Easy Equation.

Number Theory Level 5

$Find\quad the\quad number\quad of\quad ordered\quad pairs\quad of\quad solutions\quad \\ where\quad x,y\quad are\quad integers\quad of\quad the\quad following\quad equation:\\ \\ \quad \quad \quad \quad \frac { 1 }{ x } \quad +\quad \frac { 1 }{ y } \quad =\quad \frac { 1 }{ 2018 }$

The answer is 17.

1 solution

Mas Mus
Apr 21, 2015

$\begin{array}{c}&\frac{1}{x}+\frac{1}{y}=\frac{1}{2018}\\ \ \large{y=2018+\frac{2018^2}{x-2018}}\end{array}$

Since $x$ and $y$ are integers, then $x-2018$ should be a divisor of $2018^2$ . The divisors of $2018^2$ are :

$\pm1, \pm2,\pm4, \pm1009, \pm2018, \pm4036, \pm1009^2, \pm4036^2, \pm2036162$ .

But, $x\text{ and }y\neq0\left(\text{or }x-2018\neq-2018\right)$ , so we have $2\times{9}-1=\boxed{17}$ pairs solution.

Nice solution but Simon's Favourite Trick would have been, I'd not say easier necessarily, but faster method to approach the problem. Anyways, now I see a new way to approach these problems. c:

Kunal Verma - 6 years, 1 month ago

Jesus!I wrote the equation the form of $(x-pq)(y-pq)=p^2q^2$ where $p,q$ were $2,1009$ respectively.Now by calculating all possible cases for $p^2q^2$ divisors I got 9 answers for $x,y$ multiplying by 2 to get results for integers I got the answer 18 but now I see where I went wrong ...... RIP 200 points! :((

Arian Tashakkor - 6 years, 1 month ago

That was the entire catch of the problem.

Kunal Verma - 6 years, 1 month ago

Easy Equation.

The answer is 17.

1 solution

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