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

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If a,b and c are consecutive positive integers and a>b>c. What can be [#permalink] New post   12 Sep 2016, 03:13
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If a,b and c are consecutive positive integers and a>b>c. What can be the value of (a^2-b^2)(b^2-c^2)?

A. 21
B. 79
C. 143
D. 231
E. 450
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C
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dhaariniv
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Re: If a,b and c are consecutive positive integers and a>b>c. What can be [#permalink] New post   12 Sep 2016, 03:26
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C. 143. From the condition that a,b, c are consecutive,it's clear that (a^2-b^2)(b^2-c^2) = (a+b)*(b+c)
Therefore, the product should have factors with a difference of 2. 143 = 11*13 hence this is the option

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Re: If a,b and c are consecutive positive integers and a>b>c. What can be [#permalink] New post   12 Sep 2016, 18:12
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As a,b and c are consecutive, cleary a-b = b-c = 1

hence (a^2-b^2)(b^2-c^2) = (a+b)*(a-b)*(b+c)*(b-c) = (a+b)*(b+c)
as a,b,c are consecutive
either a ans c will be odd or they will be even.
Now (a+b) and (b+c) both will be odd with a difference of 2.
only option that satisfies this is C.
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Re: If a,b and c are consecutive positive integers and a>b>c. What can be [#permalink] New post   10 Feb 2018, 06:06
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(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

So answer should be C.
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Re: If a,b and c are consecutive positive integers and a>b>c. What can be [#permalink] New post   13 Sep 2016, 15:54
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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
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If a,b and c are consecutive positive integers and a>b>c. What can be [#permalink] New post   Updated on: 10 Feb 2018, 06:20
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.
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Re: If a,b and c are consecutive positive integers and a>b>c. What can be [#permalink] New post   27 Feb 2022, 03:16
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Re: If a,b and c are consecutive positive integers and a>b>c. What can be [#permalink]
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