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x and y are positive integers such that x^25^y = 10,125

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x and y are positive integers such that x^25^y = 10,125 [#permalink] New post 14 Sep 2017, 06:52
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x and y are positive integers such that \(x^25^y = 10,125\)

Quantity A
Quantity B
\(x^2\)
\(5^y\)


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Re: x and y are positive integers such that x^25^y = 10,125 [#permalink] New post 28 Sep 2017, 08:37
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This is a tricky one.
If 10,125 is reduced to prime factors it is equal to \(5^2*3^4 = 5^2*9^2\) thus, one could be tempted of stating x = 9, y = 3 so that column B is greater. This is actually true. The problem is that this is not the only way to write 10,125 as x^2*5^y. Indeed, if we take x = 45 and y = 1, we get 10,125 too but in this case column A is greater.

Thus, our answer is D!
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Re: x and y are positive integers such that x^25^y = 10,125 [#permalink] New post 28 Sep 2017, 12:12
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Carcass wrote:
x and y are positive integers such that \(x^25^y = 10,125\)

Quantity A
Quantity B
\(x^2\)
\(5^y\)


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.


This question begs for some prime factorization
10,125 = (5)(5)(5)(3)(3)(3)(3)

From this, we can see a few options for x² and 5^y
Here are two possible cases:

case 1: x = 45 and y = 1.
So, (x²)(5^y) = (45²)(5^1)
= [(3)(3)(5)]²(5)
= (3)(3)(3)(3)(5)(5)(5)
= 10,125
For this case, we get:
Quantity A: 45²
Quantity B: 5
Quantity A is clearly greater

case 2: x = 9 and y = 3.
So, (x²)(5^y) = (9²)(5^3)
= [(3)(3)]²(5)(5)(5)
= (3)(3)(3)(3)(5)(5)(5)
= 10,125
For this case, we get:
Quantity A: 81
Quantity B: 125
Quantity B is clearly greater

Answer:
[Reveal] Spoiler:
D


Cheers,
Brent
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Re: x and y are positive integers such that x^25^y = 10,125   [#permalink] 28 Sep 2017, 12:12
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