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If x and y are integers, is xy divisible by x^2?
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09 Jan 2017, 03:26
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Re: If x and y are integers, is xy divisible by x^2?
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09 Jan 2017, 09:12
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?
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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?
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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?
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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.
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Re: If x and y are integers, is xy divisible by x^2?
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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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Re: If x and y are integers, is xy divisible by x^2?
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10 Jan 2017, 22:57
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 ... lynumber/https://www.veritasprep.com/blog/2010/1 ... tsquares/
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Re: If x and y are integers, is xy divisible by x^2?
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16 Jan 2017, 05:12
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?
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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?
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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)
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