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.

2005 Indonesia MO (Problems)
Preceded by Problem 4	1 • 2 • 3 • 4 • 5 • 6 • 7 • 8	Followed by Problem 6
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2005 Indonesia MO Problems/Problem 5

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