1985 AMC 8 Problem 22

Below is the professionally curated solution for Problem 22 of the 1985 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1985 AMC 8 solutions, or check the answer key.

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Concepts:basic probabilitymultiplication principle

Difficulty rating: 1000

22.

Assume every 77-digit whole number is a possible telephone number except those that begin with 00 or 1.1. What fraction of telephone numbers begin with 99 and end with 0?0?

163\dfrac{1}{63}

180\dfrac{1}{80}

181\dfrac{1}{81}

190\dfrac{1}{90}

1100\dfrac{1}{100}

Solution:

The first digit is one of 88 allowed digits (22 through 99), so 18\dfrac18 of the numbers begin with 9.9. The last digit is any of 1010 digits, so 110\dfrac{1}{10} end in 0.0.

These conditions are independent, so the fraction is 18110=180.\dfrac18 \cdot \dfrac{1}{10} = \dfrac{1}{80}.

Thus, the correct answer is B .

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