AMC 10 A

Number theory Difficulty 3.9 AMC 10/12 Find the answer United States

What is the least value of $n$ such that $n!$ is a multiple of 2024?

Pick one

Solution

Because the prime factorization of $2024$ is $2^3 \cdot 11 \cdot 23$ , it follows that $n!$ is a multiple of $2024$ if and only if $n \ge 23$ . Therefore $23$ is the least value of $n$ such that $n!$ is a multiple of $2024$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.