=2$ ($y$ is even) and $g_3=1$ ($x$ is not a multiple of 3). | |
| $v_p(x)$ must be a multiple of $2/2=1$ for all $p$, so $x$ is any integer. | |
| $v_p(y)$ must be a multiple of $3/1=3$ for all $p$, so $y$ is a perfect cube ($y=m^3$). | |
| $y=m^3$ is even $\implies m$ is even. | |
| $x$ is not a multiple of 3. | |
| We seek to minimize $x+y = x + m^3$ with $x \not\equiv 0 \pmod 3$ and $m$ even. | |
| Smallest $m$ is 2, so smallest $y=2^3=8$. Smallest $x$ not a multiple of 3 is 1. | |
| $(x, y) = (1, 8)$. $x+y=1+8=9$. $(1,8)$ satisfies the case conditions ($8$ is even, $1$ is not a multiple of 3). $1^8=1$ is a square, $8^1=8$ is a cube. Valid. | |
| Case 4: $g_2=2$ ($y$ is even) and $g_3=3$ ($x$ is a multiple of 3). | |
| $v_p(x)$ must be a multiple of $2/2=1$ for all $p$, so $x$ is any integer. | |
| $v_p(y)$ must be a multiple of $3/3=1$ for all $p$, so $y$ is any integer. | |
| $x$ is a multiple of 3. | |
| $y$ is even. | |
| We seek to minimize $x+y$ with $x \equiv 0 \pmod 3$ and $y$ is even. | |
| Smallest $x$ is 3. Smallest $y$ is 2. | |
| $(x, y) = (3, 2)$. $x+y=3+2=5$. $(3,2)$ satisfies the case conditions ($2$ is even, $3$ is a multiple of 3). $3^2=9$ is a square, $2^3=8$ is a cube. Valid. | |
| Comparing the minimum values found in each case: 2, 10, 9, 5. The overall minimum possible value of $x+y$ is 2. | |
| The minimum value of $x+y$ occurs in Case 1 with $(x,y)=(1,1)$, giving $x+y=2$. | |
| Final check: $x=1, y=1$. $x^y = 1^1 = 1$, which is a perfect square ($1^2$). $y^x = 1^1 = 1$, which is a perfect cube ($1^3$). Both conditions are satisfied for positive integers $x=1, y=1$. $x+y = 1+1=2$. | |
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