# 1984 AHSME Problems/Problem 21

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## Problem

The number of triples $(a, b, c)$ of positive integers which satisfy the simultaneous equations

$ab+bc=44$

$ac+bc=23$

is

$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 2 \qquad \mathrm{(D) \ }3 \qquad \mathrm{(E) \ } 4$

## Solution

We can factor the second equation to get $c(a+b)=23$, so we see that $c$ must be a factor of $23$, and since this is prime, $c=1$ or $c=23$. However, if $c=23$, then $a+b=1$, which is impossible for the field of positive integers. Therefore, $c=1$ for all possible solutions. Substituting this into the original equations gives

$ab+b=44$

and

$a+b=23$.

From the second equation, $a=23-b$, and substituting this into the first equation yields $b(23-b)+b=44$, or $b^2-24b+44=0$. Factoring this gives $(b-2)(b-22)=0$, so $b=2$ or $b=22$. Both of these yield integer solutions for $a$, giving $a=21$ or $a=1$, respectively. Therefore, the only solutions are $(21, 2, 1)$ and $(1, 22, 1)$, yielding $2$ solutions, $\boxed{\text{C}}$.