2024 AMC 10B Problem 8

1. List the divisors of $42$: $1, 2, 3, 6, 7, 14, 21, 42$. There are $8$ of them.
2. Pair each divisor with the one that multiplies with it to give $42$:
$$1 \cdot 42, \quad 2 \cdot 21, \quad 3 \cdot 14, \quad 6 \cdot 7.$$
3. Each pair multiplies to $42$ and there are $4$ pairs, so
$$N = 42 \cdot 42 \cdot 42 \cdot 42.$$
4. Only the last digit matters, and the last digit of a product depends only on the last digits of the numbers being multiplied. So follow just the last digit, which starts at $2$:
- $2 \cdot 2 = 4$
- $4 \cdot 2 = 8$
- $8 \cdot 2 = 16$, last digit $6$.
5. The units digit of $N$ is $6$.

The answer is $\boxed{\textbf{(D)}\ 6}$.

Divisors of a number come in pairs that multiply to the number itself, so the product of all divisors is that number raised to half the count of divisors. For last-digit questions, keep only the last digit at every step. And a last digit of $0$ is impossible here because nothing in the product has a factor of $5$.

Just multiply the last digits of all eight divisors, keeping only the last digit each time. The last digits are $1, 2, 3, 6, 7, 4, 1, 2$. Running product: $1$, then $2$, then $6$, then $36 \to 6$, then $42 \to 2$, then $8$, then $8$, then $16 \to 6$. The units digit is $6$.