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If a^5+7a+2b+b^2 is even and a and b are integers, is

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If a^5+7a+2b+b^2 is even and a and b are integers, is  [#permalink]

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New post 08 Feb 2018, 18:28
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Question Stats:

56% (01:55) correct 44% (02:40) wrong based on 57 sessions

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If a^5 + 7a + 2b + b^2 is even and a and b are integers, is a even integer?

(1) a + 2b is even.
(2) a + b is even.

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Re: If a^5+7a+2b+b^2 is even and a and b are integers, is  [#permalink]

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New post 08 Feb 2018, 18:56
chetan2u wrote:
If a^5+7a+2b+b^2 is even and a and b are integers, is a even integer?
(1) a+2b is even.
(2) a+b is even.


From Statement 1: We have \(a + 2b = Even,\) 2b will be always even irrespective of b = odd/even. Hence, a = Even.
From Statement 2: We have \(a + b = Even.\)
I. a= Odd, b = Odd
\(a^5 + 7a + 2b + b^2 = Even\)---> b = Even(But as per our assumption b has to be Odd). So, this is ruled out.

II. a = Even, b = Even
\(a^5 + 7a + 2b + b^2 = Even\)
The only case possible is a = Even.

Hence, D.
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Re: If a^5+7a+2b+b^2 is even and a and b are integers, is  [#permalink]

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New post 08 Feb 2018, 21:50
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chetan2u wrote:
If a^5 + 7a + 2b + b^2 is even and a and b are integers, is a even integer?

(1) a + 2b is even.
(2) a + b is even.


In a^5 + 7a + 2b + b^2, since a & b are integers:
2b is always going to be even.
a^5 + 7a will also always be even (if a is odd, we get odd+odd = even; and if a is even, we get even+even = even)
Since a^5 + 7a + 2b is always even, for entire sum to be even, b^2 also must be even. This means b is even.
Thus we know that 'b' is even for sure, we need to establish whether 'a' is even or not.

Statement 1

a+2b is even, 2b is anyway even; so for this sum to be even, 'a' also must be even. Sufficient.


Statement 2

a+b is even. We have already established that 'b' is even, so for this sum to be even, 'a' also must be even. Sufficient.


Hence D answer
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Re: If a^5+7a+2b+b^2 is even and a and b are integers, is &nbs [#permalink] 08 Feb 2018, 21:50
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If a^5+7a+2b+b^2 is even and a and b are integers, is

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