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 350,000 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: are a, b, and c consecutive integers? [#permalink]
22 Oct 2009, 22:59

4

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

SMAbbas wrote:

If a, b, and c are positive integers, with a < b < c, are a, b, and c consecutive integers?

(1) 1/a – 1/b = 1/c

(2) a + c = b2 – 1

IMO D,

Please verify my answer

If a, b, c are consecutive and a<b<c, then must be true that: b=a+1 and c=a+2.

(1) \frac{1}{a}-\frac{1}{b}=\frac{1}{c} --> (b-a)c=ab --> (a+1-a)(a+2)=a(a+1) --> a^2=2 --> a^2=2 as a is positive integer this equation has no positive integer roots --> a, b, c, are not consecutive positive integers. Sufficient.

(2) a+c=b^2-1 --> a+a+2=(a+1)^2-1 --> 2a+2=a^2+2a+1-1 --> a^2=2. The same here: a, b, c, are not consecutive positive integers. Sufficient.

Re: If a, b, and c are positive integers, with a < b < c, [#permalink]
24 Oct 2013, 06:05

1

This post received KUDOS

SMAbbas wrote:

If a, b, and c are positive integers, with a < b < c, are a, b, and c consecutive integers?

(1) 1/a – 1/b = 1/c

(2) a + c = b^2 – 1

Hey guys - The way I did it was

(1) 1/a – 1/b = 1/c ---> 1/a = 1/b + 1/c

LCM of 'b' and 'c' would never be smaller than b or c because these numbers are co-prime which means that they do not share any common integer factor other than 1.

Now since c>b>a then they following the logic above they can't be consecutive integers. Sufficient

(2) a + c = b^2 – 1 ---> a+c = (b+1)(b-1)

Now if a,b,c were consecutive integers it will be true that (a+c)/2 is equal to b So we would have that a+c = 2b

Now if we replace a+c = 2b in the first equation we would have that

2b = b^2-1

And when trying to solve this quadratic we would realize that b is not an integer, since we would have to use the formula So it contradicts the information that a,b,c are integers. Hence a,b,c cannot be consecutive integers

Insuff

Answer (D)

Please let me know whether this makes sense. If it does, throw me some Kudos!

Cheers J

Last edited by jlgdr on 24 Oct 2013, 07:05, edited 1 time in total.

Re: If a, b, and c are positive integers, with a < b < c, [#permalink]
24 Oct 2013, 06:41

Expert's post

jlgdr wrote:

SMAbbas wrote:

If a, b, and c are positive integers, with a < b < c, are a, b, and c consecutive integers?

(1) 1/a – 1/b = 1/c

(2) a + c = b^2 – 1

Hey guys - The way I did it was

(1) 1/a – 1/b = 1/c ---> 1/a = 1/b + 1/c

LCM of 'b' and 'c' would never be smaller than b or c because these numbers are co-prime which means that they do not share any common integer factor other than 1.

Now since c>b>a then they following the logic above they can't be consecutive integers. Sufficient

(2) a + c = b^2 – 1 ---> a+c = (b+1)(b-1)

Now if a,b,c were consecutive integers it will be true that (a+c)/2 is equal to b So we would have that a+c = 2b

Now if we replace a+c = 2b in the first equation we would have that

2b = b^2-1

And when trying to solve this quadratic we would realize that b is not an integer, since we would have to use the formula So it contradicts the information that a,b,c are integers. Hence a,b,c cannot be consecutive integers

Insuff

Answer (E)

Please let me know whether this makes sense. If it does, throw me some Kudos!

Re: If a, b, and c are positive integers, with a < b < c, [#permalink]
24 Oct 2013, 07:06

Bunuel wrote:

jlgdr wrote:

SMAbbas wrote:

If a, b, and c are positive integers, with a < b < c, are a, b, and c consecutive integers?

(1) 1/a – 1/b = 1/c

(2) a + c = b^2 – 1

Hey guys - The way I did it was

(1) 1/a – 1/b = 1/c ---> 1/a = 1/b + 1/c

LCM of 'b' and 'c' would never be smaller than b or c because these numbers are co-prime which means that they do not share any common integer factor other than 1.

Now since c>b>a then they following the logic above they can't be consecutive integers. Sufficient

(2) a + c = b^2 – 1 ---> a+c = (b+1)(b-1)

Now if a,b,c were consecutive integers it will be true that (a+c)/2 is equal to b So we would have that a+c = 2b

Now if we replace a+c = 2b in the first equation we would have that

2b = b^2-1

And when trying to solve this quadratic we would realize that b is not an integer, since we would have to use the formula So it contradicts the information that a,b,c are integers. Hence a,b,c cannot be consecutive integers

Insuff

Answer (E)

Please let me know whether this makes sense. If it does, throw me some Kudos!

Re: If a, b, and c are positive integers, with a < b < c, [#permalink]
04 Feb 2014, 05:24

I assumed a,b,c to be consecutive and substituted b-1,b,b+1 in each statement independently.

(1) gives b^2=2b+1 and b is not an integer hence my assumption that a,b,c are consecutive cannot be true. sufficient. (2) also gives b^2 = 2b+1. same as (1) hence sufficient.

Re: are a, b, and c consecutive integers? [#permalink]
14 May 2014, 03:40

Bunuel wrote:

SMAbbas wrote:

If a, b, and c are positive integers, with a < b < c, are a, b, and c consecutive integers?

(1) 1/a – 1/b = 1/c

(2) a + c = b2 – 1

IMO D,

Please verify my answer

If a, b, c are consecutive and a<b<c, then must be true that: b=a+1 and c=a+2.

(1) \frac{1}{a}-\frac{1}{b}=\frac{1}{c} --> (b-a)c=ab --> (a+1-a)(a+2)=a(a+1) --> a^2=2 --> a^2=2 as a is positive integer this equation has no positive integer roots --> a, b, c, are not consecutive positive integers. Sufficient.

(2) a+c=b^2-1 --> a+a+2=(a+1)^2-1 --> 2a+2=a^2+2a+1-1 --> a^2=2. The same here: a, b, c, are not consecutive positive integers. Sufficient.

Answer D.

Hi Bunuel,

Thanks for the answer. One question - what is the meaning of "this eq has no positive integer root ?". It is because the eq has A = +- 2 ?

Re: are a, b, and c consecutive integers? [#permalink]
14 May 2014, 05:35

Expert's post

gauravsoni wrote:

Bunuel wrote:

SMAbbas wrote:

If a, b, and c are positive integers, with a < b < c, are a, b, and c consecutive integers?

(1) 1/a – 1/b = 1/c

(2) a + c = b2 – 1

IMO D,

Please verify my answer

If a, b, c are consecutive and a<b<c, then must be true that: b=a+1 and c=a+2.

(1) \frac{1}{a}-\frac{1}{b}=\frac{1}{c} --> (b-a)c=ab --> (a+1-a)(a+2)=a(a+1) --> a^2=2 --> a^2=2 as a is positive integer this equation has no positive integer roots --> a, b, c, are not consecutive positive integers. Sufficient.

(2) a+c=b^2-1 --> a+a+2=(a+1)^2-1 --> 2a+2=a^2+2a+1-1 --> a^2=2. The same here: a, b, c, are not consecutive positive integers. Sufficient.

Answer D.

Hi Bunuel,

Thanks for the answer. One question - what is the meaning of "this eq has no positive integer root ?". It is because the eq has A = +- 2 ?

We are given that a is a positive integer, while from a^2=2, a=-\sqrt{2} or a=\sqrt{2}. Hence our assumption that a, b, and c were consecutive integers was wrong. _________________

Re: are a, b, and c consecutive integers? [#permalink]
14 May 2014, 06:29

gauravsoni wrote:

Bunuel wrote:

SMAbbas wrote:

If a, b, and c are positive integers, with a < b < c, are a, b, and c consecutive integers?

(1) 1/a – 1/b = 1/c

(2) a + c = b2 – 1

IMO D,

Please verify my answer

If a, b, c are consecutive and a<b<c, then must be true that: b=a+1 and c=a+2.

(1) \frac{1}{a}-\frac{1}{b}=\frac{1}{c} --> (b-a)c=ab --> (a+1-a)(a+2)=a(a+1) --> a^2=2 --> a^2=2 as a is positive integer this equation has no positive integer roots --> a, b, c, are not consecutive positive integers. Sufficient.

(2) a+c=b^2-1 --> a+a+2=(a+1)^2-1 --> 2a+2=a^2+2a+1-1 --> a^2=2. The same here: a, b, c, are not consecutive positive integers. Sufficient.

Answer D.

Hi Bunuel,

Thanks for the answer. One question - what is the meaning of "this eq has no positive integer root ?". It is because the eq has A = +- 2 ?

We are given that a is a positive integer, while from a^2=2, a=-\sqrt{2} or a=\sqrt{2}. Hence our assumption that a, b, and c were consecutive integers was wrong.[/quote]

hey guys, A metallurgist but currently working in a NGO and have scheduled my GMAT in December for second round .....u know. I read some but valuable blogs on this...

First of all, thank you for all the interest in pursuing graduate studies in MBA as well as the Rotman School of Management. I have received a lot of...

Today, 1st year Rotman students had a great simulation event hosted by Scotiabank, one of Canada’s best and largest banks. Attended by entire Rotman 1st year students, the...