2003 AMC 10A Problem 25

By definition $n = 100q + r$. Rewrite this as
$$n = 99q + (q + r).$$
Because $99 = 9 \cdot 11$, the term $99q$ is a multiple of $11$, so $n$ and $q + r$ leave the same remainder when divided by $11$. Hence $11 \mid q + r$ exactly when $11 \mid n$.

So we count five-digit multiples of $11$. The smallest is $11 \cdot 910 = 10010$ (since $11 \cdot 909 = 9999$ is too small) and the largest is $11 \cdot 9090 = 99990$. The multipliers run from $910$ to $9090$, giving
$$9090 - 910 + 1 = 8181$$
values of $n$.

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

Splitting a number as $100q + r$ and noticing that $100 \equiv 1 \pmod{11}$ collapses the condition to divisibility of $n$ itself. Whenever a problem mixes the "front" and "back" of a number, check what the split point ($100$, $1000$, ...) is congruent to modulo the target divisor; a residue of $1$ or $-1$ makes the front and back combine simply.

Equivalently, $q + r = \overline{abc} + \overline{de} \equiv (a - b + c) + (-d + e) \pmod{11}$, the alternating digit sum of $n = \overline{abcde}$, and the alternating-sum test says this is $0$ mod $11$ exactly when $11 \mid n$. Count as above, or as $\lfloor 99999/11 \rfloor - \lfloor 9999/11 \rfloor = 9090 - 909 = 8181$.

• Counting the multiples of 11 from 11910 to 119090 as 9090 - 910 = 8180, forgetting the +1, which is choice (A).
• Trying to count pairs $(q, r)$ directly with $900$ choices of $q$ and $9$ or $10$ choices of $r$ each; this works but is easy to get slightly wrong.