1. [4 points] Find the number of eight-digit numbers, the product of whose digits equals 700. The answer should be presented as an integer.
Answer: 2520.

Solution. Since $700=7 \cdot 2^{2} \cdot 5^{2}$, the sought numbers can consist of the following digits: (a) two twos, two fives, one seven, and three ones, or (b) one four, two fives, one seven, and four ones. We will calculate the number of variants in each case.

(a) First, we choose two places out of eight for the twos ( $C_{8}^{2}=\frac{8!}{2!6!}$ ways), then two places out of the remaining six for the fives ( $C_{6}^{2}=\frac{6!}{4!2!}$ ways), and then one place out of the remaining four for the seven ( $C_{4}^{1}=4$ ways). Finally, the remaining places are occupied by ones. By the rule of product, we get $C_{8}^{2} \cdot C_{6}^{2} \cdot 4=\frac{8!}{3!2!2!}=1680$ ways.

(b) Reasoning similarly, we find that the number of ways in this case is $\frac{8!}{2!4!}=840$.

Finally, we get $840+1680=2520$ ways.