2015/16 AHSMC Part 1 Problem 14

The numbers of two digits which are written as a product of four prime numbers are the following: $$\begin{align*} 2\cdot 2\cdot 2\cdot 2 &= 16, & 2\cdot 2\cdot 2\cdot 3 &= 24, & 2\cdot 2\cdot 2\cdot 5 &= 40,\\ 2\cdot 2\cdot 2\cdot 7 &= 56, & 2\cdot 2\cdot 2\cdot 11 &= 88, & 2\cdot 2\cdot 3\cdot 3 &= 36,\\ 2\cdot 2\cdot 3\cdot 5 &= 60, & 2\cdot 2\cdot 3\cdot 7 &= 84, & 2\cdot 3\cdot 3\cdot 3 &= 54,\\ 2\cdot 3\cdot 3\cdot 5 &= 90, & 3\cdot 3\cdot 3\cdot 3 &= 81. \end{align*}$$ We conclude that the number that satisfies the conditions in the problem should have all its digits equal to $8$. The answer is (E).