Let $n$ be a $5$-digit number, and let $q$ and $r$ be the quotient and the remainder, respectively, when $n$ is divided by $100$. For how many values of $n$ is $q+r$ divisible by $11$? 
$\mathrm{(A) \ } 8180\qquad \mathrm{(B) \ } 8181\qquad \mathrm{(C) \ } 8182\qquad \mathrm{(D) \ } 9000\qquad \mathrm{(E) \ } 9090$
Simple Solution
$11|(q+r)$ implies that $11|(99q+q+r)$ and therefore $11|(100q+r)$, so $11|n$. Then, $n$ can range from $10010$ to $99990$ for a total of $\boxed{8181\Rightarrow \mathrm{(B)}}$ numbers.
<3

Solution 1
When a $5$-digit number is divided by $100$, the first $3$ digits become the quotient, $q$, and the last $2$ digits become the remainder, $r$. 
Therefore, $q$ can be any integer from $100$ to $999$ inclusive, and $r$ can be any integer from $0$ to $99$ inclusive. 
For each of the $9\cdot10\cdot10=900$ possible values of $q$, there are at least $\left\lfloor \frac{100}{11} \right\rfloor = 9$ possible values of $r$ such that $q+r \equiv 0\pmod{11}$. 
Since there is $1$ "extra" possible value of $r$ that is congruent to $0\pmod{11}$, each of the $\left\lfloor \frac{900}{11} \right\rfloor = 81$ values of $q$ that are congruent to $0\pmod{11}$ have $1$ more possible value of $r$ such that $q+r \equiv 0\pmod{11}$. Another way to think about it is the number of possible values of q when r, the remainder, is $0$. In this case, q itself has to be a multiple of $11$. $\left\lfloor \frac{999}{11} = 90 \right\rfloor$. Then we'll need to subtract $9$ from $90$ since we only want multiples of $11$ greater than $100$ $(90-9=81)$
Therefore, the number of possible values of $n$ such that $q+r \equiv 0\pmod{11}$ is $900\cdot9+81\cdot1=8181 \Rightarrow B$.
~ Minor Edit by PlainOldNumberTheory

Solution 2
Let $n$ equal $\overline{abcde}$, where $a$ through $e$ are digits. Therefore,
$q=\overline{abc}=100a+10b+c$
$r=\overline{de}=10d+e$
We now take $q+r\bmod{11}$:
$q+r=100a+10b+c+10d+e\equiv a-b+c-d+e\equiv 0\bmod{11}$
The divisor trick for 11 is as follows:
"Let $n=\overline{a_1a_2a_3\cdots a_x}$ be an $x$ digit integer. If $a_1-a_2+a_3-\cdots +(-1)^{x-1} a_x$ is divisible by $11$, then $n$ is also divisible by $11$."
Therefore, the five-digit number $n$ is divisible by 11. The 5-digit multiples of 11 range from $910\cdot 11$ to $9090\cdot 11$. There are $8181\Rightarrow \mathrm{(B)}$ divisors of 11 between those inclusive.

Note
The part labeled "divisor trick" actually follows from the same observation we made in the previous step: $10\equiv (-1)\pmod{11}$, therefore $10^{2k}\equiv 1$ and $10^{2k+1}\equiv (-1)$ for all $k$. 
For a $5-$digit number $\overline{abcde}$ we get $\overline{abcde}\equiv a\cdot 1 + b\cdot(-1) + c\cdot 1 + d\cdot(-1) + e\cdot 1 = a-b+c-d+e$, as claimed.
Also note that in the "divisor trick" we want to assign the signs backward - if we make sure that the last sign is a $+$, the result will have the same remainder modulo $11$ as the original number.

Solution 3
Since $q$ is a quotient and $r$ is a remainder when $n$ is divided by $100$. So we have $n=100q+r$. Since we are counting choices where $q+r$ is divisible by $11$, we have $n=99q+q+r=99q+11k$ for some $k$. This means that $n$ is the sum of two multiples of $11$ and would thus itself be a multiple of $11$. Then we can count all the five-digit multiples of $11$ as in Solution 2. (This solution is essentially the same as Solution 2, but it does not necessarily involve mods and so could potentially be faster.)

Solution 4
Defining $q$ and $r$ in terms of floor functions and $n$ would yield: $q=\left \lfloor \frac{n}{100} \right \rfloor$ and $r=n-100 \left \lfloor \frac{n}{100} \right \rfloor$. Since $q+r \equiv 0\pmod{11}$, $\left \lfloor \frac{n}{100} \right \rfloor + n-100 \left \lfloor \frac{n}{100} \right \rfloor \equiv 0\pmod{11}$. Simplifying gets us $n-99 \left \lfloor \frac{n}{100} \right \rfloor\equiv 0\pmod{11} \rightarrow n \equiv 0\pmod{11}$ ($99 \left \lfloor \frac{n}{100} \right \rfloor\equiv 0\pmod{11}$ is always true since floor function always yields an integer, and 99 is divisible by 11 w/o any remainder). After we come to this conclusion, it becomes easy to solve the rest of the problem ($\left \lfloor \frac{n}{99999} \right \rfloor - \left \lfloor \frac{n}{10000} \right \rfloor$). 
~hw21