Problem 683

Answer (D): The prime factorization of $42$ is $2 \cdot 3 \cdot 7$. A divisor of $42$ is determined by choosing whether or not to include each prime as a factor of the divisor, so there are $2^3 = 8$ divisors. Furthermore, the divisors can be formed into $4$ pairs each of whose product is $42$: $1 \cdot 42$, $2 \cdot 21$, $3 \cdot 14$, and $6 \cdot 7$. Therefore $N = 42^4$. The value of the requested units digit of $N$ is the same as the units digit of $2^4$, which is $6$.