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# If x and y are integers, is xy divisible by x^2?

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Math Expert
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If x and y are integers, is xy divisible by x^2? [#permalink]

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09 Jan 2017, 03:26
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If x and y are integers, is xy divisible by x^2?

(1) x divides into y^2 with no remainder.
(2) x is a prime.
[Reveal] Spoiler: OA

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Re: If x and y are integers, is xy divisible by x^2? [#permalink]

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09 Jan 2017, 07:44
Bunuel wrote:
If x and y are integers, is xy divisible by x^2?

(1) x divides into y^2 with no remainder.
(2) x is a prime.

(1) x is a multiple of y^2

min $$x=0$$. In this case we'll get $$\frac{0}{0}$$, which is undefined.

next $$x=y^2$$, $$x^2 = y^4$$

$$\frac{xy}{x^2} = \frac{y^3}{y^4} = \frac{1}{y}$$

y=1 divisible, y>1 not.

Insufficient

(2) No information about y. Insufficient.

(1) & (2) Prime number is only divisilble by $$1$$ and itself, but surely not $$itself^2$$ or any multiple of $$itself^2$$. Only possible option is y=1. Sufficient
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Re: If x and y are integers, is xy divisible by x^2? [#permalink]

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09 Jan 2017, 09:12
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Bunuel wrote:
If x and y are integers, is xy divisible by x^2?

(1) x divides into y^2 with no remainder.
(2) x is a prime.

x and y are integers.
Is xy divisible by x^2?

x^2 has two x's. xy already has one x. For xy to be divisible by x^2, now all we need is another x to be a factor of y.
If x is a factor of y, xy will be divisible by x^2.

(1) x divides into y^2 with no remainder.
x is a factor y^2 does not mean that x is a factor of y. For example, if x = 4 and y = 2, x is a factor of y^2 but not a factor of y.
Not sufficient.

(2) x is a prime.
x may or may not a factor of y. Not sufficient.

Using both, we know that x is prime and it is a factor of y^2. In that case, it must be a factor of y because whichever prime factors are in y, only those prime factors will be in y^2 too. So x must be a factor of y.
Sufficient.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 21 Apr 2016 Posts: 196 Re: If x and y are integers, is xy divisible by x^2? [#permalink] ### Show Tags 09 Jan 2017, 20:04 VeritasPrepKarishma wrote: we know that x is prime and it is a factor of y^2. In that case, it must be a factor of y because whichever prime factors are in y, only those prime factors will be in y^2 too. So x must be a factor of y. I know that's the truth from childhood when I learned how to check whether a number is prime. But now I'm struggling to figure out how. Could you please point me to any resource that proves this. Thanks! Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7988 Location: Pune, India Re: If x and y are integers, is xy divisible by x^2? [#permalink] ### Show Tags 10 Jan 2017, 03:18 manhasnoname wrote: VeritasPrepKarishma wrote: we know that x is prime and it is a factor of y^2. In that case, it must be a factor of y because whichever prime factors are in y, only those prime factors will be in y^2 too. So x must be a factor of y. I know that's the truth from childhood when I learned how to check whether a number is prime. But now I'm struggling to figure out how. Could you please point me to any resource that proves this. Thanks! I am not sure what you are struggling with... this - whichever prime factors are in y, only those prime factors will be in y^2 too? If yes, say $$y = 2^3 * 3 * 7$$ What is $$y^2$$? It is $$y^2 = 2^6 * 3^2 * 7^2$$ Can y^2 have 5 in it given that y doesn't have it? Every prime factor that y^2 has, has a power of at least 2 in it. In y, its power will be at least 1. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: If x and y are integers, is xy divisible by x^2? [#permalink]

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10 Jan 2017, 16:39
VeritasPrepKarishma wrote:
manhasnoname wrote:
VeritasPrepKarishma wrote:
we know that x is prime and it is a factor of y^2. In that case, it must be a factor of y because whichever prime factors are in y, only those prime factors will be in y^2 too. So x must be a factor of y.

I know that's the truth from childhood when I learned how to check whether a number is prime.

But now I'm struggling to figure out how. Could you please point me to any resource that proves this. Thanks!

I am not sure what you are struggling with...
this - whichever prime factors are in y, only those prime factors will be in y^2 too?

If yes, say $$y = 2^3 * 3 * 7$$
What is $$y^2$$? It is
$$y^2 = 2^6 * 3^2 * 7^2$$

Can y^2 have 5 in it given that y doesn't have it? Every prime factor that y^2 has, has a power of at least 2 in it. In y, its power will be at least 1.

To check whether a number N is prime, it is enough to test whether any of the prime numbers less than or equal to sqrt(N) is a factor of N or not. I was trying to prove to myself why and how this is true.
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Joined: 16 Oct 2010
Posts: 7988
Location: Pune, India
Re: If x and y are integers, is xy divisible by x^2? [#permalink]

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10 Jan 2017, 22:57
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manhasnoname wrote:
To check whether a number N is prime, it is enough to test whether any of the prime numbers less than or equal to sqrt(N) is a factor of N or not. I was trying to prove to myself why and how this is true.

I have explained this in detail here:
http://www.veritasprep.com/blog/2010/12 ... ly-number/
https://www.veritasprep.com/blog/2010/1 ... t-squares/
_________________

Karishma
Veritas Prep | GMAT Instructor
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews BSchool Forum Moderator Joined: 12 Aug 2015 Posts: 2494 GRE 1: 323 Q169 V154 Re: If x and y are integers, is xy divisible by x^2? [#permalink] ### Show Tags 16 Jan 2017, 05:12 1 This post received KUDOS Amazing Question. Given Data-> x and y are integers. We are asked if xy/x^2 is an integer or not => y/x must be an integer too. So we can rephrase the Question to is y divisible by x i.e--> y/x is an integer or not. Statement 1-> Case 1-> x=1 y=1 YES x=4 y=2 NO Hence not sufficient. Statement 2=>x is a prime x=3 y=3 YES x=3 y=13 NOT Hence not sufficient. Combing the two statements it gets interesting. Property in action -> X and X^n always have the exact same prime factors. Hence as x is a prime factor of y^2 => x must be a prime factor of y too. Thus y/x must be an integer. Hence C. _________________ Getting into HOLLYWOOD with an MBA The MOST AFFORDABLE MBA programs! STONECOLD's BRUTAL Mock Tests for GMAT-Quant(700+) Average GRE Scores At The Top Business Schools! Senior Manager Joined: 27 May 2012 Posts: 455 Re: If x and y are integers, is xy divisible by x^2? [#permalink] ### Show Tags 10 Mar 2018, 08:12 VeritasPrepKarishma wrote: Bunuel wrote: If x and y are integers, is xy divisible by x^2? (1) x divides into y^2 with no remainder. (2) x is a prime. x and y are integers. Is xy divisible by x^2? x^2 has two x's. xy already has one x. For xy to be divisible by x^2, now all we need is another x to be a factor of y. If x is a factor of y, xy will be divisible by x^2. (1) x divides into y^2 with no remainder. x is a factor y^2 does not mean that x is a factor of y. For example, if x = 4 and y = 2, x is a factor of y^2 but not a factor of y. Not sufficient. (2) x is a prime. x may or may not a factor of y. Not sufficient. Using both, we know that x is prime and it is a factor of y^2. In that case, it must be a factor of y because whichever prime factors are in y, only those prime factors will be in y^2 too. So x must be a factor of y. Sufficient. Answer (C) Hi Karishma, When it says x divides into y^2 does it mean y^2 is divisible by x or vice versa. I solved it correctly assuming the former. So if y = 4 then x can be 2,4,8 and 16 . If you can please share your insight , we would be grateful. Thank you. _________________ - Stne Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7988 Location: Pune, India Re: If x and y are integers, is xy divisible by x^2? [#permalink] ### Show Tags 11 Mar 2018, 23:57 stne wrote: VeritasPrepKarishma wrote: Bunuel wrote: If x and y are integers, is xy divisible by x^2? (1) x divides into y^2 with no remainder. (2) x is a prime. x and y are integers. Is xy divisible by x^2? x^2 has two x's. xy already has one x. For xy to be divisible by x^2, now all we need is another x to be a factor of y. If x is a factor of y, xy will be divisible by x^2. (1) x divides into y^2 with no remainder. x is a factor y^2 does not mean that x is a factor of y. For example, if x = 4 and y = 2, x is a factor of y^2 but not a factor of y. Not sufficient. (2) x is a prime. x may or may not a factor of y. Not sufficient. Using both, we know that x is prime and it is a factor of y^2. In that case, it must be a factor of y because whichever prime factors are in y, only those prime factors will be in y^2 too. So x must be a factor of y. Sufficient. Answer (C) Hi Karishma, When it says x divides into y^2 does it mean y^2 is divisible by x or vice versa. I solved it correctly assuming the former. So if y = 4 then x can be 2,4,8 and 16 . If you can please share your insight , we would be grateful. Thank you. "x divides into y^2" means y^2 / x is an integer. It means x is a factor of y^2. So yes, if y = 4, y^2 = 16. Factors of y^2 are 1, 2, 4, 8, 16 and x can take any of these values (provided other conditions are met) _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: If x and y are integers, is xy divisible by x^2?   [#permalink] 11 Mar 2018, 23:57
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# If x and y are integers, is xy divisible by x^2?

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