# 2005 Indonesia MO Problems/Problem 5

## Problem

For an arbitrary real number $x$, $\lfloor x\rfloor$ denotes the greatest integer not exceeding $x$. Prove that there is exactly one integer $m$ which satisfy $m-\left\lfloor \frac{m}{2005}\right\rfloor=2005$.

## Solution

First we will show that $m = 2006$ is a solution. Then we will show that there are no other integers that are a solution.

To find the solution, note that the absolute value of $\left\lfloor \frac{m}{2005}\right\rfloor$ is usually smaller than $m$, so $m$ should be close to $2005$. Plugging in $m = 2005$ results in $m-\left\lfloor \frac{m}{2005}\right\rfloor = 2004$. Trying out $m = 2006$ results in $m-\left\lfloor \frac{m}{2005}\right\rfloor = 2005$, so that value is a solution.

To show that there are no other integers that are a solution, we will consider two cases: one where $m$ is less than $2006$ and one where $m$ is greater than $2006$.

Case 1: $m < 2006$

The highest integer $m$ that meets the conditions is $2005$, and in that condition, $m-\left\lfloor \frac{m}{2005}\right\rfloor=2004$.

Assume that $m - \left \lfloor \frac{m}{2005} \right \rfloor < 2005$. Let $m = 2005a + b$, where $a,b$ are integers and $b$ is less than $2005$.

If $b = 0$, then $m-\left\lfloor \frac{m}{2005}\right\rfloor = 2005a - \left\lfloor \frac{2005a}{2005}\right\rfloor = 2004a$. Additionally, \begin{align*} m-1-\left\lfloor \frac{m-1}{2005}\right\rfloor &= 2005a - 1 - \left\lfloor \frac{2005a - 1}{2005}\right\rfloor \\ &= 2005a - 1 - \left\lfloor \frac{2005(a-1) + 2004}{2005} \right\rfloor \\ &= 2005a - 1 - a + 1 \\ &= 2004a. \end{align*} If $b > 0$, then $m-\left\lfloor \frac{m}{2005}\right\rfloor = 2005a + b - \left\lfloor \frac{2005a+b}{2005}\right\rfloor = 2004a + b$. Additionally, \begin{align*} m-1-\left\lfloor \frac{m-1}{2005}\right\rfloor &= 2005a + b - 1 - \left\lfloor \frac{2005a + b - 1}{2005}\right\rfloor \\ &= 2005a + b - 1 - a \\ &= 2004a + b - 1. \end{align*} For both cases, $m-1-\left\lfloor \frac{m-1}{2005}\right\rfloor \le m-\left\lfloor \frac{m}{2005}\right\rfloor$, so by induction, there are no integers less than $2006$ that satisfy the equation.

Case 2: $m > 2006$

The lowest integer $m$ that meets the conditions is $2007$, and in that condition, $m-\left\lfloor \frac{m}{2005}\right\rfloor=2006$.

Assume that $m - \left \lfloor \frac{m}{2005} \right \rfloor > 2005$. Once again, let $m = 2005a + b$, where $a,b$ are integers and $b$ is less than $2005$.

If $b = 2004$, then $m-\left\lfloor \frac{m}{2005}\right\rfloor = 2005a + 2004 - \left\lfloor \frac{2005a+2004}{2005}\right\rfloor = 2004a + 2004$. Additionally, \begin{align*} m+1-\left\lfloor \frac{m+1}{2005}\right\rfloor &= 2005a + 2005 - \left\lfloor \frac{2005a +2005}{2005}\right\rfloor \\ &= 2005a + 2005 - a - 1 \\ &= 2004a + 2004 \end{align*} If $b < 2004$, then $m-\left\lfloor \frac{m}{2005}\right\rfloor = 2005a + b - \left\lfloor \frac{2005a+b}{2005}\right\rfloor = 2004a + b$. Additionally, \begin{align*} m+1-\left\lfloor \frac{m+1}{2005}\right\rfloor &= 2005a + b + 1 - \left\lfloor \frac{2005a + b + 1}{2005}\right\rfloor \\ &= 2005a + b + 1 - a \\ &= 2004a + b + 1 \end{align*} For both cases, $m+1-\left\lfloor \frac{m+1}{2005}\right\rfloor \ge m-\left\lfloor \frac{m}{2005}\right\rfloor$, so by induction, there are no integers greater than $2006$ that satisfy the equation.

Therefore, there is only one integer that satisfies the original equation.