Egyptian Floor Fractions

First off, note that the number $x$ can be expressed as $a+b$ , where $a$ is the integer part and $b$ is the fractional part.

We will then attack this problem by cases:

$0 \le x < \dfrac{1}{5}$

$\dfrac{1}{5}\le x < \dfrac{1}{3}$

$\dfrac{1}{3}\le x < \dfrac{2}{5}$

$\dfrac{2}{5}\le x < \dfrac{1}{2}$

$\dfrac{1}{2}\le x < \dfrac{3}{5}$

$\dfrac{3}{5}\le x < \dfrac{2}{3}$

$\dfrac{2}{3}\le x < \dfrac{4}{5}$

$\dfrac{4}{5}\le x < 1$

Case 1: plugging in and simplifying we have $\dfrac{1}{a}=\dfrac{1}{2a}+\dfrac{1}{3a}+\dfrac{1}{5a}$ which is never true.

Case 2: plugging in and simplifying we have $\dfrac{1}{a}=\dfrac{1}{2a}+\dfrac{1}{3a}+\dfrac{1}{5a+1}$ This simplifies to $\dfrac{a-1}{a(5a-1)}=0$ which yields $a=1$ .

Thus, any $x$ such that $x\in \left[\dfrac{6}{5},\dfrac{4}{3}\right)$ satisfies the equation.

Case 3, etc, All later cases give non-integral solutions for $a$ (I don't want to type up all the straightforward calculations) so we can only have $x\in \left[\dfrac{6}{5},\dfrac{4}{3}\right)$ . Thus $m=\dfrac{6}{5}$ and $M=\dfrac{4}{3}$ .

Thus $M+m=\dfrac{38}{15}$ and our answer is $38+15=\boxed{53}$ .