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If a, b and c are different real numbers such that a+1/b = b+1/c = c+1

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If a, b and c are different real numbers such that a+1/b = b+1/c = c+1  [#permalink]

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New post 19 Jun 2019, 17:53
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Question Stats:

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If a, b and c are different real numbers such that \(a+\frac{1}{b}\) = \(b+\frac{1}{c}\) = \(c+\frac{1}{a}\) , which of the following is a possible value of the a*b*c?

A. 1
B. 2
C. 3
D. 4
E. 8
Director
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Posts: 554
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Concentration: Entrepreneurship, Marketing
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If a, b and c are different real numbers such that a+1/b = b+1/c = c+1  [#permalink]

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New post Updated on: 19 Jun 2019, 18:58
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nick1816 wrote:
If a, b and c are different real numbers such that \(a+\frac{1}{b}\) = \(b+\frac{1}{c}\) = \(c+\frac{1}{a}\) , which of the following is a possible value of the a*b*c?

A. 1
B. 2
C. 3
D. 4
E. 8


**Edited

OK! Not sure a GMAT question or not? Definitely took me > 2 min

Given,
\(a+\frac{1}{b}\) = \(b+\frac{1}{c}\)
--> \(a - b\) = \(\frac{1}{c}\) - \(\frac{1}{b}\) = \(\frac{(b - c)}{bc}\) ....... (1)

\(b+\frac{1}{c}\) = \(c+\frac{1}{a}\)
--> \(b - c\) = \(\frac{1}{a}\) - \(\frac{1}{c}\) = \(\frac{(c - a)}{ac}\) ....... (2)

\(a+\frac{1}{b}\) = \(c+\frac{1}{a}\)
--> \(a - c\) = \(\frac{1}{a}\) - \(\frac{1}{b}\) = \(\frac{(b - a)}{ab}\) ....... (3)

Multiply (1), (2) & (3)
--> \((a - b)*(b - c)*(a - c)\) = (\(\frac{(b - c)}{bc}\))*(\(\frac{(c - a)}{ac}\))*(\(\frac{(b - a)}{ab}\))
--> \((a - b)*(b - c)*(a - c)\) = \(\frac{(b - c)*(c - a)*(b - a)}{(abc)^2}\)

The above equation is only possible when (abc)^2 = 1
--> a*b*c can be 1 or -1

So, possible value = 1

IMO Option A

Pls Hit kudos if you like the solution

Originally posted by Dillesh4096 on 19 Jun 2019, 18:18.
Last edited by Dillesh4096 on 19 Jun 2019, 18:58, edited 1 time in total.
Director
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Re: If a, b and c are different real numbers such that a+1/b = b+1/c = c+1  [#permalink]

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New post 19 Jun 2019, 18:27
a, b and c are different, though your answer might be correct. ;)

Dillesh4096 wrote:
nick1816 wrote:
If a, b and c are different real numbers such that \(a+\frac{1}{b}\) = \(b+\frac{1}{c}\) = \(c+\frac{1}{a}\) , which of the following is a possible value of the a*b*c?

A. 1
B. 2
C. 3
D. 4
E. 8


We can clearly see that if each of a, b & c is 1, the given equation is satisfied

So, definitely a*b*c can be 1

IMO Option A

Pls Hit kudos if you like the solution

Posted from my mobile device
Director
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Joined: 20 Jul 2017
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If a, b and c are different real numbers such that a+1/b = b+1/c = c+1  [#permalink]

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New post 19 Jun 2019, 18:30
nick1816 wrote:
a, b and c are different, though your answer might be correct. ;)

Dillesh4096 wrote:
nick1816 wrote:
If a, b and c are different real numbers such that \(a+\frac{1}{b}\) = \(b+\frac{1}{c}\) = \(c+\frac{1}{a}\) , which of the following is a possible value of the a*b*c?

A. 1
B. 2
C. 3
D. 4
E. 8


We can clearly see that if each of a, b & c is 1, the given equation is satisfied

So, definitely a*b*c can be 1

IMO Option A

Pls Hit kudos if you like the solution

Posted from my mobile device


Oh Damn! Lemme redo :lol:
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Re: If a, b and c are different real numbers such that a+1/b = b+1/c = c+1  [#permalink]

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New post 25 Jun 2019, 19:59
I think it must have something related with reciprocals and tahts why is 1, but idk how to solve it.
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Re: If a, b and c are different real numbers such that a+1/b = b+1/c = c+1   [#permalink] 25 Jun 2019, 19:59
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