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GRE 1: Q170 V170
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Re: x and y are positive integers such that x^25^y = 10,125 [#permalink]
1
Factor 10,125 to its prime factors: 10,125 = 3^4*5^3
.
So, x25y = 3^4*5^3
.
In order to have 53 on the right side, there have to be three factors of 5 on the
left side. All three could be in the 5y
term (i.e., y could equal 3). Or, one of the
5’s could be in the 5y
term, and two of the 5’s in the x2 term (i.e., y could
equal 1 and x could have a single factor of 5).
In order to have 34 on the right side, x2 must have 34 = (32)2 as a factor. In
other words, x must have 32 as a factor, because 32
is certainly not a factor of
5. Thus, x is a multiple of 9.
In one case, Quantity A is greater. In the other, Quantity B is greater. The
relationship cannot be determined from the information given.
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Re: x and y are positive integers such that x^25^y = 10,125 [#permalink]
1
Start by finding prime factors
10125 = 3^4 * 5^3

From this, we can make combinations of these values, like:
10125 = 9^2 * 5^3 --> in this case, the second term is greater
10125 = 45^2 * 5^1 --> in this case, the first term is greater

So the answer is D.
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Re: x and y are positive integers such that x^25^y = 10,125 [#permalink]
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Re: x and y are positive integers such that x^25^y = 10,125 [#permalink]
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