2024 AMC 10A Problem 5

Factor $2024 = 2^3 \cdot 11 \cdot 23.$ The prime $23$ is the bottleneck: for $23$ to divide $n!,$ we need $n \ge 23.$ At $n = 23,$ the product $23!$ already has $23,$ $11,$ and plenty of factors of $2,$ so $23!$ is divisible by $2024.$ The least value is $23.$ Thus, D is the correct answer.