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Re: If a, b, and c are all integers, is ab+bc+ca [#permalink]

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16 Sep 2012, 01:01

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navigator123 wrote:

If a, b, and c are all integers, is ab+bc+ca+(a*a) odd?

(1) a is odd. (2) (b+c) is odd. - From HULT free tests

From Question stem : a^2+ab+ac+bc =(a+c)(a+b). product of 2 numbers. St1: Insufficient:Let say a is odd, c even & b even, so (a+b) =odd & (a+b)=odd , So (a+b)(a+c) = odd, which gives YES Let say a is odd, c odd & b even, So (a+b)= odd & (a+c)=even, So (a+b)(a+c)= even, which gives NO.

St 2: Sufficient: if b+c = odd, any one of b & c should be even and another odd. so a can be either odd or even. Let say b=O & C=E and A=O (a+b)=E, (a+c) = E, hence (a+b)(a+c)= E, which gives NO Now Let say b=O & C=E and A=E, (a+b)=O, (a+c)=E, Hence (a+b)(a+c)= E, which gives NO So any one of the factors (a+b) or (a+c) will be either E or O which will always give an Even number.
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If a, b, and c are all integers, is ab+bc+ca+a2 odd?

(1) a is odd. (2) (b+c) is odd.

Looking at the statements, I would try to club b and c together.

Given Expression: a(b+c) + a^2 + bc There are 3 terms: a(b+c), a^2, bc

(1) a is odd. a^2 is certainly odd. But we don't know anything about the other two terms. Say b and c are both even. Then 2 terms (a(b+c) and bc) are even and one (a^2) is odd so sum is odd. Say b and c are both odd. Then 2 terms (a^2 and bc) are odd and one (a(b+c)) is even so sum is even. Not sufficient.

(2) (b+c) is odd. If b+c is odd, it means one of b and c is odd and the other is even. So bc will be even Now if a is odd, two terms are odd (a(b+c) and a^2) while the third term (bc) is even. So sum will be even. If a is even, all three terms are even so sum will be even. In any case, the sum will be even so this statement alone is sufficient.

Re: If a, b, and c are all integers, is ab+bc+ca+a2 odd? [#permalink]

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20 Feb 2014, 00:18

mhknair wrote:

If a, b, and c are all integers, is ab+bc+ca+a2 odd?

(1) a is odd. (2) (b+c) is odd.

Simpler: Regroup to get \((ab + ac) + bc + a^2\). Since a is odd, we know \(\underbrace{ab + ac}_{odd} + bc + \underbrace{a^2}_{odd}\). Next, since b+c is odd we know either b or c is odd and the other is even -- so bc must be even. Odd + even + odd is even.

Alternatively: Regroup to get \((a^2 + ab + ac) + bc\). Since a and (b+c) are odd, we know \(a+b+c\) is even and \(\underbrace{a(a+b+c)}_{even} + bc\). Next, since b+c is odd we know b or c is odd and the other is even -- so bc must be even. Even + even is even.

Re: If a, b, and c are all integers, is ab+bc+ca [#permalink]

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21 Sep 2016, 00:20

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Re: If a, b, and c are all integers, is ab+bc+ca+a^2 odd? [#permalink]

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14 Jan 2017, 06:50

Is \(ab+bc+ca+a^2\) odd = a(b + c) + bc + a^2 = a(b + c + a) + bc

St1: a is odd. --> No info about b and c. Clearly insufficient.

St2: (b+c) is odd. --> Possible when one value is odd and the other value is even. Hence if b + c is odd then b *c is even.

Consider a(b + c + a) + bc --> If a is even then the expression is Even + Even = Even. If a is odd then the expression is --> Odd*(Odd + Odd) + Even = Even + Even = Even. Thus, \(ab+bc+ca+a^2\) is always even. Sufficient.

Re: If a, b, and c are all integers, is ab+bc+ca+a^2 odd? [#permalink]

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14 Jan 2017, 06:54

roastedchips wrote:

If a, b, and c are all integers, is \(ab+bc+ca+a^2\) odd?

(1) a is odd.

(2) (b+c) is odd.

STEM

a(b+c+a) + bc = ??

from 1

clearly insuff

from 2

b+c = odd thus one of them is even and the other is odd

now in a(b+c+a) , we know b+c = odd thus if a is odd the b+c+a = even and thus a(b+c+a) is even too and if a is even then a(a+b+c) is even too , i,e, in all cases it is even bc = even ( since b+c = odd) thus a(b+c+a) +bc is always even and the answer to the question is definite no

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