# 2002 AIME II Problems/Problem 8

## Problem

Find the least positive integer $k$ for which the equation $\left\lfloor\frac{2002}{n}\right\rfloor=k$ has no integer solutions for $n$. (The notation $\lfloor x\rfloor$ means the greatest integer less than or equal to $x$.)

## Solution

Note that if $\frac{2002}n - \frac{2002}{n+1}\leq 1$, then either $\left\lfloor\frac{2002}{n}\right\rfloor=\left\lfloor\frac{2002}{n+1}\right\rfloor$, or $\left\lfloor\frac{2002}{n}\right\rfloor=\left\lfloor\frac{2002}{n+1}\right\rfloor+1$. Either way, we won't skip any natural numbers.

The smallest $n$ such that $\frac{2002}n - \frac{2002}{n+1} > 1$ is $n=45$. (The inequality simplifies to $n(n+1)>2002$, which is easy to solve by trial, as the solution is obviously $\simeq \sqrt{2002}$.)

We can now compute: $$\left\lfloor\frac{2002}{45}\right\rfloor=44$$ $$\left\lfloor\frac{2002}{44}\right\rfloor=45$$ $$\left\lfloor\frac{2002}{43}\right\rfloor=46$$ $$\left\lfloor\frac{2002}{42}\right\rfloor=47$$ $$\left\lfloor\frac{2002}{41}\right\rfloor=48$$ $$\left\lfloor\frac{2002}{40}\right\rfloor=50$$

From the observation above (and the fact that $\left\lfloor\frac{2002}{2002}\right\rfloor=1$) we know that all integers between $1$ and $44$ will be achieved for some values of $n$. Similarly, for $n<40$ we obviously have $\left\lfloor\frac{2002}{n}\right\rfloor > 50$.

Hence the least positive integer $k$ for which the equation $\left\lfloor\frac{2002}{n}\right\rfloor=k$ has no integer solutions for $n$ is $\boxed{049}$.