2003 AMC 10A Problem 25

Attempt Problem 25 of the 2003 AMC 10A 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 10A solutions, or check the answer key.

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25.

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

Answer: B

Concepts:divisibility modular arithmetic counting integers in a range

Difficulty rating: 2070

Small hint:

Write

n = 100q + r,

so

n = (q + r) + 99q

Big hint:

Since

99q

is divisible by

11,

q + r

is divisible by

11

exactly when

n

is

Solution:

Write

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

Since

99q

is a multiple of

11,

q + r

is divisible by

11

if and only if

n

is.

The

5

-digit multiples of

11

satisfy

10000 \le n \le 99999,

and there are

\begin{aligned} &\left\lfloor\dfrac{99999}{11}\right\rfloor - \left\lfloor\dfrac{9999}{11}\right\rfloor \\ &= 9090 - 909 \\ &= 8181. \end{aligned}

Thus, the correct answer is B.