2024 AMC 10A Problem 5

$n!$ means $1 \cdot 2 \cdot 3 \cdots n$. To see which $n$ works, first find out what $2024$ is made of.

1. Break $2024$ into primes:
$$2024 = 8 \cdot 253 = 2 \cdot 2 \cdot 2 \cdot 11 \cdot 23.$$
(To split $253$, try small primes: $253 = 11 \cdot 23$.)
2. The largest prime, $23$, is the one that decides things, so look at it first. A prime like $23$ can only divide $n!$ if one of the numbers $1, 2, \ldots, n$ is a multiple of $23$. The first such number is $23$ itself. So $n$ must be at least $23$.
3. Now check that $n = 23$ really works. The product $23!$ contains the factors $2$ and $4$ (which give $2 \cdot 2 \cdot 2$), the factor $11$, and the factor $23$. So it is a multiple of $2 \cdot 2 \cdot 2 \cdot 11 \cdot 23 = 2024$.

The least value is $n = 23$.

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

A prime number divides $n!$ exactly when that prime is one of the numbers $1$ to $n$. So the largest prime factor of the target sets the smallest possible $n$, and the smaller factors are usually covered along the way (just check them). It also pays to know the prime factors of the contest year: $2024 = 2^3 \cdot 11 \cdot 23$.

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 $\boxed{\textbf{(D)}\ 23}$.