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e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 2943
x, y are positive integers. Find the number of even factors of 4x^2  [#permalink]

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3
18 00:00

Difficulty:   95% (hard)

Question Stats: 27% (02:46) correct 73% (02:53) wrong based on 356 sessions

### HideShow timer Statistics Question 3 of the e-GMAT Primes Trio: 3 Questions on Number of factors and prime factors

$$x, y$$ are positive integers. Find the number of even factors of $$4x^2$$

I. $$x^3 – y^3 + 3xy$$ is odd and x is a prime.
II. $$x^{(x+y)}*y^{3x} +x^{3y}$$ is odd.

Here is a fresh question from e-GMAT. Go ahead and give it a shot! Regards,
The e-GMAT Quant Team

P.S.: Solutions with clarity of thought and elegance will get kudos! Here is an official question which tests similar concept.
http://gmatclub.com/forum/if-n-4p-where ... 44781.html

Hope this helps. _________________

Originally posted by EgmatQuantExpert on 05 May 2015, 02:17.
Last edited by EgmatQuantExpert on 07 Aug 2018, 02:55, edited 7 times in total.
e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 2943
Re: x, y are positive integers. Find the number of even factors of 4x^2  [#permalink]

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4
6
Here is the solution for this question. Steps 1 & 2: Understand Question and Draw Inferences

We are given that x and y are positive integers. The question asks us to find the number of even factors of $$4x^2$$

The Number of Even Factors of a number = Total Number of Factors – Number of Odd Factors

To find the total number of factors we need to know the powers of each prime factor.

To find the number of odd factors we need to know the powers of every odd prime factor.

So essentially we need to know the prime factorization of $$4*x^2$$

In other words, we need to know the prime factorization of $$x^2$$ or $$x$$ to be able to answer the given question. Step 3: Analyze Statement 1: $$x^3 – y^3 + 3xy$$ is odd and $$x$$ is a prime.

This statement says $$x$$ is a prime. So the prime factorization of $$4*x^2$$ will be $$2^2 * x^2$$

However, we still don’t know whether $$x$$ is even or odd.

If $$x$$ is even, then $$x = 2$$ which gives us $$4*x^2$$ as $$2^4$$ and since this expression doesn’t have any odd prime factors, the total number of even factors are same as the total number of factors. (which will be $$5$$ in this case.)

However, if $$x$$ is an odd prime, then $$4*x^2$$ will be $$2^2 * x^2$$.

In this case, the total number of factors will be $$(2+1)*(2+1) = 9$$

And the number of odd factors will be $$(2+1) = 3$$

This gives us the number of even factors = $$9 – 3 = 6$$

So we need to know whether $$x$$ is even or odd to determine a unique answer.

We will try to use the other piece of information ” $$x^3 – y^3 + 3xy$$ is odd” to determine the even-odd nature of $$x$$.

We know that a positive integral power doesn’t affect the even-odd nature. Similarly, multiplication by a odd number won’t affect the even-odd nature.

Therefore if $$x^3 – y^3 + 3xy$$ is odd then $$x – y + xy$$ must also be odd. So let us now see the various cases where $$x – y + xy$$ must be odd. As you can see, $$x – y + xy$$ is odd when $$x$$ is odd and also in a case when $$x$$ is even.

So we cannot determine whether $$x$$ is even or odd.

Therefore statement 1 is not sufficient to arrive at a unique answer.

Step 4: Analyze Statement 2: $$x^{(x+y)}*y^{3x} +x^{3y}$$ is odd.

Since positive integral powers do not affect the even-odd nature, statement 2 essentially tells us that $$xy + x$$ is odd. Which means $$x*(y+1)$$ is odd. So let us now see the various cases where $$x*(y + 1)$$ must be odd. As you can see, “$$x(y + 1)$$ is odd” implies that “$$x$$ is odd”.

However, $$x$$ need not be a prime. In other words, $$x$$ could be written as $$x = p^a * q^b * r^c$$ … where $$p$$, $$q$$, $$r$$, etc. are primes and $$a$$, $$b$$, $$c$$ are non-negative integers. Therefore without knowing these values we cannot determine the prime factorization of $$x$$ and therefore cannot arrive at a unique answer.

Therefore statement 2 is not sufficient.

Step 5: Analyze Both Statements Together (if needed):

From statement 1, we have: $$x$$ is a prime and we need to know if $$x$$ is even or odd to arrive at a unique answer.

From statement 2, we have: $$x$$ is odd.

Therefore both statements together are sufficient.

Hope this helps. Regards,
Krishna
_________________

Originally posted by EgmatQuantExpert on 05 May 2015, 02:18.
Last edited by EgmatQuantExpert on 08 May 2015, 07:18, edited 1 time in total.
##### General Discussion
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GMAT 1: 690 Q49 V34 Re: x, y are positive integers. Find the number of even factors of 4x^2  [#permalink]

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1
Q : x,y are positive integers. Find the number of even factors of 4$$x^{2}$$

from the question stem we need to find out factors of 4$$x^{2}$$ and then pick up even factors , essentially we need two things value of x (is it a composite number) and weather x is odd/even.

4$$x^{2}$$ == > $$2^{2}$$$$x^{2}$$

Note: if x is odd , even factors will be less (only the ones with multiple of 2 will be counted)

Stmt I. $$x^{2}$$-$$y^{3} +3xy$$ is odd and x is a prime.

x is prime so that means (2, x) are the only prime factors of $$4x^{2}$$ = $$2^{2}$$$$x^{2}$$ ==> can have (2+1)(2+1) = 9 factors
how many are even ?
x is prime can be 2, 3 (if its odd i.e prime $$\neq{2}$$ ) , then we can figure out (even factors out of 9 factors)

$$x^{2}$$-$$y^{3}$$ +3xy = odd

x - y + 3xy = odd (power does not matter in calculation odd/even nature)

x -> even , then y -> odd for above equation to be true
x -> odd , then y -> even for above equation to be true

x can be odd prime or even prime, so Not Sufficient

Stmt II. $$x^{x+y}$$∗$$y^{3x}$$+$$x^{3y}$$ is odd.

this stmt only tells us about the even/odd nature (so won't be sufficient in determine the actual answer)

$$x^{x+y}$$∗$$y^{3x}$$+$$x^{3y}$$ = odd.

xy + x = odd (power does not matter in calculation odd/even nature)

x(y + 1) = odd

from above x is odd (x can't be even, y can't be odd)

x is odd but can be (3, 9, 25) so Not Sufficient

combining Stmt I. and Stmt II.

x is odd , and x is prime = Sufficient
4$$x^{2}$$ = $$2^{2}$$$$x^{2}$$ ==> can have (2+1)(2+1) = 9 factors

x is prime and x is odd i.e $$x\neq{2}$$

total even factors = 2,4,2x,4x,2x^2,4x^2...

Ans : C
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Re: x, y are positive integers. Find the number of even factors of 4x^2  [#permalink]

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i highly doubt a gmat question would be like this. I personally find e-gmat questions waaay tougher than the gmat hard questions...lack of feedback from users just confirms this..
my 2 cents on solving this one..although I got it wrong the first time...because I was spending way to much to solve it >3 mins (~4 min).

EgmatQuantExpert wrote:
$$x, y$$ are positive integers. Find the number of even factors of $$4x^2$$

I. $$x^3 – y^3 + 3xy$$ is odd and x is a prime.
II. $$x^{(x+y)}*y^{3x} +x^{3y}$$ is odd.

x is prime.
so x=even or x=odd.
y - even or odd.
x -even y even
we have 4 cases:
even-even+even=even out.
even-odd+even=odd - works
odd-even+odd = even - out.
odd-odd+odd = odd. - works

x can be even or odd.
since we need to find the prime factorization of x, we cannot solve the question.

2.expression is odd.
we have few cases:
x - odd or even
y - odd or even

x - odd, y even.
odd*even+odd - odd - works
x-odd, y-odd.
odd*odd+odd - even, out.
x-even, y-even.
even*even+even - even - out.
x-even, y-odd
even*odd+even - even, out.

we clearly see that x is odd, and y is even.
alone is not sufficient.

1+2
x is prime, and it is odd.
thus, x^2 will have only 1 prime factor.
suppose x=3.
then, 4x^2 will be broken down to 2^2*3^2, or total number of factors 3*3=9.
odd factors - 2.
even factors - 7.

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GMAT 1: 660 Q49 V30 Re: x, y are positive integers. Find the number of even factors of 4x^2  [#permalink]

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x,y are positive integers. Find the number of even factors of 4x^2

I. x^3–y^3+3xy is odd and x is a prime.
II. x^(x+y)∗y^(3x)+x^(3y) is odd.

Statement 1 "

X is prime. so, X can be even or odd

x^3–y^3+3xy is odd

Can not find , whether X is even or odd

Insufficient

Statement 2

x^(x+y)∗y^(3x) + x^(3y) is odd.

statement 2 can be odd , in 2 ways

1st way

x^(3y) is even & x^(x+y)∗y^(3x) is odd

x^(3y) is even , then x is even.

As x is even x^(x+y)∗y^(3x) can not be odd.

so we get for x^(x+y)∗y^(3x) + x^(3y) to be odd.

x^(3y) is odd & x^(x+y)∗y^(3x) is even

or x is odd

Insufficient ; different values of odd (X) will result in different even factors for 4x^2

combing st 1 & st 2

X is odd & prime

we get total even factors = 2

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Re: x, y are positive integers. Find the number of even factors of 4x^2  [#permalink]

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_________________ Re: x, y are positive integers. Find the number of even factors of 4x^2   [#permalink] 27 Dec 2018, 09:07
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