2024 AMC 10A Problem 5

Attempt Problem 5 of the 2024 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2024 AMC 10A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

What is the least value of

n

such that

n!

is a multiple of

2024?

11

21

22

23

253

Factor

2024 = 2^3 \cdot 11 \cdot 23

The largest prime factor is

23,

so it must appear as a factor in

n!

Factor

2024 = 2^3 \cdot 11 \cdot 23.

The prime

23

is the bottleneck: for

23

to divide

n!,

we need

n \ge 23.

n = 23,

the product

23!

already has

23,

11,

and plenty of factors of

2,

23!

is divisible by

2024.

The least value is

23.

Thus, D is the correct answer.

Problem 5 in other years