Product of the distinct digits of a natural number is 60.
How many such numbers are possible?
60 cannot be written as a product of two single digit
numbers. SO, the number in question should either have 3 or more digits.
The digits could be 345 or 265
Digits being 345 - there are 3! such numbers
There are six numbers for each of these outlines. So, there
are 3! + 3! = 12 three-digit numbers
The digits could be 1345, 1265 or 2235.
Digits being 1345 - there are 4! such numbers
Digits being 1265 - there are 4! such numbers
Digits being 2235 - this is not possible as digits have to be distinct.
So, there are totally 24 + 24 four-digit numbers
possible. 48 four-digit numbers.
Total number of numbers = 12 + 48 = 60.
(this post has been modified to plug error)
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