Difference between revisions of "2009 AMC 8 Problems/Problem 22"
(→Solution) |
Mathmagicops (talk | contribs) (→Solution (Super Fast!)) |
||
| Line 10: | Line 10: | ||
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 <math>9*9*9=729</math> such numbers. However, we over counted by one; 0 is not between 1 and 1000, so there are <math>\boxed{\textbf{(D)}\ 728}</math> numbers. | 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 <math>9*9*9=729</math> such numbers. However, we over counted by one; 0 is not between 1 and 1000, so there are <math>\boxed{\textbf{(D)}\ 728}</math> numbers. | ||
| + | |||
| + | |||
| + | Note: Something like 020 just counts as 20, but 000 is not between 1 and 1000, as stated above. | ||
==Solution 2 (Easy Casework)== | ==Solution 2 (Easy Casework)== | ||
Revision as of 23:46, 16 January 2022
Problem
How many whole numbers between 1 and 1000 do not contain the digit 1?
Solution (Super Fast!)
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 
such numbers. However, we over counted by one; 0 is not between 1 and 1000, so there are 
numbers.
Note: Something like 020 just counts as 20, but 000 is not between 1 and 1000, as stated above.
Solution 2 (Easy Casework)
We consider the 3 cases, where the number is 1,2 or 3 digits.
Case 1 : The number has a 
digit. Well, 1-9 all work except for , so 
numbers, for numbers that are 1 digit.
Case 2 : The number is has 
digits. Consider the two digit number. Since the tens digit can't be 0 or 1, we have 8 choices for the tens digit. For the ones digit, we have 9, since the ones digit can't be a 1. This gives us a total of 
two-digit numbers.
Case 3 : The number has 
digits. Using similar logic to Case 2, we have 8*9*9 choices for the third number.
We add the cases up getting 
which gives us 
numbers total.
See Also
| 2009 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
