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If a,b and c are consecutive positive integers and a>b>c. What can be [#permalink]
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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^2b^2)(b^2c^2)? A. 21 B. 79 C. 143 D. 231 E. 450
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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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12 Sep 2016, 03:26
C. 143. From the condition that a,b, c are consecutive,it's clear that (a^2b^2)(b^2c^2) = (a+b)*(b+c) Therefore, the product should have factors with a difference of 2. 143 = 11*13 hence this is the option Sent from my Lenovo A6020a46 using GMAT Club Forum mobile app



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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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12 Sep 2016, 18:12
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As a,b and c are consecutive, cleary ab = bc = 1
hence (a^2b^2)(b^2c^2) = (a+b)*(ab)*(b+c)*(bc) = (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]
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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^2b^2)(b^2c^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]
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Updated on: 10 Feb 2018, 06:20
The easiertosee algebraic solution would be: b a=b+1 c=b1 (ab)*(a+b)*(b+c)*(bc)=(b+1b)*(b+1+b)*(b+b1)*(bb+1)=(2b+1)*(2b1)=4b^21= 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^21=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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10 Feb 2018, 06:06
(a^2b^2)(b^2c^2)=(ab)(a+b)(bc)(b+c) As a>b>c and a, b,c are consecutive integers (ab)=1 and (bc)=1 (a^2b^2)(b^2c^2)=(a+b)(b+c) As a,b,c are consecutive integers let us assume in the form x+1,x,x1respectively So (a+b)(b+c)=(2x+1)(2x1)=4x^21. Out of the given options only when 4x^21=143,we get x to be an integer
So answer should be C.




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