2024 AMC 10B Problem 8

Question

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

A number with

d

divisors has divisor product equal to itself raised to the

\tfrac{d}{2}

power

42 = 2 \cdot 3 \cdot 7

has

8

divisors, so

N = 42^4;

track only the units digit

Since

42 = 2 \cdot 3 \cdot 7

has

(1+1)^3 = 8

divisors, we can pair each divisor with its complement, so

N = 42^{\frac{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.

Problem 8 in other years