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Posts: 3,651
Updated on: 04 May 2020, 07:53
Originally posted by EgmatQuantExpert on 14 Apr 2020, 08:12.
Last edited by Bunuel on 04 May 2020, 07:53, edited 2 times in total.
Last edited by Bunuel on 04 May 2020, 07:53, edited 2 times in total.
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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28% (02:31) correct
73%
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wrong
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If p and k are non-negative distinct integers, then is \(p^{18} + 18^{pk}\) odd?
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- (1) \(p^k\) is even
(2) \(p^{k + 1}\) is even
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ArunSharma12
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Updated on: 14 Apr 2020, 20:56
Originally posted by ArunSharma12 on 14 Apr 2020, 08:53.
Last edited by ArunSharma12 on 14 Apr 2020, 20:56, edited 1 time in total.
Last edited by ArunSharma12 on 14 Apr 2020, 20:56, edited 1 time in total.
EgmatQuantExpert
Given: p,k \(\ge0\)
statement 1:\(p^k\) is even
when p = 0 & k > 0
\(p^{18} + 18^{pk}\) = even + 1 = odd
when p > 0 & even then k > 0
\(p^{18} + 18^{pk}\) = even + even = even
not sufficient
statement 2:\(p^(k + 1)\) is even
when p = 0 & k \(\ge\) 0
\(p^{18} + 18^{pk}\) = even + 1 = odd
when p > 0 & even k = 0
\(p^{18} + 18^{pk}\) = even + 1 = odd
when p > 0, even & k > 0
\(p^{18} + 18^{pk}\) = even + even = even
not sufficient
combining both statements,
when p = 0 & k > 0
\(p^{18} + 18^{pk}\) = even + 1 = odd
when p > 0, even & k > 0
\(p^{18} + 18^{pk}\) = even + even = even
not sufficient
Ans: E
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If p and k are non-negative distinct integers, then is \(p^{18} + 18^{pk}\) odd?
[list](1) \(p^k\) is even
(2) \(p^(k + 1)\) is even
1) p^k is even. So, we can assume that p is an even integer(including 0) and k >0. If p is 0, then p^18 +18^pk will be 0 + 1=1, odd. But if p is non-zero even, then p^18 +18^pk will be even. Not sufficient.
2) In this case also p can assume the value 0. not sufficient.
Together, p can be any even value including 0, which will make the sum 1. Other even values will add to an even integer. Not sufficient.
E is the answer.
[list](1) \(p^k\) is even
(2) \(p^(k + 1)\) is even
1) p^k is even. So, we can assume that p is an even integer(including 0) and k >0. If p is 0, then p^18 +18^pk will be 0 + 1=1, odd. But if p is non-zero even, then p^18 +18^pk will be even. Not sufficient.
2) In this case also p can assume the value 0. not sufficient.
Together, p can be any even value including 0, which will make the sum 1. Other even values will add to an even integer. Not sufficient.
E is the answer.
EgmatQuantExpert
EgmatQuantExpert,
Please check the OA
Statement 1 : P=0 and k=1 , \(p^{18} + 18^{pk}\) = 0+1 =1 Odd, so ans is yes
\(\hspace{8mm} \) P=2 \(and\hspace{1mm}\)k=2 then \(p^{18} + 18^{pk}\) = even+even =even , so Ans is No.
INSUFF.
Hence A cannot be the answer.
In fact P=0 and K=1 and P=2 k=2 satisfy both statements , but give different Answers. In the current format ans. should be E.
_________________
- Stne
Its simple. Answer is A
p and k are non-negative distinct integers
Given p>= 0 and k>=0
p^18= a
18^(pk) = b
a+b = odd if ( a=odd, b = even ) and if ( a = even, b = odd )
a+b = even if ( a=even, b = even ) and if ( a = odd, b = odd )
b can be odd only if pk = 0 .
(1) p^k is even
This means that p = even, k not equal to 0 . ie k= 1,2,3,4,.....
p^18 is even
18^pk is even
Overall Even
SUFFICIENT
(2) p^(k+1) is even
This means that p is even, k = 0,1,2,3,4,...
p^18 is even
18^(pk) = 1 or even. Dont know
NOT SUFFICIENT
Hence Answer A
p and k are non-negative distinct integers
Given p>= 0 and k>=0
p^18= a
18^(pk) = b
a+b = odd if ( a=odd, b = even ) and if ( a = even, b = odd )
a+b = even if ( a=even, b = even ) and if ( a = odd, b = odd )
b can be odd only if pk = 0 .
(1) p^k is even
This means that p = even, k not equal to 0 . ie k= 1,2,3,4,.....
p^18 is even
18^pk is even
Overall Even
SUFFICIENT
(2) p^(k+1) is even
This means that p is even, k = 0,1,2,3,4,...
p^18 is even
18^(pk) = 1 or even. Dont know
NOT SUFFICIENT
Hence Answer A
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ArunSharma12
I too feel that it is E. If it was mentioned in the stem that p was positive, then A would suffice. Bunuel, is my understanding correct?
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How can 0 be considered as even number.
Answer is a clear A
According to the cyclic rule.
None of the odd number power to any number --> leads to an even number.
Answer is a clear A
According to the cyclic rule.
None of the odd number power to any number --> leads to an even number.
NischalP
Dear NischalP,
You should have googled it, before being so sure.According to my knowledge ZERO is in fact an EVEN number.
https://www.britannica.com/story/is-zer ... odd-number
Hope it's clear.
OA is clearly wrong ! Hope Moderator looks into this.
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NischalP
I suppose you still didn't do any research.
Would you call this a random link also.
https://gmatclub.com/forum/is-zero-even-84800.html
Additionally instead of asking me to check with my school teacher, you should perhaps check your source, from where exactly did you come to know that 0 is not even. Please be a little respectful while replying, Thank you.
_________________
- Stne
NischalP
0 is an even number. There is not discussion about it.
ZERO:
1. 0 is an integer.
2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.
3. 0 is neither positive nor negative integer (the only one of this kind).
4. 0 is divisible by EVERY integer except 0 itself.
Check more here: https://gmatclub.com/forum/number-proper ... 74996.html
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Bunuel
Dear Moderator,
Thank you for clearing this.
Can you please look into the OA, unless many of us missed something , correct answer seems to be E. Thank you.
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- Stne
stne
Yes, the answer to the question posted is E. There might be some formatting error or a typo in the text.
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EgmatQuantExpert
Asked: If p and k are non-negative distinct integers, then is \(p^{18}+18^{pk}\) odd?
p^{18} is even/odd if p is even/odd.
18^{pk} is odd if one of p/k is 0, otherwise is even.
Q. Is \(p + 2^{pk}\) odd?
If p and k are non-negative distinct integers; then both p & k are not 0, but one of them may be 0.
(1) \(p^k\) is even
If p is even ; p^k is even when \(k \neq 0 \)
But p may be 0
NOT SUFFICIENT
(2) \(p^{k+1}\) is even
If p is even ; p^k is even when \(k+1 \neq 0 \)
k may be 0.
NOT SUFFICIENT
(1) + (2)
(1) \(p^k\) is even
If p is even ; p^k is even when \(k \neq 0 \)
But p may be 0
(2) \(p^{k + 1}\) is even
If p is even ; p^k is even when \(k+1 \neq 0 \)
k may be 0.
Either p or k may be 0
NOT SUFFICIENT
IMO E
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Bunuel
What is 0 to the power 0??
desertEagle
0^0, in some sources equals to 1, some mathematicians say it's undefined. But you won't need this for the GMAT because the case of 0^0 is not tested on the GMAT.
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