20) Consider all 8-digit numbers formed using each of the digits $1,2,3,4,6,7,8,9$ exactly once. Assuming we multiply them all together, what is the unit digit of the resulting product?
(A) 0
(B) 1
(C) 4
(D) $6 \quad(\mathbf{E}) 8$
20) The answer is (D). Among the numbers of this type (which are 8 ! $=1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8=40320$ ) there are 7 ! with the last digit 1, 7 ! with the last digit $2, \ldots$ and 7 ! with the last digit 9. The unit digit of their product will therefore be the same as that of the number $(1 \times 2 \times 3 \times 4 \times 6 \times 7 \times 8 \times 9)^{7}$. Since the number in parentheses has 6 as its last digit, every power of it will still have 6 as the unit digit.

Second Solution For each final digit, there is the same number of numbers. If now one pairs a number ending in 1 with one ending in 6, one ending in 2 with one ending in 3, one ending in 4 with one ending in 9, and one ending in 7 with one ending in 8, one gets four numbers ending in 6. Therefore, the final product is the product of many (i.e., $4 \cdot 7$ !) numbers that end in 6, and thus ends in 6.