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

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Bunuel
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If x and y are integers, is xy divisible by x^2? [#permalink] New post   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.
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C
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KarishmaB
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If x and y are integers, is xy divisible by x^2? [#permalink] New post   Updated on: 09 Aug 2023, 03:54
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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.
Answer (C)

Check this video: https://youtu.be/DxIH8rjhpKY

Originally posted by KarishmaB on 09 Jan 2017, 09:12.
Last edited by KarishmaB on 09 Aug 2023, 03:54, edited 1 time in total.
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Re: If x and y are integers, is xy divisible by x^2? [#permalink] New post   09 Jan 2017, 07:44
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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.


(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] New post   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!
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Re: If x and y are integers, is xy divisible by x^2? [#permalink] New post   10 Jan 2017, 03:18
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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.
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Re: If x and y are integers, is xy divisible by x^2? [#permalink] New post   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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If x and y are integers, is xy divisible by x^2? [#permalink] New post   Updated on: 17 Oct 2022, 02:18
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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.


Check out the blog posts on the link given in my signature below.

Originally posted by KarishmaB on 10 Jan 2017, 22:57.
Last edited by KarishmaB on 17 Oct 2022, 02:18, edited 1 time in total.
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Re: If x and y are integers, is xy divisible by x^2? [#permalink] New post   16 Jan 2017, 05:12
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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.
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Re: If x and y are integers, is xy divisible by x^2? [#permalink] New post   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.
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Re: If x and y are integers, is xy divisible by x^2? [#permalink] New post   11 Mar 2018, 23:57
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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)
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Re: If x and y are integers, is xy divisible by x^2? [#permalink] New post   03 Feb 2020, 11:53
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Re: If x and y are integers, is xy divisible by x^2? [#permalink]
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