6. Find the number of integers $n$ such that
$$
1+\left\lfloor\frac{100 n}{101}\right\rfloor=\left\lceil\frac{99 n}{100}\right\rceil .
$$
Answer: 10100 Consider $f(n)=\left\lceil\frac{99 n}{100}\right\rceil-\left\lfloor\frac{100 n}{101}\right\rfloor$. Note that $f(n+10100)=\left\lceil\frac{99 n}{100}+99 \cdot 101\right\rceil-$ $\left\lfloor\frac{100 n}{101}+100^{2}\right\rfloor=f(n)+99 \cdot 101-100^{2}=f(n)-1$. Thus, for each residue class $r$ modulo 10100, there is exactly one value of $n$ for which $f(n)=1$ and $n \equiv r(\bmod 10100)$. It follows immediately that the answer is 10100 .