2003 AMC 10A Problem 8

Attempt Problem 8 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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8.

What is the probability that a randomly drawn positive factor of 6060 is less than 7?7?

110\dfrac{1}{10}

16\dfrac{1}{6}

14\dfrac{1}{4}

13\dfrac{1}{3}

12\dfrac{1}{2}

Answer: E
Concepts:factorbasic probability
Difficulty rating: 1250
Small Hint:

List all positive divisors of 6060

Big Hint:

Count how many of the 1212 divisors are less than 77

Solution:

The factors of 6060 are 1,1, 2,2, 3,3, 4,4, 5,5, 6,6, 10,10, 12,12, 15,15, 20,20, 30,30, and 60.60.

Six of these twelve factors are less than 7,7, so the probability is 612=12.\dfrac{6}{12} = \dfrac{1}{2}.

Thus, the correct answer is E.

Problem 7#7
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