Given the equation $\sum_{k=11}^{99} \left[x + \frac{k}{100} \right] = 765$ find $[10 \, x]$. 
Given the equation $\sum_{k=11}^{99} \left[x + \frac{k}{100} \right] = 765$ find $[10 \, x]$.

I have tried this problem in many ways and I think it uses the identity 
$$[x]+\left[x+\frac{1}{n}\right]+\left[x+\frac{2}{n}\right]+\left[x+\frac{3}{n}\right]+.....+\left[x+\frac{n-1}{n}\right]= [nx]$$
Could someone please help me
and keep in mind x is a real number and $[\cdot]$ denotes the greatest integer function
 A: We start with the equation $765 = \sum_{i=1}^{89}\lfloor x + \frac{10+i}{100} \rfloor$
The first observation is that all fractions are between $0.11$ and $0.99$. If $x$ were an integer, none of the added terms would add something to the sum because they would be eliminated by the floor function.
Hence $8 < x < 9$ since $712 = 8 * 89 < 765 < 9 * 89 = 801$.
So we'll need $765 - 712 = 53$ terms that lift $\lfloor x + \frac{10+i}{100} \rfloor$ from $8$ to $9$
Obviously that will be the last $53$ terms. Hence all terms with $i > 36$.
This means that $x + \frac{10 + 36}{100} < 9$ and $x + \frac{10 + 37}{100} \geq 9$. In other words $8.54 > x \geq 8.53$.
This means that $\lfloor 10x \rfloor = 85$.
A: This is a well known identity known as Hermite's identity. A simple proof is here (Although, Hermite's original proof relied on identities involving exponents)
https://en.wikipedia.org/wiki/Hermite%27s_identity 
For general setup, we can also do the following. 
\begin{eqnarray*}
S &=& \sum_{k=m+1}^{n-1}{\left\lfloor x+\frac{k}{n}  \right\rfloor} \\
&=& l  \left\lfloor x \right\rfloor + (n-1-m) \left( 1+ \left\lfloor x \right\rfloor\right) \\
\end{eqnarray*}
\begin{eqnarray*}
  \left\lfloor x \right\rfloor  &\ge&  \frac{S}{n-m-1} \\
  &=& \left\lceil  \frac{S}{n-m-1}\right\rceil 
 \end{eqnarray*}
Using this, we can solve $l$,
\begin{eqnarray*}
l &=&  (n-m-1)  \left( 1+ \left\lfloor x \right\rfloor\right) -S
\end{eqnarray*}
Now observe that,
\begin{eqnarray*}
\left\lfloor x+ \frac{k}{n}  \right\rfloor &=&  \begin{cases}  \lfloor x \rfloor, &0\le k \le m+l \\  1+ \lfloor x \rfloor, &k > m+l \end{cases}.
\end{eqnarray*}
For any $\epsilon \in [0,1]$, s.t., $q=n\epsilon, q \in \mathbb{Z}_{+}$, 
\begin{eqnarray*}
\left\lfloor x+ \frac{k}{\epsilon n}  \right\rfloor &=&  \begin{cases}  \lfloor x \rfloor, &0\le k \le \lfloor \epsilon(m+l) \rfloor \\  1+ \lfloor x \rfloor, &k > \lfloor \epsilon (m+l) \rfloor \end{cases}.
\end{eqnarray*}
Now then the desired partial Hermite sum, $S_{\epsilon} (x)= \lfloor \epsilon n x \rfloor =  \sum_{k=0}^{n\epsilon-1}{\left\lfloor x+\frac{k}{n \epsilon}  \right\rfloor}$, will be, 
\begin{eqnarray*}
S_{\epsilon} (x) &=&  \sum_{k=0}^{n\epsilon-1}{\left\lfloor x+\frac{k}{n \epsilon}  \right\rfloor} \\
&=&\left(1+ \left\lfloor \epsilon (m+l) \right\rfloor \right) \lfloor r \rfloor + \left(1+ \lfloor r \rfloor  \right) \left(n-1- \left\lfloor \epsilon (m+l) \right\rfloor \right) \\
&=&  \left(1+ \left\lfloor \epsilon (m+l) \right\rfloor \right)  \left\lceil  \frac{S}{n-m-1}\right\rceil  + \left(1+ \left\lceil  \frac{S}{n-m-1}\right\rceil \right) \left(n-1- \left\lfloor \epsilon (m+l) \right\rfloor \right) 
\end{eqnarray*}
The running example has $S=765,\epsilon=0.1, n=100, m=10$. Substitution will give $l=36,  \lfloor x \rfloor =8$ , and $S_{\epsilon} (x) = \lfloor 10x \rfloor =85$, matching what Ronald arrived at.
Note: (1+floor(0.1*(10+36)))*8+(1+8)floor(0.1(100-1-(10+36)))=85.
Note: I was merely trying to generalize this problem. Hermite's identity is quite useful and several partial sums like these are possible, by simple manipulations.
On the fun side, it is easy to play with different $\epsilon$ and $m$ values. As a special case, when $m=0$ and $\epsilon=1$, we can get $S_{0}$,the standard Hermite identity. Here is a short matlab code snippet
function [Se,xfloor,l]= partial_hermite_puzzle(S,m,n,epsilon)
% Se -- Partial Hermite sum (0 to epsilon*n-1)
% S  -- Hermite Sum
xfloor=ceil((S-n+m+1)/(n-m-1));
l=-S+(n-m-1)*(xfloor+1);
Se=(1+floor(epsilon*(m+l)))*xfloor+(1+xfloor)*floor(epsilon*(n-1-(m+l)));

 
