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03 Jul 2018, 11:08
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
34% (01:28) correct
66%
(01:09)
wrong
based on 139
sessions
History
Date
Time
Result
Not Attempted Yet
03 Jul 2018, 11:45
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Is \(q^r + 1\) an odd number?
\(q^r+1\) odd means \(q^r\) is even
Easy to wrong by believing q as even would be sufficient----
NO we have to know
a) q and r are positive integers
b) q is even
(1) q + r is even
3+(-1) is even
but \(q^r+1=3^{-1}+1=\frac{1}{3}+1\).. not odd
2+2 is even.. yes \(q^r+1\) is odd
insuff
(2) q is even
nothing about r
insuff
Combined
q is2 and r is 0...2^0+1=2.... Even
wish 2 and r is -2.....2^-2+1=5/2.... fraction
q is 2 and r is 4..... 2^4+1=17...odd
Insufficient
E
\(q^r+1\) odd means \(q^r\) is even
Easy to wrong by believing q as even would be sufficient----
NO we have to know
a) q and r are positive integers
b) q is even
(1) q + r is even
3+(-1) is even
but \(q^r+1=3^{-1}+1=\frac{1}{3}+1\).. not odd
2+2 is even.. yes \(q^r+1\) is odd
insuff
(2) q is even
nothing about r
insuff
Combined
q is2 and r is 0...2^0+1=2.... Even
wish 2 and r is -2.....2^-2+1=5/2.... fraction
q is 2 and r is 4..... 2^4+1=17...odd
Insufficient
E
03 Jul 2018, 15:40
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Bunuel
we are asked to determine whether \(q^r\) + 1 = odd or not.
Statement 1 :
q + r = even . This fact can be proved by 2 ways.
1. even + even = even . 2 +2 = 4
2. odd + odd = even . 3 + 1 = 4
If both q and r are even , we will always get odd number but the scenario is different when q and r are odd.
NOT sufficient.
Statement 2 :
This statement is a trap. anyone can say that if q is even \(q^r\) + 1 will be odd. But we don't sufficient information about r. Thus this statement becomes fruitless. r could be anything. so, NOT sufficient .
combining both statements:
if q is even r has to be even if we consider statement 1. even or odd can't be fraction .
So, both statements together are sufficient.
Thus the best answer is C.
