2024 AMC 10A Problem 5

Factor: $2024 = 8\cdot 253 = 2^3\cdot 11\cdot 23$.

For $n!$ to be divisible by the prime $23$, some factor $1, 2, \dots, n$ must be a multiple of $23$, so $n \ge 23$. Conversely $23!$ contains the factors $2, 4$ (giving $2^3$), $11$ and $23$, so it is a multiple of $2024$.

The answer is $\boxed{\textbf{(D)}\ 23}$.

A prime $p$ divides $n!$ exactly when $p \le n$, so the largest prime factor of the target sets the minimum $n$; the smaller prime powers are then automatically covered (check them only if the largest prime is small). Knowing the factorization of the current year is a recurring AMC convenience.

• Stopping at 2024 = 8·253 and answering 11 or 253 without factoring 253 = 11·23.
• Thinking $22!$ might work because it has plenty of factors of $2$ and $11$; it has no factor of $23$.