2006 SMT/Algebra Problems/Problem 6

Problem

Let $a, b, c$ be real numbers satisfying:

\[ab-a=b+119\] \[bc-b=c+59\] \[ca-c=a+71\]

Determine all possible values of $a+b+c$.

Solution

From the first equation, we have $a(b-1)=b+119\implies a=\frac{b+119}{b-1}$. Plugging this into the third equation, we get $c\left(\frac{b+119}{b-1}\right)-c=\frac{b+119}{b-1}+71$. Multiplying both sides by $b-1$, we get $c(b+119)-c(b-1)=b+119+71(b-1)\implies 120c=72b+48\implies c=\frac{3b+2}{5}$.


Now we plug that into the second equation. We have $b\left(\frac{3b+2}{5}\right)-b=\frac{3b+2}{5}+59$. Getting rid of the fractions, we have $b(3b+2)-5b=3b+2+295\implies 3b^2-6b-297=0\implies b^2-2b-99=0$. We can factor that as $(b-11)(b+9)=0$, so $b=11$ or $b=-9$.


If $b=11$, then $a=\frac{11+119}{11-1}=13$ and $c=\frac{3\cdot11+2}{5}=7$, so $a+b+c=13+11+7=31$.


If $b=-9$, then $a=\frac{-9+119}{-9-1}=-11$ and $c=\frac{3\cdot-9+2}{5}=-5$, so $a+b+c=-11-9-5=-25$.


Therefore, the possible values of $a+b+c$ are $\boxed{31 \text{ and } -25}$.

Solution 2

We can rearrange the equations as follows:

$ab-a-b = 119$

$bc-b-c = 71$

$ca-c-a = 59$

Then, using Simon's Favorite Factoring Trick we get:

$(a-1)(b-1) = 120$

$(b-1)(c-1)= 72$

$(c-1)(a-1) = 60$

Multiplying the three equations together yields $[(a-1)(b-1)(c-1)]^2 = 120 \cdot 72 \cdot 60 = 518400 \leadsto (a-1)(b-1)(c-1) = \pm 720$

If $(a-1)(b-1)(c-1) = +720$, then dividing this equation by the factored equations yields:

$a-1 = \tfrac{720}{72} = 10 \leadsto a = 11$

$b-1 = \tfrac{720}{60} = 12 \leadsto b = 13$

$c-1 = \tfrac{720}{120} = 6 \leadsto c = 7$

and $a+b+c = 11+13+7 = 31$

If $(a-1)(b-1)(c-1) = -720$, then dividing this equation by the factored equations yields:

$a-1 = \tfrac{-720}{72} = -10 \leadsto a = -9$

$b-1 = \tfrac{-720}{60} = -12 \leadsto b = -11$

$c-1 = \tfrac{-720}{120} = -6 \leadsto c = -5$

and $a+b+c = -9-11-5 = -25$

Thus, $a+b+c = \boxed{31 \text{ \ or \ } -25}$.

See Also

2006 SMT/Algebra Problems

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