2024 AMC 10B Problem 8

Since $42 = 2 \cdot 3 \cdot 7$, it has $2 \cdot 2 \cdot 2 = 8$ divisors: $1, 2, 3, 6, 7, 14, 21, 42$.

Pair each divisor $d$ with $\frac{42}{d}$: $(1, 42), (2, 21), (3, 14), (6, 7)$. Each pair multiplies to $42$, and there are $4$ pairs, so
$$N = 42^4.$$
The units digit of $42^4$ is the units digit of $2^4 = 16$, namely $6$.

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

The product of all divisors of $n$ is always $n^{\tau(n)/2}$, where $\tau(n)$ counts the divisors, because divisors come in complementary pairs. Units digits depend only on units digits, so a big power collapses to a small one. Note that $5$ never divides $42$, so a final digit of $0$ is impossible from the start.

Multiply the units digits directly: $1 \cdot 2 \cdot 3 \cdot 6 \cdot 7 \cdot 4 \cdot 1 \cdot 2$. Step by step mod $10$: $1, 2, 6, 6, 2, 8, 8, 6$. The units digit is $6$.

• Assuming the product is a multiple of 10 (answering 0) because 42 is even and the list is long, forgetting that no divisor supplies a factor of 5.
• Missing a divisor (often $14$ or $21$), which changes both the count and the final digit.