6. How many five-digit positive integers have the product of their digits equal to 1000 ?
A 10
B 20
C 28
D 32
E 40
Solution
E

The prime factorisation of $1000=2^{3} \times 5^{3}$. To obtain a factor of $5^{3}$ in the product of the digits, three of the digits must be 5 since that is the only one-digit multiple of 5 . The other two digits must have product $2^{3}=8$. This can be obtained by using the digits 8 and 1 , or the digits 2 and 4. Using 8 and 1 , there are five choices for where to place the 8 and then four choices for where to place the 1 , hence $5 \times 4=20$ possibilities. Similarly there are 20 possibilities using 2 and 4 , hence 40 possibilities overall.