8. Ans: 853
First observe that $\left\lfloor x+\frac{k}{00}\right\rfloor=\lfloor x\rfloor$ or $\lfloor x\rfloor+1$ for $11 \leq k \leq 99$. Since there are 89 terms on the left-hand side of the equation and $89 \times 8<765<89 \times 9$, we deduce that $\lfloor x\rfloor=8$. Now suppose $\left\lfloor x+\frac{k}{100}\right\rfloor=8$ for $11 \leq k \leq m$ and $\left\lfloor x+\frac{k}{100}\right\rfloor=9$ for $m+1 \leq k \leq 99$. Then
$$
8(m-10)+9(99-m)=765,
$$
which gives $m=46$. Therefore $\left\lfloor x+\frac{46}{100}\right\rfloor=8$ and $\left\lfloor x+\frac{47}{100}\right\rfloor=9$, which imply $8 \leq x+\frac{46}{100}<9$ and $9 \leq x+\frac{47}{100}<10$ respectively. The inequalities lead to $7.54 \leq x<8.54$ and $8.53 \leq x<9.53$. Consequently, we see that $8.53 \leq x<8.54$. Hence we conclude that $[100 x]=853$.
8. If $x$ is a real number that satisfies
$$
\left\lfloor x+\frac{11}{100}\right\rfloor+\left\lfloor x+\frac{12}{100}\right\rfloor+\cdots+\left\lfloor x+\frac{99}{100}\right\rfloor=765
$$
find the value of $\lfloor 100 x\rfloor$. Here $\lfloor a\rfloor$ denotes the largest integer $\leq a$.