Reciprocal Sum Equation

Consider multiplying both sides by $50ab$ . This gets the equation $50a+50b=ab$ . Take all terms to one side and factor to get: $ab-50a-50b=0$ , so $(a-50)(b-50) = 2500$ . The factors of 2500 are 1, 2, 4, 5, 10, 20, 25, 50, 100, 125, 250, 500, 625, 1200, and 2500. There are 15 factors.

Since a is less than or equal to b, 50 will occur twice among (a-50) and (b-50), so there are 8 pairs.

Since $a,b>0$ , we can multiply throughout by $50ab$ to obtain $ab-50a-50b=0$ . We complete the factorization to get $50^2 = ab-50a-50b+50^2 = (a-50)(b-50)$ . If $(a-50) < 0$ , then $(b-50) < 0$ since their product is positive. However this implies that $\frac{1}{a} + \frac{1}{b} > \frac{1}{50}$ . Thus, we have $a-50 > 0$ and $b-50 > 0$ .

By the divisor theorem , $50^2 = 2^2 5^4$ has $(2 + 1)(4 + 1) = 15$ factors. Let $d$ be a divisor of $50^2$ then $\frac{50^2}{d}$ is also a divisor. Since $50^2 = 50\times 50$ , then we have one case where $d = \frac{50^2}{d}$ and the other $14$ divisors can be paired up as $\left(d , \frac{50^2}{d}\right)$ , subject to $d < \frac {50^2}{d}$ . Thus we can set $a - 50 = d \Rightarrow a = d + 50$ and $b - 50 = \frac{50^2}{d} \Rightarrow b = 50 + \frac{50^2}{d}$ , where $d \leq 50$ is a divisor of $50^2$ . Therefore we have $1$ pair where $a = b$ and $7$ pairs where $a < b$ , making a total of $8$ pairs.

I don't want to say more than that, if you like this problem then you might also like

Project Euler Problem 108