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WE 1: 7 Yrs in Automobile (Commercial Vehicle industry)

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

WE: Securities Sales and Trading (Investment Banking)

Re: If a, b, and c are all integers, is ab + bc + ca + a^2 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 + 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

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

This question can be solved by TESTing VALUES. By extension, if you recognize the Number Properties involved, then you'll only have to TEST 'odds' vs. 'evens' (and NOT have to work through multiple iterations of each option).

We're told that A, B and C are all INTEGERS. We're asked if (A)(B) + (B)(C) + (C)(A) + A^2 is ODD. This is a YES/NO question.

1) A is ODD

With this Fact, we have to consider whether B and C are odds, evens or a mix of the two...

IF... A = 1, then we have.... (1)(B) + (B)(C) + (C)(1) + (1)^2

IF... B=0 and C=0.... (1)(0) + (0)(0) + (0)(1) + (1)^2 = 1 and the answer to the question is YES. IF... B=1 and C=1.... (1)(1) + (1)(1) + (1)(1) + (1)^2 = 4 and the answer to the question is NO. Fact 1 is INSUFFICIENT

2) (B+C) is ODD

With this Fact, we know that ONE of these two variables is ODD and the other is EVEN. The "A" can be odd or even...

IF... B=0 and C=1, then we have.... (A)(0) + (0)(1) + (1)(A) + (A)^2

IF.... A=0.... (0)(0) + (0)(1) + (1)(0) + (0)^2 = 0 and the answer to the question is NO. IF.... A=1.... (1)(0) + (0)(1) + (1)(1) + (1)^2 = 2 and the answer to the question is NO.

IF... B=1 and C=0, then we have.... (A)(1) + (1)(0) + (0)(A) + (A)^2

IF.... A=0.... (0)(1) + (1)(0) + (0)(1) + (0)^2 = 0 and the answer to the question is NO. IF.... A=1.... (1)(1) + (1)(0) + (0)(1) + (1)^2 = 2 and the answer to the question is NO. The answer to the question is ALWAYS NO. Fact 2 is SUFFICIENT