2024 AMC 10B Problem 8

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.