2003 AMC 10B Problem 25

Below is the professionally curated solution for Problem 25 of the 2003 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 10B solutions, or check the answer key.

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Concepts:divisibility digits modular arithmetic

Difficulty rating: 1400

25.

How many distinct four-digit numbers are divisible by $3$ and have $23$ as their last two digits?

$27$

$30$

$33$

$81$

$90$

Solution:

Write the number as $\overline{ab23}.$ It is divisible by $3$ when $a+b+2+3=a+b+5$ is divisible by $3,$ that is, when $a+b\equiv 1 \pmod 3.$

The two-digit prefix $\overline{ab}$ ranges over the $90$ values from $10$ to $99,$ and exactly one third of them satisfy this, giving $\dfrac{90}{3}=30.$

Thus, the correct answer is B.

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