2009 AMC 8 Problem 13 — Solution | LIVE by Po-Shen Loh

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2009 AMC 8 Problem 13

Attempt Problem 13 of the 2009 AMC 8 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 2009 AMC 8 solutions, or check the answer key.

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

13.

A three-digit integer contains one of each of the digits $1,$ $3,$ and $5.$ What is the probability that the integer is divisible by $5$ ?

$\dfrac{1}{6}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

$\dfrac{5}{6}$

Answer: B

Concepts:basic probability divisibility

Difficulty rating: 930

Solution:

The number is equally likely to end in $1$ , $3$ , or $5$ .

It is divisible by $5$ only if the last digit is $5$ , which happens with probability $\dfrac{1}{3}$ .

Thus, B is the correct answer.

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