The product of the digits of a four-digit number is 810 . If none of the digits is repeated, the sum of the digits is
(A) 18
(B) 19
(C) 23
(D) 25
(E) 22
Suppose that the four digit number has digits $a b c d$, ie. the product $a b c d=810$.

We must determine how to write 810 as the product of 4 different digits, none of which can be 0 . So we must start by factoring 810 , as $810=81 \times 10=3^{4} \times 2 \times 5$.

So one of the digits must have a factor of 5 . But the only non-zero digit having a factor of 5 is 5 itself, so 5 is a digit of the number.

Now we need to find 3 different digits whose product is $3^{4} \times 2$.

The only digits with a factor of 3 are 3,6 , and 9 , and since we need 4 factors of 3 , we must use each of these digits (the 9 contributes 2 factors of 3 ; the others contribute 1 each). In fact, $3 \times 6 \times 9=3^{4} \times 2=162$.

Therefore, the digits of the number are $3,5,6$, and 9 , and so the sum of the digits is 23 .

ANSWER: (C)