find a,b,c

[Albert]

$a,b,c \in N$

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

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

$find :a,b,c$

 

[mathworker]

Since modulo is "zero" there is no remainder.
\(\displaystyle \frac{(ab-1)(bc-1)(ca-1)}{abc}\) is not a fraction
\(\displaystyle abc-a-b-c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}\) is not a fraction
So the little terms must sum it up to zero or 1 so,
\(\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}=0\)(let)
\(\displaystyle \frac{ab+bc+ca-1}{abc}=0\)
\(\displaystyle ab+bc+ca=1\) which certainly can't be true
If,
\(\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}=1\)
\(\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1+\frac{1}{abc}\)
But as \(\displaystyle a,b,c>1\) \(\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) 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?

 

[mathbalarka]

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!

 

[mathworker]

yeah I got it

.....has a maxmum value of 1.....

max value is not 1 \(\displaystyle its \frac{13}{12}\)

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But,I guess (2,3,5) is the only solution....

 

[Albert]

But,I guess (2,3,5) is the only solution....

yes (2,3,5) is the only solution..,but how to get the answer ?

 

[mathworker]

In,
\(\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1+\frac{1}{abc}\)
L.H.S to be greater than one (a,b) should be (2,3) substituting them rest is linear equation
\(\displaystyle \frac{1}{2}+\frac{1}{3}+\frac{1}{c}=1+\frac{1}{6c}\)
\(\displaystyle c=5\)

 

[Albert]

In,
\(\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1+\frac{1}{abc}\)
L.H.S to be greater than one (a,b) should be (2,3) substituting them rest is linear equation
\(\displaystyle \frac{1}{2}+\frac{1}{3}+\frac{1}{c}=1+\frac{1}{6c}\)
\(\displaystyle c=5\)

very nice solution