If x and y are positive integers, and 4x^2=3y, then which of the follo

\(4x^2=3y\)
--> \(x^2 = 3y/4\)
--> \(x\) 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

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

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

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

Posted from my mobile device

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.

\(4x^2 = 3y\)

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

Stepping back and looking at the equation and what it implies:

(Number value on left side) must equal (number value on right side)

Which means, the values on each side must have the same Prime Factorization, given that X and Y must be positive integers.

4 * (x)^2 = 3 * y

Since 4 is NOT divisible by 3, the other term on the LEFT side (i.e., (x)^2) MUST be divisible by 3 in order for the condition of X and Y as a positive integer to remain true.

The MINIMUM value we must make X is 3

Inserting 3 in for X:

4 (3)^2 = 3 * y

Divide each side of the equation by 3

4 * 3 = y

Therefore, the MINIMUM value of Y must be 12 (which makes sense since the coefficient 3 on the right side is not divisible by 4)

Minimum values:

x = 3
y = 12

(X)^2 = (3)^2 = 9
(Y)^2 = (12)^2 = 144
XY = 3 * 12 = 36

At the MINIMUM Possible values of (x, y) in order for the equation to be satisfied and the variables to be positive integers, all three terms MUST be divisible by 9

E: I, II, and III

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