2003 AMC 12A Problem 18

Attempt Problem 18 of the 2003 AMC 12A below, then check your answer against the professionally curated solution 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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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

q + r

divisible by

11?

8180

8181

8182

9000

9090

n=100q+r=(q+r)+99q

Since

99

is a multiple of

11,

q+r\equiv n\pmod{11}

Since

n=100q+r=(q+r)+99q

and

99

is divisible by

11,

we have

q+r\equiv n\pmod{11}.

(q+r)

is a multiple of

11

exactly when

n

is a multiple of

11.

Among the

5

-digit numbers, the count of multiples of

11

\left\lfloor\dfrac{99999}{11}\right\rfloor

-\left\lfloor\dfrac{9999}{11}\right\rfloor

=9090-909=8181.

Thus, the correct answer is B.

Problem 18 in other years