2025 SSMO Team Round Problems/Problem 7

Problem

Let $n$ be the largest integer such that $20!^{25!} + 25!^{20!}$ is divisible by $2025^n$ . The value of $n$ can be written in the form $a\cdot b!,$ where $a$ and $b$ are positive integers and $b$ is maximized. Find $a+b$ .

Solution

As usual, for a prime $p$ and a positive integer $n$ , let $v_p(n)$ denote the exponent of $p$ in the prime factorization of $n$ . First, we find the value of $v_3(20!^{25!} + 25!^{20!})$ . Note that $v_3(20!^{25!}) = v_3(20!)\cdot 25! = 8\cdot 20!$ and $v_3(25!^{20!}) = v_3(25!)\cdot 20! = 10\cdot 25!$ . Since $8\cdot 25! > 10\cdot 20!$ , we have $v_3(20!^{25!} + 25!^{20!}) = v_3(25!)\cdot 20! = 10\cdot 20!$ . Using an analogous approach, we find that $v_5(20!^{25!} + 25!^{20!}) = 6\cdot 20!$ .

Recall that $2025 = 3^4 \cdot 5^2$ . Therefore, \begin{align*} n &= \min\{\lfloor \tfrac{10\cdot 20!}{4} \rfloor, \lfloor \tfrac{6\cdot 20!}{2} \rfloor\} \\ &= \min\{50\cdot 19!, 3\cdot 20!\} \\ &= 50\cdot 19!. \end{align*} Since $20$ does not divide $50$ , we cannot express $50\cdot 19!$ as the product of a positive integer with a factorial greater than $19!$ . The answer is $50+19 = \boxed{69}$ .

~Sedro

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2025 SSMO Team Round Problems/Problem 7

Problem

Solution