If x and y are positive integers such that 2x2

Question

If  x  and  y  are positive integers such that  2x^2 = 3y^5 , what is the smallest possible value of  x + y ?

(A) 54
(B) 72
(C) 108
(D) 114
(E) 162

agrasan · Answer

We can write:
x = ((3/2)*(y^5))^1/2

For x to be an integer, y has to be a form of (3*k)*(2*c)
But k and c has to be minimum because we have to find minimum value of x+y

y = 3*2
x = ((3^6)*(2^4))^1/2 = 108

Min x+y = 114

suntprovident · Answer

Bunuel .  kevincan  request for a simpler explanation.

Thanks

kevincan · Answer

This is simpler albeit longer

Notice that if x and y are integers, the left side is an even number and the right side is a multiple of 3.
Thus x must be a multiple of 3 and y a multiple of 2.
Because y is a multiple of 2, the right side is a multiple of 2^5. Thus x^2 is a multiple of 2^4, and so x is a multiple of 2^2. At a minimum, each side is a multiple of 2^5.
Likewise, because x is a multiple of 3, the right side of a multiple of 3^2. Thus y^5 is a multiple of 3, and so y is a multiple of 3 and y^5 of 3^5. At a mimimum, the right side (and thus the left side) is a multiple of 3^6

What do we know now? x must be a multiple of 4 and of 3^3 and y is a multiple of 2 and 3.

The smallest possible values of x and y are 108 and 6 respectively

adewale223 · Answer

2X^2 = 3Y^5

X^2 = 3Y^5/2

X = Y^2(3/2Y)^1/2

For X to be an integer, the first value of Y is 6. Then X will be 108

So minimum value of X + Y = 114

If x and y are positive integers such that 2x2

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If x and y are positive integers such that 2x2

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