Last visit was: 19 Jul 2024, 05:48 It is currently 19 Jul 2024, 05:48
Toolkit
GMAT Club Daily Prep
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

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.

# If a,b and c are consecutive positive integers and a>b>c. What can be

SORT BY:
Tags:
Show Tags
Hide Tags
RC & DI Moderator
Joined: 02 Aug 2009
Status:Math and DI Expert
Posts: 11475
Own Kudos [?]: 34437 [16]
Given Kudos: 322
Intern
Joined: 21 Jun 2016
Posts: 2
Own Kudos [?]: 2 [2]
Given Kudos: 2
Intern
Joined: 18 Jun 2015
Posts: 21
Own Kudos [?]: 30 [3]
Given Kudos: 7
Manager
Joined: 12 Jun 2015
Posts: 70
Own Kudos [?]: 71 [2]
Given Kudos: 104
Re: If a,b and c are consecutive positive integers and a>b>c. What can be [#permalink]
1
Kudos
1
Bookmarks
Since a,b and c are consecutive positive integers and a>b>c, we can write them as (x + 1), x and (x - 1) respectively
(a^2-b^2)(b^2-c^2) can be simplified into (2x + 1)(2x - 1) = 4x^2 - 1

using the options to find the product,
only C. 143 satisfactorily gives us x as an integer
Intern
Joined: 02 Dec 2017
Posts: 2
Own Kudos [?]: 0 [0]
Given Kudos: 19
If a,b and c are consecutive positive integers and a>b>c. What can be [#permalink]
The easier-to-see algebraic solution would be:
b
a=b+1
c=b-1
(a-b)*(a+b)*(b+c)*(b-c)=(b+1-b)*(b+1+b)*(b+b-1)*(b-b+1)=(2b+1)*(2b-1)=4b^2-1=
As all three numbers are integers, we just through putting numbers in the given variants can find out what is the value of B, and hence the answer solution to the given equation.
4b^2-1=143 b^2=36 b=6. Only 143 fits the condition of the numbers being integer, the other answer solutions do not. SO the answer is 143.

Originally posted by Sona168 on 10 Feb 2018, 05:45.
Last edited by Sona168 on 10 Feb 2018, 06:20, edited 1 time in total.
Intern
Joined: 31 Jan 2018
Posts: 13
Own Kudos [?]: 2 [2]
Given Kudos: 0
Re: If a,b and c are consecutive positive integers and a>b>c. What can be [#permalink]
1
Kudos
1
Bookmarks
(a^2-b^2)(b^2-c^2)=(a-b)(a+b)(b-c)(b+c)
As a>b>c and a, b,c are consecutive integers (a-b)=1 and (b-c)=1
(a^2-b^2)(b^2-c^2)=(a+b)(b+c)
As a,b,c are consecutive integers let us assume in the form x+1,x,x-1respectively
So (a+b)(b+c)=(2x+1)(2x-1)=4x^2-1.
Out of the given options only when 4x^2-1=143,we get x to be an integer