2021 AMC 12A Fall Problem 8

Below is the professionally curated solution for Problem 8 of the 2021 AMC 12A Fall, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 12A Fall solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:least common multipleprime factorization

Difficulty rating: 1440

8.

Let MM be the least common multiple of all the integers 1010 through 30,30, inclusive. Let NN be the least common multiple of M,32,33,34,35,36,37,38,39,M, 32, 33, 34, 35, 36, 37, 38, 39, and 40.40. What is the value of NM?\dfrac{N}{M}?

11

22

3737

7474

28862886

Solution:

M=lcm(10,,30)M = \operatorname{lcm}(10, \ldots, 30) contains 242^4 (from 1616), 333^3 (from 2727), 525^2 (from 2525), 7,7, and every prime up to 29.29.

Among 32,,40,32, \ldots, 40, the only new contributions are 32=25,32 = 2^5, which raises the power of 22 from 242^4 to 25,2^5, and the new prime 37.37. Everything else factors into primes and powers already in M.M.

Therefore NM=237=74.\dfrac{N}{M} = 2 \cdot 37 = 74.

Thus, the correct answer is D.

Problem 8 in Other Years