Question 313391
How many three digit numbers are such that the product of their digits is 120?


Since the product of these numbers end with a 0 (zero), and the digits must be from 1 – 9 (can’t be 0, because the product would be 0), then one of the 3 digits must be a 5, since the only digit that yields a product with a units digits of 0 is 5. 

Dividing 120 by 5, we get 24. 


Now, the factors of 24, between 1 & 9, are 8 & 3, and 6 & 4. This means that the digits that multiply to 120 are 583 and 564. Since there are 6 {{{_[3]P[3]}}}, or {{{3!}}} ways that we can arrange the digits in 583 and 6 {{{_[3]P[3]}}}, or {{{3!}}} ways that we can arrange the digits in 564, there are 12 (6 + 6) ways that we can arrange both sets of digits. This means that there are {{{highlight_green(12)}}} 3-digit numbers that can be formed whose product is 120.