Hi All,
In the below question, can someone please tell....
what will sum be:
a) Sum = n + k*n...
where k = odd ?
or
b) Sum = k* n
where k = odd.
===============================================
An odd number added to itself an odd number of times yields:
a) an odd number.
b) an even number.
c) a prime number.
d) a positive number.
e) a perfect square.
OA = A
Odd number Properties
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Answer is A because
a) k = odd, n = even or odd
even test: even + (even*odd) = even + even = even
odd test: odd + (odd*odd) = odd + odd = even
Therefore the sum must be even. Sufficient.
b) k = odd, n = even or odd
even test: odd*even = even
odd test: odd*odd = odd
Therefore the sum can either be odd or even. Insufficient.
A
a) k = odd, n = even or odd
even test: even + (even*odd) = even + even = even
odd test: odd + (odd*odd) = odd + odd = even
Therefore the sum must be even. Sufficient.
b) k = odd, n = even or odd
even test: odd*even = even
odd test: odd*odd = odd
Therefore the sum can either be odd or even. Insufficient.
A
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Hi tim0thy,tim0thy wrote:Answer is A because
a) k = odd, n = even or odd
even test: even + (even*odd) = even + even = even
odd test: odd + (odd*odd) = odd + odd = even
Therefore the sum must be even. Sufficient.
b) k = odd, n = even or odd
even test: odd*even = even
odd test: odd*odd = odd
Therefore the sum can either be odd or even. Insufficient.
A
Probably you misread the question....this is not a DS question....it is a problem solving question...
Thanks
Mohit
Sorry Mohit I didn't see the question. But here's my attempt to explain why it's A.
an odd number added to itself an odd number of times would be odd+odd+odd, or odd+odd+odd+odd+odd etc., rewritten as 3*odd or 5*odd, or it can be generalized to (2k+1)*odd for any integer k. Because 2k+1 will always be odd for any integer k, and because we are multiplying it by another odd number, by rule odd*odd will always be odd. Therefore answer is A.
Hope this helps.
an odd number added to itself an odd number of times would be odd+odd+odd, or odd+odd+odd+odd+odd etc., rewritten as 3*odd or 5*odd, or it can be generalized to (2k+1)*odd for any integer k. Because 2k+1 will always be odd for any integer k, and because we are multiplying it by another odd number, by rule odd*odd will always be odd. Therefore answer is A.
Hope this helps.
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Hi tim0thy,tim0thy wrote: an odd number added to itself an odd number of times would be odd+odd+odd, or odd+odd+odd+odd+odd etc., rewritten as 3*odd or 5*odd, or it can be generalized to (2k+1)*odd for any integer k.
isn't this the case of odd added to itself "even" no. of times ?
basically what I interpret is that the number itself is fixed....and in that we are adding the number odd no. of times...
i.e. S = x + k*x
where k = odd....
and x is already given to be odd....
so sum should be S = odd + odd = even....
please tell what is the flaw in my approach ?
Thanks
Mohit
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the flaw is only in the interpretation of the question. timothy is correct.
an odd number added to itself an odd number of times=>odd times the odd number=odd*odd=odd
an odd number added to itself an odd number of times=>odd times the odd number=odd*odd=odd
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Hi Scooby,scoobydooby wrote:the flaw is only in the interpretation of the question. timothy is correct.
an odd number added to itself an odd number of times=>odd times the odd number=odd*odd=odd
Thanks.
But I am really not able to understand how I am interpreting wrongly...
e.g. No. = n = 1.
Now If we add n to n say 1 time..i.e. odd time....
then isn't n + 1* n is the correct interpretation......
I am not able to understand why are we ignoring the base "n" to which we are adding something....
Just one similar example help let people understand my doubt:
What is the value of 2 added to itself...isn't it 4...then why the same logic does not apply here.... ?
Can someone please help.
Thanks
Mohit
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it is Even. If n is even than even+(odd*even) = even+even=evengoelmohit2002 wrote:Hi All,
In the below question, can someone please tell....
what will sum be:
a) Sum = n + k*n...
where k = odd ?
If n is odd than odd+(odd*odd)= odd+odd=even
it depends on N. If n is odd than odd*odd =odd and If n is even than odd*even=evenor
b) Sum = k* n
where k = odd.
This question is really ambiguous. I doubt the source of the question===============================================
An odd number added to itself an odd number of times yields:
a) an odd number.
b) an even number.
c) a prime number.
d) a positive number.
e) a perfect square.
OA = A
I think it should be even coz an x odd number is added to itself with odd*x. This is what I can interpret from the question.
Thanks,
Quant-Master[/b]
https://gmat-quants.blocked - My Blog Updated almost daily with new quant fundas. Find collection of quants question in my blog
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Thanks QM....quant-master wrote:it is Even. If n is even than even+(odd*even) = even+even=evengoelmohit2002 wrote:Hi All,
In the below question, can someone please tell....
what will sum be:
a) Sum = n + k*n...
where k = odd ?
If n is odd than odd+(odd*odd)= odd+odd=evenit depends on N. If n is odd than odd*odd =odd and If n is even than odd*even=evenor
b) Sum = k* n
where k = odd.This question is really ambiguous. I doubt the source of the question===============================================
An odd number added to itself an odd number of times yields:
a) an odd number.
b) an even number.
c) a prime number.
d) a positive number.
e) a perfect square.
OA = A
I think it should be even coz an x odd number is added to itself with odd*x. This is what I can interpret from the question.
Thanks,
Quant-Master[/b]
This is a question taken from nova book...please see the same at the following link. What do you think of...should we happily forget about this question
https://books.google.co.in/books?id=prEE ... q=&f=false