Finding roots of an equation wich involves floor function I'm trying to solve this equation
$$
\left \lfloor{x +\frac{1}{100}}\right \rfloor + \left \lfloor{x +\frac{2}{100}}\right \rfloor + ... + \left \lfloor{x +\frac{223}{100}}\right \rfloor = 521
$$
I haven't faced until now problems like this one... How can I find $x$? 
Rather than a specific solution, I'm looking for the intuition to use in order to set up a solution to problems like this one.
 A: First, look at how many terms there are, then think about how each term may contribute to the sum. The target sum is $521$, there are 223 summands, and they are adjacent integers. Since $521/223$ is between $2$ and $3$, some summands must be $2$ and some must be $3$. Then, consider the following case.
If 100 of the summands are $2$ and 100 of the summands are $3$, then 200 of the summands add up to $500$. There are 23 number left, and $21$ more is needed to reach $521$. This is impossible even if all the 23 numbers are $1$. What this tells us is, however, that this is close to the correct solution, so let us stick with 100 copies of $2$.
With 100 of the summands fixed, we are left with 123 summands. The task is to choose the right number of $1$'s and $3$'s that will contribute to the remaining value, which is $321$.
If we pick $n$ summands to be $1$, then $123 - n$ of the summands will be $3$, and their sum will be $n + 3(123 - n) = 369 - 2n$. We want this to equal $321$, so we solve
\begin{align}
369 - 2n & = 321 \\
\therefore n & = 24.
\end{align}
To find $x$, note that we want $\lfloor x + \frac{24}{100} \rfloor$ to be $1$ and $\lfloor x + \frac{25}{100} \rfloor$ to be $2$. We can pick any $x \in \left[\frac{175}{100}, \frac{176}{100}\right)$.
A: Let $n - (i-1)/100 > x \ge n - i/100$
So if $i \le 23$ then $[x + j/100] = n - 1$ for $j \le i$.  $[x + j/100] = n$ for $i < j \le 100 + i$. $[x + j/100] = n +1$ for $100 + i < j \le 200 + i$ and $[x + j/100] = n + 2$ for $200 + i < j \le 223$.
So $521 = 223n - i + 100 + (23 - i)2 = 223n + 146 - 3i\implies 375 = 223(2) - 3i \implies i = 43$. Which is impossible. So $i > 23$.
So $[x + j/100] = n - 1$ for $j \le i$.  $[x + j/100] = n$ for $i < j \le 100 + i$. $[x + j/100] = n +1$ for $100 + i < j \le 223$
So $521 = 223n - i + 123 -i$ so $398 = 223(2) - 2i$ so $i = 24$ so 
$1.75 \le x < 1.76$.
Unless I made a mistake which I probably did. $[x + 1/100]$ to $[x + 24/100]$ = 1.  So those add up to 24.
$[x + 25/100]$ to $[x + 124/100] = 2$.  So those add up to 200 so 224 running total..
$[x + 125/100]$ to $[x + 223/100] = 3$. So those add up to 297 so 521 total.  Phew!  No mistake.  I usually make some indexing mistake.
