Each of the integers 226 and 318 have digits whose product is 24 . How many three-digit positive integers have digits whose product is 24 ?
(A) 4
(B) 18
(C) 24
(D) 12
(E) 21
Each of the integers 226 and 318 have digits whose product is 24 . How many three-digit positive integers have digits whose product is 24 ?
(A) 4
(B) 18
(C) 24
(D) 12
(E) 21

## Solution

First, we determine all of the possible ways to write 24 as the product of single-digit numbers.

(i) $24=1 ¥ 4 ¥ 6$

(ii) $24=1 ¥ 3 ¥ 8$

(iii) $24=2 ¥ 3 ¥ 4$

(iv) $24=2 ¥ 2 ¥ 6$

The cases numbered (i), (ii) and (iii) each give 6 possible arrangements. For example, if we consider $24=1 ¥ 4 ¥ 6$, the 6 possibilities are then $146,164,416,461,614$, and 641 . So for cases (i), (ii) and (iii), we have a total of 18 possibilities.

For the fourth case, there are only 3 possibilities, which are 226, 262 and 622.

In total there are $18+3=21$ possibilities.

ANSWER: (E)