Integers

Number Theory Level 5

Find $(x , y , z) \in \mathbb Z^{+}$ and $x < y < z$ such that:

$\frac{1}{x} - \frac{1}{xy} - \frac{1}{xyz} = \frac{19}{97} .$

If $(x_1 , y_1 , z_1) , (x_2 , y_2 , z_2) , \ldots , (x_n , y_n , z_n)$ are all the solutions, then enter your answer as $\displaystyle \sum_{i = 1}^{ n} (x_i + y_i + z_i)$ .

This is part of the set My Problems and THRILLER

The answer is 151.

1 solution

Patrick Corn
Nov 29, 2017

Multiply through by $97xyz$ to get $97yz - 97z - 97 = 19xyz,$ or $z(97y-97-19xy) = 97.$ So $z=1$ or $97.$

If $z=1$ then $x<y<z$ is impossible (although I should point out that it's not hard to show in this case that the only solution in positive integers is $x=5, y = 97$ ).

If $z=97$ then we get $97y-97-19xy=1,$ or $y(97-19x)=98.$ It's not hard to check manually in the range $1 \le x \le 5$ ( $x$ any larger would make the left side negative) to see that the only solution is $x=5,y=49.$

So the only solution satisfying the given conditions is $x=5,y=49,z=97,$ so the answer is $\fbox{151}.$

Integers

The answer is 151.

1 solution

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