Algebra Question by Mithil Shah # 1

$\begin{aligned} \frac 1x - \frac 1{xy} - \frac 1{xyz} & = \frac {19}{97} \\ \frac {yz-z-1}{xyz} & = \frac {19}{97} \end{aligned}$

Since 97 in the RHS is a prime, this means that one of $x$ , $y$ and $z$ must be 97 or its multiple.

Now consider the following:

$\begin{aligned} \frac {yz-z-1}{xyz} & = \frac {19}{97} \\ 1 - \frac 1y - \frac 1{yz} & = \frac {19}{97}x \end{aligned}$

Since the LHS $< 1$ , $\implies 1 \le x \le 5$ .

Assuming $z=97$ , then $x < y < 97$ :

$\begin{aligned} \frac {yz-z-1}{xyz} & = \frac {19}{97} \\ 97y - 97-1 & = 19xy \\ (97-19x)y & = 98 \\ \implies y & = \frac {98}{97-19x} \end{aligned}$

And $y=49$ , the only integral solution when $x=5$ . Therefore, $x+y+z = 5+49+97=\boxed{151}$ .