I have lost my Ceiling!

For any nonnegative integer $a$ , we have $\lfloor \tfrac{x}{99} \rfloor \,=\, \lfloor \tfrac{x}{101} \rfloor \,=\, a$ provided that $99a \; \le \; x \; < \; 99(a+1) \hspace{2cm} 101a \; \le \; x \; < \; 101(a+1)$ and these inequalities are only simultaneously possible provided that $101a \; \le \; x \; < \; 99(a+1)$ Thus we must have $2a < 99$ , and hence $0 \le a \le 49$ . For any particular value of $a$ , we must have $101a \le x \le 99a + 98$ , and hence there are $(99a + 98) - 101a + 1 \,=\, 99 - 2a$ possible values of $x$ associated with it. Thus the total number of nonnegative integers $x$ such that $\lfloor \tfrac{x}{99} \rfloor \,=\, \lfloor \tfrac{x}{101} \rfloor$ is $\sum_{a=0}^{49}(99 - 2a) \; =\; 99 \times 50 - 49\times50 \; = \; 2500$ Since we are only interested in positive integer solutions, we must discard the case $x=0$ , and so it follows that there are $\boxed{2499}$ positive solutions.

$floor \frac{x}{99} = floor \frac{x}{101}$ $x=99k+r_{1} , r_{1}<99$ $=101k+r_{2} , r_{2}<101$

$2k+r_{2}-r_{1}=0$

$r_{1}=2k+r_{2} (1)$

Since $r_{1}=0$ implies that $x=0$ , we can eliminate this possibility. For each value of $r_{1}$ , there are $floor \frac{r_{1}}{2}+1$ nonnegative integers $k$ that satisfy (1). Therefore the solution S is: $S=\displaystyle\sum_{i=1}^{98}(floor \frac{i}{2}+1)$

$=\displaystyle\sum_{i=1}^{98}floor\frac{i}{2} +98$

$=\displaystyle\sum_{i=2}^{97}floor\frac{i}{2} +98$

$=2\times\displaystyle\sum_{i=1}^{48}k +147$

$=48\times49+147$

$=2499$