My Floor is Broken

Write floor function as $\frac {m}{2005}-$ { $\frac{m}{2005}$ } i.e. in terms of fractional part. You will get $\\$ $\frac {2004m}{2005}+$ { $\frac{m}{2005}$ } $=2005$ . $\\$ Let $m=2005x$ . Equation reduces to $\\$ $2004x+$ { $x$ } $=2005$ . Or { $x$ } $=2005-2004x$ . $\\$ Since fractional part lies between 0 and 1, we get $\\$ $0 \leq 2005-2004x<1$ $\\$ Multiplying by $-1$ , we get $\\$ $-1<2004x-2005 \leq 0$ . $\\$ Adding $2005$ , we get $\\$ $2004 <2004x \leq 2005$ . $\\$ Dividing by $2004$ , we get $\\$ $1<x \leq \frac{2005}{2004}$ . $\\$ Substituting $x= \dfrac{m}{2005}$ , $\\$ we get $m \in (2005,\frac {2005^{2}}{2004}]$ . Which is approximately equal to $2006.0004$ . Only integer in the range is $2006$ . Thus $2006$ is the answer.

We note that

$\left\lfloor \dfrac{m}{2005} \right\rfloor=\begin{cases} 0 &\quad m < 2005\\ 1 &\quad 2005 \leq m < 4010\end{cases}$

Therefore,

$m - \left\lfloor \dfrac{m}{2005} \right\rfloor=\begin{cases} m &\quad m < 2005\\ m-1 &\quad 2005 \leq m < 4010\end{cases}$

Now, if $m<2005$ , we have $m = 2005$ , which is contradicting itself

If $2005 \leq m < 4010$ , we have $m-1 = 2005$ , which gives $m=\boxed{2006}$

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Formal solution

$m - \left\lfloor \dfrac{m}{2005} \right\rfloor = 2005\\ m - \left( \dfrac{m}{2005} - \left\{\dfrac{m}{2005}\right\} \right) = 2005\\ \dfrac{2004}{2005}m + \left\{\dfrac{m}{2005}\right\} = 2005\\ \left\{\dfrac{m}{2005}\right\} = 2005 - \dfrac{2004}{2005}m$

Now, we know that the fractional part function, $\left\{\dfrac{m}{2005}\right\}$ has this range:

$0 \leq \left\{\dfrac{m}{2005}\right\} < 1$

This implies that

$0 \leq 2005 - \dfrac{2004}{2005}m < 1\\ -2005 \leq -\dfrac{2004}{2005}m < -2004\\ 2005 \geq \dfrac{2004}{2005}m > 2004\\ 2004 < \dfrac{2004}{2005}m \leq 2005\\ 2005 < m \leq \dfrac{2005^2}{2004} \approx 2006.0005$

Now, we know that $m$ must be an integer to yield the value $2005$ , therefore the only possible value of $m = \boxed{2006}$

We first off note that $2005 = m - \left \lfloor \frac{m}{2005} \right \rfloor \leq m \\\implies m \geq 2005$

So, we try $m=2005$ (which doesn't work) and then $m=\boxed{2006}$ (which works).