Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

16 Sep 2012, 02:01

1

This post received KUDOS

2

This post was BOOKMARKED

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.
_________________

Regards SD ----------------------------- Press Kudos if you like my post. Debrief 610-540-580-710(Long Journey): http://gmatclub.com/forum/from-600-540-580-710-finally-achieved-in-4th-attempt-142456.html

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]

Show Tags

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.

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

Show Tags

21 Sep 2016, 01:20

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

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

Show Tags

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.

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

Show Tags

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

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

Show Tags

05 Apr 2017, 08:21

Top Contributor

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

There’s something in Pacific North West that you cannot find anywhere else. The atmosphere and scenic nature are next to none, with mountains on one side and ocean on...

This month I got selected by Stanford GSB to be included in “Best & Brightest, Class of 2017” by Poets & Quants. Besides feeling honored for being part of...

Joe Navarro is an ex FBI agent who was a founding member of the FBI’s Behavioural Analysis Program. He was a body language expert who he used his ability to successfully...