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05 Jul 2018, 05:47
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D
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Difficulty:
95%
(hard)
Question Stats:
41% (02:12) correct
59%
(01:54)
wrong
based on 290
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If x and y are positive integers, then what is the value of \((–1)^x + (–1)^y + (–1)^x*(–1)^y\) ?
(1) x does not have a prime factor greater than 1.
(2) y is a common factor of all prime numbers between 101 to 202
M36-98
(1) x does not have a prime factor greater than 1.
(2) y is a common factor of all prime numbers between 101 to 202
M36-98
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12 Nov 2021, 09:03
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Bunuel
Official Solution:
If x and y are positive integers, then what is the value of \((–1)^x + (–1)^y + (–1)^x*(–1)^y\) ?
(1) \(x\) does not have a prime factor greater than 1.
The above means that \(x\) must be 1. All other numbers are either composite or prime.
In this case: \((–1)^x + (–1)^y + (–1)^x*(–1)^y=(–1)^{1} + (–1)^{y} + (–1)^{1}*(–1)^{y}=-1+(–1)^{y}-(–1)^{y}=-1\). Sufficient.
(2) \(y\) is a common factor of all prime numbers between 101 to 202
The only common factor those integer have is 1, so \(y=1\).
In this case: \((–1)^x + (–1)^y + (–1)^x*(–1)^y=(–1)^{x} + (–1)^{1} + (–1)^{x}*(–1)^{1}=(–1)^{x}-1-(–1)^{x}=-1\). Sufficient.
Answer: D
General Discussion
05 Jul 2018, 07:17
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Bunuel
Given, x and y are positive integers
St1:- x does not have a prime factor greater than 1
So, x=1 implies\((-1)^x=(-1)^1=-1\)
when y=even, (-1)^y=1
So, \((–1)^x + (–1)^y + (–1)^x*(–1)^y\)=-1+1+(-1)(1)=-1
when y=odd, \((-1)^y=-1\)
So, \((–1)^x + (–1)^y + (–1)^x*(–1)^y\)=-1+(-1)+(-1)(-1)=-1
hence sufficient.
St2:-y is a common factor of all prime numbers between 101 to 202
So, y=1 implies \((-1)^y=(-1)^1=-1\)
when x=even, \((-1)^x=1\)
So, \((–1)^x + (–1)^y + (–1)^x*(–1)^y\)=1+(-1)+1(-1)=-1
when x=odd, (-1)^x=-1
So, \((–1)^x + (–1)^y + (–1)^x*(–1)^y\)=-1+(-1)+(-1)(-1)=-1
hence sufficient.
Ans. D
