+0

# i need help asap!

0
80
1

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

Apr 11, 2020
edited by ueshui  Apr 11, 2020
edited by ueshui  Apr 11, 2020

#1
+1

The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24

We know that we can't have a two digit factor in one of our three digit numbers(it can't fill in one digit), so we are left with:

1,2,3,4,6,8

Now, we can just bash out unique cases:

138

146

226

234

These are all the unique cases, for which you can reverse these to form any other arrangements that work(do you see the way I wrote it out systematically?) Notice that I write these numbers such that they're in ascending order, with the digits of each 3 digit number going in ascending order(in other words, the digit on the right of a given spot is either equal or greater in value than that digit). We now multiply out the number of ways we can arrange these to get our final answer.

138 = 3! = 6 ways

146 = 3! = 6 ways

226 = 3!/2! = 3 ways

234 = 3! = 6 ways

add all these up, and we get:

6+6+3+6 = 21 numbers, meaning our answer is E)

Apr 11, 2020