2011/12 AHSMC Part 1 Problem 5

Since $1000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5$, the digits can only be $1$, $2$, $4$, $5$ and $8$. Three of them must be $5$s and they can be placed among the six digits in $\binom{6}{3} = 20$ ways. The product of the other three digits is $8$, and they are $(1,1,8)$, $(1,2,4)$ or $(2,2,2)$. They can be placed in $3$, $6$ and $1$ ways respectively. Hence the total number of six-digit millenium numbers is $20(3+6+1)=200$. The answer is (E).