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Is \((x+y-3)^2\) divisible by 4, given that x and y are integers?
1) \(x = 1\); \(m - n = 1\), where \(m\) and \(n\) are the number of distinct prime factors of \(2y\) and \(y\) respectively.
2) \(x^2 +y = x^3 + 3x\)
1) \(x = 1\); \(m - n = 1\), where \(m\) and \(n\) are the number of distinct prime factors of \(2y\) and \(y\) respectively.
2) \(x^2 +y = x^3 + 3x\)
GMATbuster's Collection
23 Sep 2019, 02:59
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Is \((x+y-3)^2\) divisible by 4?
or Is \((x+y-3)\) divisible by 2?
or, Is \((x+y-3)\) even?
or \(x+y\) = odd?
Statement 1: x = 1,
we need to find whether y is odd or even.
Hence, y is ODD.
So, x + y = odd+odd = even,
Hence, it is sufficient to answer that \((x+y-3)^2\) is not divisible by 4.
SUFFICIENT
Statement 2:
We know that having a positive integral exponent doesn't change the even or odd nature of any number
Also, multiplication by odd number doesn't change the even or odd nature of any number.
Hence,
if x = odd
\(x^2 +y = x^3 + 3x\)
or, odd +y = odd +odd
or y = odd
Hence (x+y) = even
if x = even
\(x^2 +y = x^3 + 3x\)
or, even +y = even +even
or y = even
Hence (x+y) = even
So, in all cases (x+y) = even
Hence, it is sufficient to answer that \((x+y-3)^2\) is not divisible by 4.
SUFFICIENT.
Hence, the answer is D
or Is \((x+y-3)\) divisible by 2?
or, Is \((x+y-3)\) even?
or \(x+y\) = odd?
Statement 1: x = 1,
we need to find whether y is odd or even.
if y is even, then y and 2y would have the same number of distinct prime factors, hence y is not even.
but \(m - n = 1\), where \(m\) and \(n\) are the number of distinct prime factors of \(2y\) and \(y\) respectivelyHence, y is ODD.
So, x + y = odd+odd = even,
Hence, it is sufficient to answer that \((x+y-3)^2\) is not divisible by 4.
SUFFICIENT
Statement 2:
We know that having a positive integral exponent doesn't change the even or odd nature of any number
if \(x\) is odd ( or even), \(x^n\) = odd ( or even), when n is a positive integer.
Also, multiplication by odd number doesn't change the even or odd nature of any number.
Hence,
if \(x\) is odd ( or even), \(x*a\) = odd ( or even), when a is an odd integer.
if x = odd
\(x^2 +y = x^3 + 3x\)
or, odd +y = odd +odd
or y = odd
Hence (x+y) = even
if x = even
\(x^2 +y = x^3 + 3x\)
or, even +y = even +even
or y = even
Hence (x+y) = even
So, in all cases (x+y) = even
Hence, it is sufficient to answer that \((x+y-3)^2\) is not divisible by 4.
SUFFICIENT.
Hence, the answer is D
gmatbusters
Is \((x+y-3)^2\) divisible by 4?
1) \(x = 1\); \(m - n = 1\), where \(m\) and \(n\) are the number of distinct prime factors of \(2y\) and \(y\) respectively.
2) \(x^2 +y = x^3 + 3x\)
1) \(x = 1\); \(m - n = 1\), where \(m\) and \(n\) are the number of distinct prime factors of \(2y\) and \(y\) respectively.
2) \(x^2 +y = x^3 + 3x\)
GMATbuster's Collection
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