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

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

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New post 09 Jan 2017, 02:26
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Re: If x and y are integers, is xy divisible by x^2?  [#permalink]

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New post 09 Jan 2017, 08:12
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2
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)
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Re: If x and y are integers, is xy divisible by x^2?  [#permalink]

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New post 09 Jan 2017, 06:44
1
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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New post 09 Jan 2017, 19:04
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]

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New post 10 Jan 2017, 02:18
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.
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Re: If x and y are integers, is xy divisible by x^2?  [#permalink]

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New post 10 Jan 2017, 15:39
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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Re: If x and y are integers, is xy divisible by x^2?  [#permalink]

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New post 10 Jan 2017, 21:57
1
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/
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Re: If x and y are integers, is xy divisible by x^2?  [#permalink]

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New post 16 Jan 2017, 04:12
1
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]

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New post 10 Mar 2018, 07:12
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]

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New post 11 Mar 2018, 22:57
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)
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Re: If x and y are integers, is xy divisible by x^2? &nbs [#permalink] 11 Mar 2018, 22:57
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