2003 AMC 12A Problem 18

Below is the professionally curated solution for Problem 18 of the 2003 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 12A solutions, or check the answer key.

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Concepts:modular arithmetic multiple

Difficulty rating: 1800

18.

Let $n$ be a $5$ -digit number, and let $q$ and $r$ be the quotient and remainder, respectively, when $n$ is divided by $100.$ For how many values of $n$ is $q + r$ divisible by $11?$

$8180$

$8181$

$8182$

$9000$

$9090$

Solution:

Since $n=100q+r=(q+r)+99q$ and $99$ is divisible by $11,$ we have $q+r\equiv n\pmod{11}.$

So $11\mid(q+r)$ exactly when $11\mid n.$

Among the $5$ -digit numbers, the count of multiples of $11$ is $\left\lfloor\dfrac{99999}{11}\right\rfloor-\left\lfloor\dfrac{9999}{11}\right\rfloor=9090-909=8181.$

Thus, the correct answer is B.

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