2009 AMC 8 Problem 22

Attempt Problem 22 of the 2009 AMC 8 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 2009 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

22.

How many whole numbers between 11 and 10001000 do not contain the digit 1?1?
512512
648648
720720
728728
800800
Answer: D
Concepts:digitsmultiplication principlecasework
Small hint:
Count by number of digits, or pad numbers with leading zeroes.
Big hint:
For three padded digits, each position has 99 choices if digit 11 is forbidden, but exclude 000.000.
Solution:
We can case on the number of digits.
There are 88 one digit numbers excluding 1.1.
There are 8⋅9=728 \cdot 9 = 72 two digit numbers that lack the digit 1.1.
There are 8⋅9⋅9=6488 \cdot 9 \cdot 9 = 648 three digit numbers that do not include 1.1.
This yields a total of 8+72+648=728 8 + 72 + 648 = 728 numbers that do not contain the digit 1.1.
Thus, D is the correct answer.
LIVE by Po-Shen Loh
Programs

LIVE Classes

Self-Paced Classes

Stars

Scholarship

About Us

Who Is Po-Shen Loh?

Email Contact

Newsletter

Jobs

Become a Star

For a more Thought Full world.

© 2026 Expii, Inc.