Hi chetan2u,
This question is a "twist" on prime factorization. Since we're limited to 5-digit numbers, we have to consider ALL of the different possible 5-digit permutations.
We're asked to name all of the 5-digit numbers whose product-of-digits is 400. We can use prime-factorization to figure out the digits, BUT this comes with a "twist".....NOT ALL of the digits have to be primes....
400 = (25)(16)
The 25 = (5)(5) and there's no way around that. Every option MUST have two 5s as two of the digits.
The 16 can be broken down in a few different ways though:
16 = (1)(2)(8)
16 = (1)(4)(4)
16 = (2)(2)(4)
Each of those options has to be accounted for.
So, we have 3 sets of 5 digits:
1) 1, 2, 8, 5, 5 --> This gives us 5!/2! arrangements = 120/2 = 60 different 5-digit numbers.
2) 1, 4, 4, 5, 5 --> This gives us 5!/(2!2!) arrangements = 120/4 = 30 different 5-digit numbers.
3) 2, 2, 4, 5, 5 --> This gives us 5!/(2!2!) arrangements = 120/4 = 30 different 5-digit numbers.
60 + 30 + 30 = 120 different 5-digit numbers
Final Answer:
GMAT assassins aren't born, they're made,
Rich
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