2009 AMC 8 Problem 22 — Solution | LIVE by Po-Shen Loh

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2009 AMC 8 Problem 22

Below is the professionally curated solution for Problem 22 of the 2009 AMC 8, 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 of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:digits multiplication principle casework

Difficulty rating: 1400

22.

How many whole numbers between $1$ and $1000$ do not contain the digit $1?$

$512$

$648$

$720$

$728$

$800$

Solution:

We can case on the number of digits.

There are $8$ one digit numbers excluding $1.$

There are $8 \cdot 9 = 72$ two digit numbers that lack the digit $1.$

There are $8 \cdot 9 \cdot 9 = 648$ three digit numbers that do not include $1.$

This yields a total of $8 + 72 + 648 = 728$ numbers that do not contain the digit $1.$

Thus, D is the correct answer.

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