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09 Jan 2017, 03:26
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C
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Difficulty:
75%
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Question Stats:
46% (01:57) correct
54%
(01:49)
wrong
based on 185
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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.
(1) x divides into y^2 with no remainder.
(2) x is a prime.
Bunuel
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 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
General Discussion
09 Jan 2017, 07:44
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Bunuel
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 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
