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If a, b, and c are integers, is (a+b)2+c(a2+b2) an even number?

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If a, b, and c are integers, is (a+b)2+c(a2+b2) an even number? [#permalink]

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New post 28 Mar 2017, 15:50
2
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A
B
C
D
E

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Question Stats:

39% (00:55) correct 61% (01:10) wrong based on 46 sessions

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If a, b, and c are integers, is \((a+b)^2+c(a^2+b^2)\) an even number?

1) a and b are odd numbers
2) c is an odd number.

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Re: If a, b, and c are integers, is (a+b)2+c(a2+b2) an even number? [#permalink]

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New post 28 Mar 2017, 20:00
What is the OE?

IMO, the answer should be C.

Statement 1 says :-
a and b are odd. Which means (a+b)^2 will be even. Now whether c(a^2+b^2) is even, depends on the value of c.[(a^2 +b^2) is even as odd plus odd is always even]. So if c is odd, the overall value of the equation will be odd, else it will be even.
Hence Statement 1 is insufficient.

Statement 2:-
C is odd.
This statement doesn't tell us anything about a and b and hence it is insufficient on its own.

Combining both we can conclude:-
(a+b)^2 to be even.
and c(a^2+b^2) to be odd.

Odd + even = odd
Hence C.

Am I wrong in my conclusion?
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Re: If a, b, and c are integers, is (a+b)2+c(a2+b2) an even number? [#permalink]

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New post 28 Mar 2017, 20:44
ziyuen wrote:
If a, b, and c are integers, is \((a+b)^2+c(a^2+b^2)\) an even number?

1) a and b are odd numbers
2) c is an odd number.


Hi ,

\(a^2+b^2\) will be same that is odd/ even as a+b would be..

So let's take TWO cases possible
A) a+b is Even, so \((a+b)^2+c(a^2+b^2)\) becomes E + c*E..
Both terms are even so total will be even, IRRESPECTIVE of c..
B) a+b is Odd, so \((a+b)^2+c(a^2+b^2)\) becomes Odd + c*Odd..
Both terms are Odd if c is odd so total will be even IRRESPECTIVE of a and b..
If c is Even, total becomes Odd+Even*Odd=Odd

Let's see the statements now..
I) a & b are odd, so a+b is Even
And TOTAL is Even as per case A.
Sufficient

II) c is odd...
If c is odd, irrespective of a and b , ans will be even as per case B.
Sufficient

D
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Re: If a, b, and c are integers, is (a+b)2+c(a2+b2) an even number?   [#permalink] 28 Mar 2017, 20:44
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