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If x and y are positive integers, and 4x^2=3y, then which of the follo
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Bunuel
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If x and y are positive integers, and 4x^2=3y, then which of the follo [#permalink] New post  01 Nov 2019, 04:39
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If x and y are positive integers, and \(4x^2=3y\), then which of the following must be a multiple of 9?

I. \(x^2\)
II. \(y^2\)
III. \(xy\)

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III


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kk47
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Re: If x and y are positive integers, and 4x^2=3y, then which of the follo [#permalink] New post  01 Nov 2019, 05:28
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Since x and y are positive integers therefore,

y = 12*n^2
And
x = 3n
Where n is a positive integer.

This implies that x^2,y^2 and xy are multiple of 9 for sure.

Therefore, answer is (E).

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Re: If x and y are positive integers, and 4x^2=3y, then which of the follo [#permalink] New post  01 Nov 2019, 06:04
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Bunuel wrote:
If x and y are positive integers, and \(4x^2=3y\), then which of the following must be a multiple of 9?

I. \(x^2\)
II. \(y^2\)
III. \(xy\)

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III


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\(4x^2=3y\)
--> \(x^2 = 3y/4\)
--> \(x\) is a multiple of 3 & \(y\) is also a multiple of 3 (only then \(x\) will be a positive integer)
--> \(x = 3p\), \(y = 3q\), for any positive integers \(p\) & \(q\)
--> \(x^2 = 9*p^2\) --> I is true
\(y^2 = 9*q^2\) --> II is true
\(x*y = 3p*3q = 9*p*q\) --> III is also true

IMO Option E
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Abhishek009
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Re: If x and y are positive integers, and 4x^2=3y, then which of the follo [#permalink] New post  01 Nov 2019, 07:02
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Bunuel wrote:
If x and y are positive integers, and \(4x^2=3y\), then which of the following must be a multiple of 9?

I. \(x^2\)
II. \(y^2\)
III. \(xy\)

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III


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Let, \(4x^2=3y = 36\)

So, \(x^2 = 9\) Or, \(x = +3 , -3\)

And, \(y = 12\)

(I) \(\frac{x^2}{9}\) = \(\frac{9}{9} = 1\)
(II) \(\frac{y^2}{9} = 12*12/9 = 16\)
(III) \(xy/9 = 12*3/9 = 4\)

Thus, Answer must be (E) I, II and III
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Re: If x and y are positive integers, and 4x^2=3y, then which of the follo [#permalink] New post  01 Nov 2019, 07:08
given
4*x^2/3y =1
x^2 has to be 3 ; and y = 12
4*3^2/3*12 = 1
in that case all given conditions are valid

I. \(x^2\)
II. \(y^2\)
III. \(xy\)

IMO E

Bunuel wrote:
If x and y are positive integers, and \(4x^2=3y\), then which of the following must be a multiple of 9?

I. \(x^2\)
II. \(y^2\)
III. \(xy\)

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III


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Kinshook
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Re: If x and y are positive integers, and 4x^2=3y, then which of the follo [#permalink] New post  28 Mar 2020, 01:49
Bunuel wrote:
If x and y are positive integers, and \(4x^2=3y\), then which of the following must be a multiple of 9?

I. \(x^2\)
II. \(y^2\)
III. \(xy\)

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III


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If x and y are positive integers, and \(4x^2=3y\), then which of the following must be a multiple of 9?
Let x = 3k
4*3^2k^2 = 3y
y = 12k^2

I. \(x^2\)
x = 3k; x^2 = 9k^2; Multiple of 9; TRUE
II. \(y^2\)
y = 12k^2; y^2 = 144k^4; Multiple of 9; TRUE
III. \(xy\)
xy = 3k * 12k^2 = 36k^3; Multiple of 9. TRUE

IMO E
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Avysar
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Re: If x and y are positive integers, and 4x^2=3y, then which of the follo [#permalink] New post  12 May 2020, 21:01
Given 4x^2=3y and x &y are positive integers.
Let's start with the positive values for x
Now x^2 is a perfect square.
The equation doesn't work when x= 1 or 2.
It works for x=3 as
4(3^2)=3*12
So x=3 and y=12
X^2 is divisible by 9
y^2= 144 divisible by 9
Xy=36 also divisible by 9
Hence all the three
E

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yahya729
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Re: If x and y are positive integers, and 4x^2=3y, then which of the follo [#permalink] New post  03 Jun 2021, 08:56
While its easy to put numbers to solve that, that's not an approach which will always work, so you've got to do this a but more intuitively.

4x^2=3y. now quite obviously 4 can't divide 3. As x and y as positive integers, 4 has to divide y, then x has to be at least 3^2, and the I is true.
then given the equation, if we square everything we'll get 16x^4=9y^2. voila - y^2 is divisible by 9. so II is true

xy is the more tricky option here. y=(4x^2)/3, so xy=(4x^3)/3. now ask - how do we know this a integer? well x and y are integers so xy is an integer. again 4 can't divide 3, so x^3 must divide that and the make x^3 at least 3^3(times whatever), so with one 3 gone xy=4*x^2*something. 3 is a factor of x so this thing is a multiple of 9.
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MathRevolution
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Re: If x and y are positive integers, and 4x^2=3y, then which of the follo [#permalink] New post  04 Jun 2021, 00:03
Expert Reply
\(4x^2 = 3y\)

This is possible when x = 3 and y = 12.

I: \(x^2 = 3^2 = 9\) - Multiple of 9
II: \(y^2 = 12^2 = 144\) - Multiple of 9
III: \(xy = 3 * 12 = 36\) - Multiple of 9


Another possible case when x = 6 and y = 48.

I: \(x^2 = 6^2 = 36\) - Multiple of 9
II: \(y^2 = 48^2\) - Multiple of 9
III: \(xy = 6 * 48\) - Multiple of 9

Options I, II, and III.

Answer E
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Re: If x and y are positive integers, and 4x^2=3y, then which of the follo [#permalink]
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