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

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Math Expert
Joined: 02 Sep 2009
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If x^(1/2) is an integer and xy^2 = 36, how many values are possible  [#permalink]

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30 Aug 2018, 23:06
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Difficulty:

85% (hard)

Question Stats:

39% (01:32) correct 61% (01:35) wrong based on 116 sessions

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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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Re: If x^(1/2) is an integer and xy^2 = 36, how many values are possible  [#permalink]

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30 Aug 2018, 23:12
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

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Re: If x^(1/2) is an integer and xy^2 = 36, how many values are possible  [#permalink]

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30 Aug 2018, 23:18
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]

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30 Aug 2018, 23:58
1
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]

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31 Aug 2018, 00:47
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.
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Re: If x^(1/2) is an integer and xy^2 = 36, how many values are possible  [#permalink]

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03 Sep 2018, 02:19
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]

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04 Sep 2018, 11:23
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.

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Re: If x^(1/2) is an integer and xy^2 = 36, how many values are possible   [#permalink] 04 Sep 2018, 11:23
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