
3. Find the number of positive integers $x$ which satisfy the condition

$$
\left[\frac{x}{99}\right]=\left[\frac{x}{101}\right]
$$

(Here $[z]$ denotes, for any real $z$, the largest integer not exceeding $z$; e.g. $[7 / 4]=1$.)


Solution: We observe that $\left[\frac{x}{99}\right]=\left[\frac{x}{101}\right]=0$ if and only if $x \in\{1,2,3, \ldots, 98\}$, and there are 98 such numbers. If we want $\left[\frac{x}{99}\right]=\left[\frac{x}{101}\right]=1$, then $x$ should lie in the set $\{101,102, \ldots, 197\}$, which accounts for 97 numbers. In general, if we require $\left[\frac{x}{99}\right]=\left[\frac{x}{101}\right]=$ $k$, where $k \geq 1$, then $x$ must be in the set $\{101 k, 101 k+1, \ldots, 99(k+1)-1\}$, and there are $99-2 k$ such numbers. Observe that this set is not empty only if $99(k+1)-1 \geq 101 k$ and this requirement is met only if $k \leq 49$. Thus the total number of positive integers $x$ for which $\left[\frac{x}{99}\right]=\left[\frac{x}{101}\right]$ is given by

$$
98+\sum_{k=1}^{49}(99-2 k)=2499
$$

[Remark: For any $m \geq 2$ the number of positive integers $x$ such that $\left[\frac{x}{m-1}\right]=\left[\frac{x}{m+1}\right]$ is $\frac{m^{2}-4}{4}$ if $m$ is even and $\frac{m^{2}-5}{4}$ if $m$ is odd.]
