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Difficulty:
75%
(hard)
Question Stats:
59% (02:53) correct
41%
(02:40)
wrong
based on 177
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If a and b are positive integers, is \(ab + b^a + 1\) odd?
(1) \((a+b)\) is odd.
(2) (\(a+1)^b + (b+2)^{(a+1)}\) is even.
(1) \((a+b)\) is odd.
(2) (\(a+1)^b + (b+2)^{(a+1)}\) is even.
22 Sep 2021, 02:44
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Bunuel
\(ab + b^a + 1\)
Case I: If b is even, then \(ab + b^a + 1=E+E+1\) = odd irrespective of the value of a.
(b:E,a:E) and (b:E,a:O) will give answer as ODD.
Case II: If b is odd, then \(ab + b^a + 1=a*O+O+1=a*O+E\) will depend on the value of a. If a is odd, answer is odd, and if a is even, answer is even.
(b:O,a:E) and (b:O,a:O) will give answer as EVEN and ODD respectively.
Let us see the statements.
(1) \((a+b)\) is odd.
This means both are of opposite property.
It can lead to (b:O,a:E), giving result as even and (b:O,a:E), giving result as odd.
Insufficient
(2) \((a+1)^b + (b+2)^{(a+1)}\) is even.
Thus (a+1) and (b+2) are both either odd or even.
This tells us that a and b are opposite in property.
Same as I above.
It can lead to (b:O,a:E), giving result as even and (b:O,a:E), giving result as odd.
Insufficient
Combined
Nothing new.
E
22 Sep 2021, 08:25
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target is \(ab + b^a + 1\) odd
#1
\((a+b)\) is odd.
possible when either of a or b is even and other is odd
so we get yes and no to target
#2
(\(a+1)^b + (b+2)^{(a+1)}\) is even.
possible when ( a+1) + ( b+2) are both odd or both are even
i.e. same as #1 either of a or b is even and other is odd
again we get yes and no to target
from 1 &2 nothing can be determined
option E is correct
#1
\((a+b)\) is odd.
possible when either of a or b is even and other is odd
so we get yes and no to target
#2
(\(a+1)^b + (b+2)^{(a+1)}\) is even.
possible when ( a+1) + ( b+2) are both odd or both are even
i.e. same as #1 either of a or b is even and other is odd
again we get yes and no to target
from 1 &2 nothing can be determined
option E is correct
Bunuel
