16. Let $a, b, c$ be positive integers such that $a b+b c-c a=0$ and $a-c=101$. Find $b$.設 $a 、 b 、 c$ 爲正整數, 其中 $a b+b c-c a=0$ 而 $a-c=101$ 。求 $b$ 。
16. 2550
16. Let $d$ be the H.C.F. of $a$ and $c$. Then we may write $a=d x$ and $c=d y$ where $x, y$ are integers with an H.C.F. of 1. Then we have
$$
b=\frac{a c}{a+c}=\frac{d x y}{x+y} \text { and } d(x-y)=101 .
$$

Since $x$ and $y$ are relatively prime, the first equation shows that $x+y$ must divide $d$. Hence $d>1$ and we know from the second equation that $d=101$ and $x-y=1$. Since 101 is prime, we must then have $x+y=101$ and thus $x=51, y=50$ and $b=\frac{101 \times 50 \times 51}{101}=2550$.