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

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

Hi tim0thy,

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.

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.

Hi tim0thy,

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

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

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

Hi Scooby,

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

goelmohit2002 wrote:Hi All,

In the below question, can someone please tell....

what will sum be:

a) Sum = n + k*n...
where k = odd ?

it is Even. If n is even than even+(odd*even) = even+even=even
If n is odd than odd+(odd*odd)= odd+odd=even

or
b) Sum = k* n
where k = odd.

it depends on N. If n is odd than odd*odd =odd and If n is even than odd*even=even

===============================================
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

This question is really ambiguous. I doubt the source of the question

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]

quant-master wrote:

goelmohit2002 wrote:Hi All,

In the below question, can someone please tell....

what will sum be:

a) Sum = n + k*n...
where k = odd ?

it is Even. If n is even than even+(odd*even) = even+even=even
If n is odd than odd+(odd*odd)= odd+odd=even

or
b) Sum = k* n
where k = odd.

it depends on N. If n is odd than odd*odd =odd and If n is even than odd*even=even

===============================================
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

This question is really ambiguous. I doubt the source of the question

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]

Thanks QM....

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