If x, y and z are positive integers, is it true that x^2 is divisible

Question

If x, y and z are positive integers, is it true that x^2 is divisible by 16?

(1) 5x = 4y
(2) 3x^2 = 8z

Not as easy as it seems...

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gmatophobia · Answer

The key point to note in this question is the fact that "x, y and z are positive  integers "

Question

Is x^2 is divisible by 16 ?

Inference : As x is an integer, the questions wants us to find if is x a multiple of 4

Statement 1

(1) 5x = 4y

As x and y are integers for the equality to hold true, x must be a multiple of 4.

Hence the statement is sufficient.

We can eliminate options A and D.

Statement 2

(2) 3x^2 = 8z

3x^2 = 2^3z

Similar to statement 1, as x and z are integers for the equality to hold true, x must be a multiple of 4.

This statement is also sufficient.

Option D

Priyanshi06 · Answer

I didn't get how statement 2 is sufficient. Could you elaborate?

ShreyashGupta188 · Answer

It's given 3x^2 = 8z

So put x= 4 then LHS becomes 3x4^2 = 16
So as 8 will divide 16 so we can say z will be 2 and LHS= RHS

Posted from my mobile device

Priyanshi06 · Answer

I am still confused, If we take X= 4 then the solution of (3*4)^2 is 144, which means z is 18. This value of X is satisfactory. But if look at the given statement 3X^2=8Z which gives us X^2=8z/9, for X^2 to be divisible by 16 X^2should be= 2^4 or 4^2 or 8*2 or 16* any positive integer except 0, and this makes second statement not sufficient. 
This is what I understood from the statement. I could be wrong.

ShreyashGupta188 · Answer

Its not (3*4)^2 , its 3*4^2

Priyanshi06 · Answer

still, that makes X^2= 8z/3, if Z=1,3,5...or such positive integers it will not fulfill the requirement of X^2 being divisible by 16

Bunuel · Answer

We are given that x is a positive integer. Now, in this case, can z be 1, 3, or 5? Can you come up with any integer value of z, such that x is an integer and not s multiple of 4?

IanStewart · Answer

The '5' and the '3' in Statements 1 and 2 are just distractions. Statement 1 tells us x is a multiple of 4, so x^2 is a multiple of 4^2, and is sufficient.

From Statement 2, we learn x^2 is divisible by 8, so by 2^3. But if x is an integer, x^2 is never  only  divisible by 2^3, because in the prime factorization of a perfect square, all of the exponents must be even. So in the prime factorization of x^2, we have an even exponent on the '2', and the exponent must be 3 or greater since x^2 is divisible by 2^3, so x^2 must be divisible by at least 2^4, and Statement 2 is sufficient.

So the answer is D.

NK1976 · Answer

------ 
 Statement 2 
x^2 = 8z/3 = Integer (as x is an integer)
Therefore for equality to hold z must be a multiple of 3......, 
z=0, x^2 = 0 ... Divisible by 16
z=3, x^2 = 8 .... Not Divisible by 16 
z=6, x^2 = 16 .... Divisible by 16
z=9, x^2 = 24 .... Not Divisible by 16 ..... Therefore Statement 2 is Not Sufficient 
 IanStewart... can you please explain where am I going wrong.....

IanStewart · Answer

From the question, notice x is an integer, so it can't be true that x^2 = 8, for example, because then x = 2√2, which is not an integer. If you're testing values here, then if you make x^2 a multiple of 8 but not of 16 (so if x^2 = 8 or 24 or 40 or 56 etc), you'll always find that x is not an integer, essentially for the reason I explained in my earlier post.

NK1976 · Answer

From the question, notice x is an integer, so it can't be true that x^2 = 8, for example, because then x = 2√2, which is not an integer. If you're testing values here, then if you make x^2 a multiple of 8 but not of 16 (so if x^2 = 8 or 24 or 40 or 56 etc), you'll always find that x is not an integer, essentially for the reason I explained in my earlier post. Thanks a ton Ian ...

RICHA1189 · Answer

If x, y and z are positive integers, is it true that x^2 is divisible by 16?

If x^2 is divisible by 16, then we have to find that is x a multiple of 4.

(1) 5x = 4y

x = 4y/ 5 , we know that x is a positive integer. Meaning y is a multiple of 5 I.e. x can be 4,8,12,16… etc. I.e. a multiple of 4

Sufficient A/D

(2) 3x^2 = 8z
3x^2 = 2^3 z
Therefore x has an even power of 2 and z is an even multiple of 3
Ex. If we take minimum value of even power to maintain the equation = 3 * 2^4 = 2^3 * 6 ( z=6)

2^4 is divisible by 16 .

Sufficient

Answer is D

bumpbot · Answer

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If x, y and z are positive integers, is it true that x^2 is divisible

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