2003 AMC 10A Problem 15 — Solution | LIVE by Po-Shen Loh

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2003 AMC 10A Problem 15

Below is the professionally curated solution for Problem 15 of the 2003 AMC 10A, 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.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:basic probability multiple counting integers in a range

Difficulty rating: 1310

15.

What is the probability that an integer in the set $\{1, 2, 3, \ldots, 100\}$ is divisible by $2$ and not divisible by $3?$

$\dfrac{1}{6}$

$\dfrac{33}{100}$

$\dfrac{17}{50}$

$\dfrac{1}{2}$

$\dfrac{18}{25}$

Solution:

Of the $100$ integers, $50$ are divisible by $2.$

Among those, the ones also divisible by $3$ are the multiples of $6,$ of which there are $16.$

So $50 - 16 = 34$ qualify, giving probability $\dfrac{34}{100} = \dfrac{17}{50}.$

Thus, the correct answer is C.

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