2021 AMC 12A Fall Problem 8

Attempt Problem 8 of the 2021 AMC 12A Fall 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 2021 AMC 12A Fall solutions, or check the answer key.

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

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,M, 32,32, 33,33, 34,34, 35,35, 36,36, 37,37, 38,38, 39,39, and 40.40. What is the value of NM?\dfrac{N}{M}?
11
22
3737
7474
28862886
Answer: D
Concepts:least common multipleprime factorization
Small hint:
MM already contains 24,33,52,7,2^4, 3^3, 5^2, 7, and all primes up to 29;29; check what 3232–4040 add
Big hint:
32=2532 = 2^5 raises the power of 2,2, and 3737 is a new prime; nothing else is new
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=2⋅37=74.\dfrac{N}{M} = 2 \cdot 37 = 74.
Thus, the correct answer is D.
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