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Re: odds and evens [#permalink]
I will go with C.

If x + y is an odd integer and x - y is also an odd integere then both x and y must be integers. If x and / or y are fractions, the above conditions will not satisfy.
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Re: odds and evens [#permalink]
ans iS C. good expln suresh!!!+1
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Re: odds and evens [#permalink]
answer by Suresh is correct but explanation is wrong.

if x + y = odd and x - y = odd implies that

2x = odd + odd = even = 2 * integer -> x is an integer

similarly 2y = odd - odd = even inetger = 2 * integer -> y is an integer

hence 2*x*5*y is an integer and (C) is the answer !!
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Re: odds and evens [#permalink]
krishan wrote:
answer by Suresh is correct but explanation is wrong.

if x + y = odd and x - y = odd implies that

2x = odd + odd = even = 2 * integer -> x is an integer

similarly 2y = odd - odd = even inetger = 2 * integer -> y is an integer

hence 2*x*5*y is an integer and (C) is the answer !!



Agreed. Thanks for pointing out.

I modified that.



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