2003 AMC 10A Problem 15

Attempt Problem 15 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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15.

What is the probability that an integer in the set {1,2,3,,100}\{1, 2, 3, \ldots, 100\} is divisible by 22 and not divisible by 3?3?

16\dfrac{1}{6}

33100\dfrac{33}{100}

1750\dfrac{17}{50}

12\dfrac{1}{2}

1825\dfrac{18}{25}

Answer: C
Concepts:basic probabilitymultiplecounting integers in a range
Difficulty rating: 1310
Solution:

Of the 100100 integers, 5050 are divisible by 2.2.

Among those, the ones also divisible by 33 are the multiples of 6,6, of which there are 16.16.

So 5016=3450 - 16 = 34 qualify, giving probability 34100=1750.\dfrac{34}{100} = \dfrac{17}{50}.

Thus, the correct answer is C.

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