Question 61
Find the number of integers $n$ such that
$$
1+\left\lfloor\frac{100 n}{101}\right\rfloor=\left\lceil\frac{99 n}{100}\right\rceil
$$
$$
1+\left\lfloor\frac{100 n}{101}\right\rfloor=\left\lceil\frac{99 n}{100}\right\rceil
$$
MathNet 2013 Question 61
Math Olympiad Vault · MathNet 2013