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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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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]
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]
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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]
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