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Difficulty:
85%
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Question Stats:
44% (02:13) correct
56%
(02:40)
wrong
based on 117
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If \(X = P*N^K + P\) where P, N and K are positive integers, is X odd?
(1) \(N + KN = 915\)
(2) \(P^{35} + 35^P\) is even
Are You Up For the Challenge: 700 Level Questions
(1) \(N + KN = 915\)
(2) \(P^{35} + 35^P\) is even
Are You Up For the Challenge: 700 Level Questions
19 Dec 2019, 03:05
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\(X = P*N^K + P\) .. All are positive integers. X odd?
(1) \(N + KN = 915\)
Only 1 Case to satisfy above statement: odd + even = odd, hence we know that N is odd and K is even due to which KN becomes even.
We do not know about P and hence INSUFFICIENT
A D / B C E
(2) \(P^{35} + 35^P\) is even
We know that \(35^P\) will always be odd irrespective of P whether it is odd or even. So, only 1 case that satisfies this equations is Odd + odd = Even. We can figure out that P is odd to make \(P^{35}\) odd . We do not know about N and K and hence INSUFFICIENT
A D / B C E
Combining both statements - we will get N is odd and K is even and P is odd ... SUFFICIENT
'C' is the winner
(1) \(N + KN = 915\)
Only 1 Case to satisfy above statement: odd + even = odd, hence we know that N is odd and K is even due to which KN becomes even.
We do not know about P and hence INSUFFICIENT
(2) \(P^{35} + 35^P\) is even
We know that \(35^P\) will always be odd irrespective of P whether it is odd or even. So, only 1 case that satisfies this equations is Odd + odd = Even. We can figure out that P is odd to make \(P^{35}\) odd . We do not know about N and K and hence INSUFFICIENT
Combining both statements - we will get N is odd and K is even and P is odd ... SUFFICIENT
'C' is the winner
Bunuel
If \(X = P*N^K + P\) where P, N and K are positive integers, is X odd?
(1) \(N + KN = 915\)
(2) \(P^{35} + 35^P\) is even
Are You Up For the Challenge: 700 Level Questions
(1) \(N + KN = 915\)
(2) \(P^{35} + 35^P\) is even
Are You Up For the Challenge: 700 Level Questions
08 Feb 2020, 08:35
Kudos
Bookmarks
…
If \(X = P*N^K + P\) where P, N and K are positive integers, is X odd?
(1) \(N + KN = 915\)
(2) \(P^{35} + 35^P\) is even
(1) \(N + KN = 915\)
(2) \(P^{35} + 35^P\) is even
pnk: positive integers
\(x=pn^k+p=p(n^k+1)\)
if p=even then ans is no;
if n^k+1=even, or if n=odd, then ans is no.
(1) \(N + KN = 915\) sufic
n+kn=odd…n(1+k)=odd…(odd*odd=odd)…n=odd, k=even;
(2) \(P^{35} + 35^P\) is even insufic
\(P^{35} + 35^P=even…(o+o=e…e+e=e)…P+o=e…P=odd\)
\(x=pn^k+p=p(n^k+1)=odd(n^k+1)=(odd,even)\)
Ans (A)
