Exercise 15. In how many ways can $10^{6}$ be written as a product $A \times B \times C$ of three natural numbers? (N.B. For example, the writings $1 \times 1000 \times 1000$ and $1000 \times 1 \times 1000$ are considered different.)
Solution to exercise 15 We seek $A=2^{a} 5^{b}, B=2^{c} 5^{d}$ such that $a+c \leqslant 6$ and $b+d \leqslant 6$. There are $1+2+\cdots+7=28$ solutions for $(a, c)$ and as many for $(b, d)$, so in total $28^{2}=784$ solutions.