Positive integers a, b, c satisfy (a b)/(a-b)=c . What is the...

$$\begin{aligned} & \because \frac{1}{b}=\frac{1}{a}+\frac{1}{c} \\ & \Rightarrow b < \min (a, c) \end{aligned}$$$Case I: When$a=c$, then$b=\frac{a}{2} \Rightarrow b \in$even then,$a+b+c=5 b \Rightarrow(a+b+c)_{\max }=95$Case II: When$a \neq c$Let$a=\lambda a_1$and$c=\lambda c_1$where$\operatorname{gcd}$\left(a_1, c_1\right)=1$ So, $\frac{1}{b}=\frac{a_1+c_1}{\lambda a_1 c_1}$ $\Rightarrow \quad b=\left(\frac{\lambda}{a_1+c_1}\right) a_1 \cdot c_1$ So, we must select $\lambda$ such that \left(a_1+c_1\right) \mid \lambda$If we check for maximum value of$(a+b+c)=99$then \left(a_1+c_1\right)$ must be a divisor of 99 . It is possible when $\left(a_1, c_1\right)=(1,2)$, then $\lambda$ is divisible by 3 . (say $\lambda=3 \mu$ ) $$\begin{aligned} & a=3 \mu, c=6 \mu \text { and } b=2 \mu \\ & \Rightarrow a+b+c=11 \mu \end{aligned}$$ If we take $\mu=9,(a+b+c)_{\max }=99$.