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# If a, b, and c are all integers, is ab+bc+ca

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

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16 Sep 2012, 01:12
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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
[Reveal] Spoiler: OA
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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, 02:01
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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? [#permalink]

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19 Feb 2014, 17:05
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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.
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Re: If a, b, and c are all integers, is ab+bc+ca+a2 odd? [#permalink]

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19 Feb 2014, 22:59
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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.

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.

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

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20 Feb 2014, 01:18
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.
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Re: If a, b, and c are all integers, is ab+bc+ca [#permalink]

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11 Aug 2015, 14:32
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

I doubt that this is a 700+, is it? I had this as the 2nd question on a GMAT Scholarship Competition Exam.
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Re: If a, b, and c are all integers, is ab+bc+ca [#permalink]

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

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14 Jan 2017, 06:49
If a, b, and c are all integers, is $$ab+bc+ca+a^2$$ odd?

(1) a is odd.

(2) (b+c) is odd.
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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, 07:50
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.

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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, 07:54
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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Re: If a, b, and c are all integers, is ab+bc+ca [#permalink]

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14 Jan 2017, 08:48
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.

Merging similar topics. Please search before posting. Thank you.
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Re: If a, b, and c are all integers, is ab+bc+ca [#permalink]

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15 Jan 2017, 06:21
If a, b, and c are all integers, is ab+bc+ca+(a*a) odd?

(1) a is odd.

Clearly Insufficient. We do not know any info about other terms. It could Even or Odd.

(2) (b+c) is odd.

Re-arrange the question:

ab+ac+bc+a^2 = a (b+c)+ bc+a^2

when (b+c) = odd.......it means one is even and other is odd........So 'bc' is always Even

Let examine the question: a (b+c)+ bc+a^2

Let a =Odd........................O*O+E+O= O+E+0= EVEN

Let a =Even.......................E *O+E+E=E+E+E=EVEN

In both cases the answer to question is NO

Sufficient

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

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05 Apr 2017, 08:21
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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.

Target question: Is ab + bc + ca + a² odd?
This is a good candidate for rephrasing the target question.
ab + bc + ca + a² = b(a + c) + a(c + a)
= (b + a)(c + a)
So, we get....
REPHRASED target question: Is (b+a)(c+a) odd?

When I SCAN the two statements, I see that it might be useful to systematically list the possible outcomes. To do this, I'll list each possible case, and plug in 0 for any EVEN integer and plug in 1 for any ODD integer. We get:
case a: a = odd, b = odd, c = odd. Here, (b+a)(c+a) = EVEN
case b: a = odd, b = odd, c = even. Here, (b+a)(c+a) = EVEN
case c: a = odd, b = even, c = odd. Here, (b+a)(c+a) = EVEN
case d: a = odd, b = even, c = even. Here, (b+a)(c+a) = ODD
case e: a = even, b = odd, c = odd. Here, (b+a)(c+a) = ODD
case f: a = even, b = even, c = odd. Here, (b+a)(c+a) = EVEN
case g: a = even, b = odd, c = even. Here, (b+a)(c+a) = EVEN
case h: a = even, b = even, c = even. Here, (b+a)(c+a) = EVEN

Statement 1: a is odd
This means we're dealing with case a, b, c, or d
For cases a, b and c, (b+a)(c+a) is EVEN
For case d, (b+a)(c+a) is ODD
Since we cannot answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: (b+c) is odd
This means we're dealing with case b, c, f or g
In ALL of these cases, (b+a)(c+a) is EVEN
This means we can be certain that (b+a)(c+a) is EVEN
Since we can answer the REPHRASED target question with certainty, statement 2 is SUFFICIENT

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Re: If a, b, and c are all integers, is ab+bc+ca   [#permalink] 05 Apr 2017, 08:21
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