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If x^(1/2) is an integer and xy^2 = 36, how many values are possible

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Bunuel
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If x^(1/2) is an integer and xy^2 = 36, how many values are possible [#permalink] New post   31 Aug 2018, 00:06
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E

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If \(\sqrt{x}\) is an integer and \(xy^2 = 36\), how many values are possible for the integer y?

(A) Two
(B) Three
(C) Four
(D) Six
(E) Eight
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E
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Re: If x^(1/2) is an integer and xy^2 = 36, how many values are possible [#permalink] New post   31 Aug 2018, 00:12
If sqrt(x) is an intger, then x can take values
X = 1,4,9.

Then xy^2 = 36.

Y can take +-3, +-2, +-6.

Y can take 6 values

D is the answer.

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Re: If x^(1/2) is an integer and xy^2 = 36, how many values are possible [#permalink] New post   31 Aug 2018, 00:18
Since both x and y are integers ...answer should be D

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Re: If x^(1/2) is an integer and xy^2 = 36, how many values are possible [#permalink] New post   31 Aug 2018, 00:58
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Bunuel wrote:
If \(\sqrt{x}\) is an integer and \(xy^2 = 36\), how many values are possible for the integer y?

(A) Two
(B) Three
(C) Four
(D) Six
(E) Eight


+1 for (E).

Given,\(xy^2 = 36\)
Since x is a perfect square , so it could be : 1, 4, 9 and 36
So, \(y^2\): 36,9,4 and 1
hence, possible values of y:+6,-6,+3,-3,+2,-2,+1,-1 (both faces of polarities)

SO, 8 values are possible for the integer y.
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Re: If x^(1/2) is an integer and xy^2 = 36, how many values are possible [#permalink] New post   31 Aug 2018, 01:47
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Bunuel wrote:
If \(\sqrt{x}\) is an integer and \(xy^2 = 36\), how many values are possible for the integer y?

(A) Two
(B) Three
(C) Four
(D) Six
(E) Eight


The possible factors of 36 are 1x36;2x18;3x12;4x9;6x6
Since \(\sqrt{x}\) is an integer, x can be 1, 4 ,9, 36
Possible values of y corresponding to x : ±6, ±3, ±2, ±1
8 values are possible.
Answer E.
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Re: If x^(1/2) is an integer and xy^2 = 36, how many values are possible [#permalink] New post   03 Sep 2018, 03:19
x has to be a perfect square, so possible values for x: 1,4,9,25,36 (No need to consider negative values, since y will always be positive!)

Also consider negative values for y, since y is squared!

1*y^2= 36 --> y=+-6
4*y^2= 36 --> y=+-3
9*y^2= 36 --> y=+-2
36*y^2-36 --> y=+-1

So a total of 8 possible values for y. (E)
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Re: If x^(1/2) is an integer and xy^2 = 36, how many values are possible [#permalink] New post   04 Sep 2018, 12:23
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Bunuel wrote:
If \(\sqrt{x}\) is an integer and \(xy^2 = 36\), how many values are possible for the integer y?

(A) Two
(B) Three
(C) Four
(D) Six
(E) Eight


Since √x is an integer and xy^2 = 36, x must be a perfect square. So x could be 1, 4, 9, or 36.

When x is 1, y^2 = 36. So y could be 6 or -6.

When x is 4, y^2 = 9. So y could be 3 or -3.

When x is 9, y^2 = 4. So y could be 2 or -2.

When x is 36, y^2 =1. So y could be 1 or -1.

Answer: E
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Re: If x^(1/2) is an integer and xy^2 = 36, how many values are possible [#permalink] New post   08 Feb 2024, 07:12
­If sqrt(x) is an intger, then x can take values
X = 1,4,9, 36

Then xy^2 = 36.

Y can take +-1, +-3, +-2, +-6

Y can take 8 values
xy^2=36
36*(+-1)^2=36
9*(+-2)^2=36
4*(+-3)^2=36
1*(+-6)^2=36

E is the answer.
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Re: If x^(1/2) is an integer and xy^2 = 36, how many values are possible [#permalink]
08 Feb 2024, 07:12
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