Equal Floor Functions

My approach was a bit mathematical rather than intuitive as Dev Sharma's.

Let, $\lfloor \frac{x}{99} \rfloor = \lfloor \frac{x}{101} \rfloor = m$

By defination,

$m\leq\frac{x}{99}<m+1$ [ $m$ is an integer]

$\implies 99m \leq x<99(m+1)$ ... ... ...(i)

Again,

$m\leq \frac{x}{101}<m+1$

$\implies 101m\leq x<101(m+1)$ ... ... ...(ii)

Both (i) and (ii) inequalities are to be true, so $101m\leq x<99(m+1)$ Notice, if $m>49$ , then $101m>99(m+1)$ ;

So, for, $m=0,1,2,3,...49$ , the number of possible non-negative values for $x$ is,

$\displaystyle \sum_{m=0}^{49} (99m+99-101m)=\displaystyle \sum_{m=0}^{49} (99-2m)=99 \times 50 - 2 \times \frac{49 \times 50}{2} = 2500$

But $0$ is not a positive integer.

So the answer is $2500-1=2499$ .

I used a simple logic as follows :

$[x/99]=[x/101]=0$ only when x is 1,2,3,..,98 and simlarly it would work till 49. So now we can easily found the following sum :

98 + 97 + 95 + ... + 1 = 2499

Decompose $x$ in base $99$ and $101$ : \[\begin{align} \left.\begin{aligned} x&\overset{!}{=}99q_1+r_1\\ x&\overset{!}{=}101q_2+r_2 \end{aligned}\right\}&&

\Rightarrow&& q_1&=\left\lfloor\frac{x}{99}\right\rfloor=\left\lfloor\frac{x}{101}\right\rfloor=q_2,&&&r_1&\in\{0;\ldots;98\},&r_2&\in\{0;\ldots;100\} \end{align}\] The two decompositions of $x$ have to be equal: $\begin{aligned} 99q_1+r_1=x&=101q_2+r_2&\underset{q_1=q_2=:q}{\Rightarrow}&&r_1&=r_2+2q,&q\geq 0 \end{aligned}$

• If $q=0$ , the remainders are equal. Excluding the case $r_1=r_2=0$ because $x$ has to be positive, the boundaries for $r_1$ yield $98$ solutions.
• If $q>0$ , the boundaries of $r_2$ depend on $q$ : $\begin{aligned} r_1&=r_2+2q\in\{0;\ldots;98\}&\Rightarrow&&1\leq q&\leq 49,&0\leq r_2&\leq 98-2q \end{aligned}$
• Sum up all cases, beware of "off-by-1 errors": $98 + \sum_{q=1}^{49}\sum_{r_2=0}^{98-2q}1=98 + \sum_{q=1}^{49}99-2q=98+49\cdot 99-\cancel{2}\cdot\frac{49(49+1)}{\cancel{2}}=\boxed{2499}$