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Re: Math problem involving powers and smallest possible value [#permalink]
24 Sep 2009, 10:11
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The answer is D: 75.
This is a question relating to prime factorability (is that a word?).
Basically, for any number, X, to be a factor of another number, Y, the prime factors of X must be present in the prime factors of Y. For example, if you reduce the number 50 = (5^2)*(2^1), it is a factor of any number which has these prime factors when reduced.
So, for 5^2 and 3^3 to be factors of this number, the number, when reduced to prime factors, must contain 5^2 and 3^3.
As seen by the above equation, we need at least one more factor of 3, as well as two more factors of 5. As a result: