2003 AMC 10A Problem 25

First see what $q$ and $r$ are with an example, then find a link between $q + r$ and $n$.

1. Example. Take $n = 12345$. Dividing by $100$ gives quotient $q = 123$ and remainder $r = 45$. So $q$ is the first three digits and $r$ is the last two, and $n = 100 \times 123 + 45$. In general,
$$n = 100q + r.$$
2. Split off $q + r$. Write $100q$ as $99q + q$:
$$n = 99q + (q + r).$$
3. Use that $99 = 9 \times 11$. The part $99q$ is always a multiple of $11$. So $n$ and $q + r$ differ by a multiple of $11$. That means: $q + r$ is a multiple of $11$ if and only if $n$ is a multiple of $11$.
(Check with the example: $q + r = 168$, and $12345 - 168 = 12177 = 99 \times 123$.)
4. So count the five-digit multiples of $11$.
- Smallest: $11 \times 909 = 9999$ has only four digits, so the first one is $11 \times 910 = 10010$.
- Largest: $11 \times 9090 = 99990$, and $11 \times 9091 = 100001$ has six digits.
- The multipliers run from $910$ to $9090$. That is
$$9090 - 910 + 1 = 8181$$
numbers. (The $+1$ counts the starting number: from $1$ to $5$ there are $5 - 1 + 1 = 5$ numbers.)

The answer is $\boxed{\textbf{(B)}\ 8181}$.

Splitting a number into a front part and a back part, $100q + r$, and noticing that $100$ is just $1$ more than a multiple of $11$ makes the front and back combine into $q + r$ with nothing left over except a multiple of $11$. Whenever a problem mixes the front and back pieces of a number, check how the place value ($10$, $100$, $1000$) sits next to the divisor. Then count carefully, remembering the $+1$ when counting from one number up to another.

Count the multiples of $11$ in a different way: from $1$ to $99999$ there are $9090$ of them (since $99999 \div 11 = 9090.8\ldots$), and from $1$ to $9999$ there are $909$ (since $9999 = 11 \times 909$). The five-digit ones are the difference, $9090 - 909 = 8181$.