2003 AMC 12A Problem 8

Attempt Problem 8 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).

What is the probability that a randomly drawn positive factor of

60

is less than

7?

\dfrac{1}{10}

\dfrac{1}{6}

\dfrac{1}{4}

\dfrac{1}{3}

\dfrac{1}{2}

60=2^2\cdot3\cdot5

has

(2{+}1)(1{+}1)(1{+}1)=12

factors

Count the factors from

1

through

6

The number

60

has

12

positive factors:

1,2,3,4,5,6,

10,12,15,20,30,60.

Six of them are less than

7,

namely

1,2,3,4,5,6,

so the probability is

\dfrac{6}{12}=\dfrac12.

Thus, the correct answer is E.

Problem 8 in other years