Each of the integers 226 and 318 have digits whose product is 24. How many three-digit positive integers have digits whose product is...
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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?
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
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
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