What are the values of a, b, and c in the following equations?

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Discussion Overview

The discussion revolves around finding natural number values for \(a\), \(b\), and \(c\) that satisfy specific equations involving modular arithmetic and fractions. The scope includes mathematical reasoning and problem-solving related to inequalities and equations.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant states that since the modulo condition implies no remainder, certain expressions must sum to zero or one, leading to the conclusion that there are no solutions for \(a\), \(b\), and \(c\).
  • Another participant proposes a specific solution of \((a, b, c) = (2, 3, 5)\) but notes that this does not exclude the possibility of other solutions.
  • A participant corrects the earlier claim about the maximum value of \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\), stating it is \(\frac{13}{12}\) instead of 1.
  • Another participant expresses uncertainty about how to derive the solution and seeks clarification on the process.
  • A later reply outlines a method to derive \(c\) by substituting values for \(a\) and \(b\), leading to the conclusion that \(c = 5\).

Areas of Agreement / Disagreement

Participants express differing views on the existence of solutions, with some asserting that \((2, 3, 5)\) is the only solution while others question the validity of this conclusion. The discussion remains unresolved regarding the uniqueness of the solution.

Contextual Notes

Participants rely on specific assumptions about the values of \(a\), \(b\), and \(c\) being greater than 1, and there are unresolved mathematical steps in the reasoning presented.

Albert1
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$a,b,c \in N$

(1) $1<a<b<c$

(2)$(ab-1)(bc-1)(ca-1) \,\, mod \,\, (abc)=0$

$find :a,b,c$
 
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Since modulo is "zero" there is no remainder.
$$\frac{(ab-1)(bc-1)(ca-1)}{abc}$$ is not a fraction
$$abc-a-b-c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}$$ is not a fraction
So the little terms must sum it up to zero or 1 so,
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}=0$$(let)
$$\frac{ab+bc+ca-1}{abc}=0$$
$$ab+bc+ca=1$$ which certainly can't be true
If,
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}=1$$
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1+\frac{1}{abc}$$
But as $$ a,b,c>1$$ $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$ has a maxmum value of 1 so there are no solutions to this one either so I arrived at "there are no such a,b,c"
Did I do something wrong?:confused:
 
I am just going to leave this one here : (a, b, c) = (2, 3, 5). (*) Can you see where you went wrong now?

(*) Note that this doesn't mean there are no other. I haven't even revealed half of the solution yet. Keep trying!
 
yeah I got it
...has a maxmum value of 1...
max value is not 1 $$its \frac{13}{12}$$

- - - Updated - - -

But,I guess (2,3,5) is the only solution...(Smirk)
 
mathworker said:
But,I guess (2,3,5) is the only solution...(Smirk)
yes (2,3,5) is the only solution..,but how to get the answer ?
 
In,
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1+\frac{1}{abc}$$
L.H.S to be greater than one (a,b) should be (2,3) substituting them rest is linear equation
$$\frac{1}{2}+\frac{1}{3}+\frac{1}{c}=1+\frac{1}{6c}$$
$$c=5$$
 
mathworker said:
In,
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1+\frac{1}{abc}$$
L.H.S to be greater than one (a,b) should be (2,3) substituting them rest is linear equation
$$\frac{1}{2}+\frac{1}{3}+\frac{1}{c}=1+\frac{1}{6c}$$
$$c=5$$
very nice solution(Yes)
 

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