Find positive integers $x,y,z$ satisfying $x<y<z$ and $\dfrac{1}{x}-\dfrac{1}{xy}-\dfrac{1}{xyz}=\dfrac{19}{97}$ .

Submit your answer as the value of $x+y+z$ .

The answer is 151.

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The equation is equivalent to $19xyz=97(yz-z-1)$ or $19xyz+97=97z\times (y-1).$

$19$ and $97$ are primes and so $97$ should divide one of $x,y,z.$ Also $z$ should divide $97$ which means that $z=97$ is prime.

From this it follows that $y$ divides $98=2\times 49=2\times 7^2;y$ being less than $z=97.$ The possibilities are $y=2,y=7,y=14,y=49.$

If $y=2,19\times 2\times x=2\times 97-98=96$ so that $x$ is not an integer.

If $y=7,19x=83$ so that $x$ is not an integer.

If $y=14,19x=90$ so that $x$ is not an integer.

If $y=49,19\times 49x=49\times 97-98. i.e.,x=5.$

Thus $x=5,y=49,z=97.$

Therefore $x+y+z=5+49+97=\boxed{151}.$