Floor function repeated addition I've come across this problem: $$\left\lfloor r + \frac{19}{100}\right\rfloor + \left\lfloor r + \frac{20}{100} \right\rfloor + \left\lfloor r + \frac{21}{100}\right\rfloor + \dots + \left\lfloor r + \frac{91}{100}\right\rfloor = 546$$ $$\text{Find}  \
\lfloor 100r\rfloor. \textit{(Source: AIME)}$$
Here is my work:
$\lfloor r + \frac{19}{100}\rfloor  = r + \frac{19}{100} - \{r + \frac{19}{100}\}$, so the figure can be restated as $\{r + \frac{19}{100}\} = r + \frac{19}{100} + a - 546$, where $a = \lfloor r + \frac{20}{100}\rfloor + \lfloor r + \frac{21}{100}\rfloor + \dots + \lfloor r + \frac{91}{100}\rfloor$. 
Because $\{r + \frac{19}{100}\}$ is the fractional part, $0 \le r + \frac{19}{100} + a - 546 < 1$, so after some more maniuplation, $545 + \frac{81}{100} \le r + a < 546 + \frac{81}{100}$. $a$ is an integer, so the fractional part of $r$ must be $\frac{81}{100}$.
$r = \lfloor r\rfloor + \{r\}$, so $\lfloor r + \frac{19}{100}\rfloor = \lfloor \lfloor r\rfloor + \frac{81}{100} + \frac{19}{100}\rfloor$ = $\lfloor \lfloor r\rfloor + 1\rfloor$. Because both of the terms inside that floor function are integers, it must equal $\lfloor r\rfloor + 1$. This same reasoning can be applied to each of the individual floor functions of the given figure's LHS, and they each turn out to be $\lfloor r\rfloor + 1$.
Therefore, $73 \lfloor r\rfloor + 73 = 546$, so $73\lfloor r\rfloor = 473$. However, this cannot be true unless there is no answer (which I assume is not the case), or unless I did something wrong in my process, because then $\lfloor r\rfloor$ is not an integer.
If you do see the solution, it would be really nice if you did not give the answer in your response! Instead, maybe some helpful hints or partial solutions would be preferred.
 A: Don't worry about the fractional part.  And don't worry about $[r]$. Find the precise value of $k$ where $[r + \frac k{100}]\ne [r+\frac {k+1}{100}]$.  That is where $r + \frac {k}{100} < m \le r + \frac {k+1}{100}$ for some integer $m$.
====== my answer below ====

 Well, what jumps at me is $0 < \frac k{100} < 1$ so all the $[r +\frac {k}{100}]$ are either one integer, call it $n$ or the next, $n+1$. 

. 

 So if $b$ of them equal $n+1$ and $(73 -b)$ of them equal $n$ we have $(73-b)n + b(n+1) = 73n +  b =546$ where $0\le b < 73$.  

.

 So as $546\equiv 35 \pmod {73}$ and $546= 7*73 + 35$ so $b=35$ and $n=7$.  

.  

So we have (I'll have to be careful not to do a fencepost error... $91-35=56$  so....)  $[r+\frac{19}{100}],....,[r+\frac{56}{100}] = 7$ and $[r+\frac {57}{100}],...,[r+\frac {91}{100}] = 8$.

.

So  $r + \frac {56}{100} < 8\le r+\frac {57}{100}$ so    

.

$100r + 56 < 800 \le 100r + 57$

.

$100r < 744 \le 100r + 1$

.

And $100r -1 < 743 \le 100r$ so $743 \le 100r < 744$

.

So $[100r] = 743$.

A: $$  \left\lfloor r + \frac{19}{100}\right\rfloor 
  + \left\lfloor r + \frac{20}{100} \right\rfloor 
  + \left\lfloor r + \frac{21}{100}\right\rfloor 
  + \dots 
  + \left\lfloor r + \frac{91}{100}\right\rfloor 
  = 546$$
First, let $r = s - \dfrac{19}{100}$. Then the sum becomes
$$  \left\lfloor s + \frac{0}{100}\right\rfloor 
  + \left\lfloor s + \frac{1}{100} \right\rfloor 
  + \left\lfloor s + \frac{2}{100}\right\rfloor 
  + \dots 
  + \left\lfloor s + \frac{72}{100}\right\rfloor 
  = 546$$
If $s$ were an integer, then you would get $73s = 546$, which has solution $s = 7 \dfrac{35}{73}$.
Noting that $7 \cdot 73 = 511$ and $8 \cdot 73 = 584$, we see that $7 < s < 8$. Now $546-511 = 35$. So where are you going to pick up that extra $35$?
