Difference between revisions of "2009 AMC 8 Problems/Problem 22"
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<math> \textbf{(A)}\ 512 \qquad \textbf{(B)}\ 648 \qquad \textbf{(C)}\ 720 \qquad \textbf{(D)}\ 728 \qquad \textbf{(E)}\ 800</math> | <math> \textbf{(A)}\ 512 \qquad \textbf{(B)}\ 648 \qquad \textbf{(C)}\ 720 \qquad \textbf{(D)}\ 728 \qquad \textbf{(E)}\ 800</math> | ||
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| + | ==Solution== | ||
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| + | Note that this is the same as finding how many numbers with up to three digits do not contain 1. | ||
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| + | Since there are 10 total possible digits, and only one of them is not allowed (1), each digit has its choice of 9 digits, for a total of 9*9*9=729 such numbers. However, we over counted by one; 000=0 is not between 1 and 1000, so there are 728 numbers. <math>\textbf{(D)}</math> | ||
Revision as of 12:20, 5 July 2012
Problem
How many whole numbers between 1 and 1000 do not contain the digit 1?
Solution
Note that this is the same as finding how many numbers with up to three digits do not contain 1.
Since there are 10 total possible digits, and only one of them is not allowed (1), each digit has its choice of 9 digits, for a total of 9*9*9=729 such numbers. However, we over counted by one; 000=0 is not between 1 and 1000, so there are 728 numbers. 
