2024 AMC 10B Problem 8

Attempt Problem 8 of the 2024 AMC 10B 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 10B solutions, or check the answer key.

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

Let $N$ be the product of all the positive integer divisors of $42.$ What is the units digit of $N?$

$0$

$2$

$4$

$6$

$8$

Answer: D

Concepts:factor counting units digit pairing and grouping

Difficulty rating: 1310

Solution:

Since $42 = 2 \cdot 3 \cdot 7$ has $(1+1)^3 = 8$ divisors, we can pair each divisor with its complement, so $N = 42^{8/2} = 42^4.$ Only the units digit matters, and $2^4 = 16$ ends in $6,$ so $42^4$ does too. Therefore, the answer is D.

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