Difference between revisions of "2016 AMC 8 Problems/Problem 17"
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===Solution 2=== | ===Solution 2=== | ||
Counting the prohibited cases, we find that there are 10 of them. This is because we start with 9,1,1 and we can have any of the 10 digits for the last digit. So our answer is <math>10^4-10=\boxed{\textbf{(D)}\ 9990}.</math> | Counting the prohibited cases, we find that there are 10 of them. This is because we start with 9,1,1 and we can have any of the 10 digits for the last digit. So our answer is <math>10^4-10=\boxed{\textbf{(D)}\ 9990}.</math> | ||
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| + | ==See Also== | ||
{{AMC8 box|year=2016|num-b=16|num-a=18}} | {{AMC8 box|year=2016|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 16:32, 21 April 2021
Problem
An ATM password at Fred's Bank is composed of four digits from 
to 
, with repeated digits allowable. If no password may begin with the sequence 
then how many passwords are possible?
Solutions
Solution 1
For the first three digits, there are 
combinations since 
is not allowed. For the final digit, any of the 
numbers are allowed. 
Solution 2
Counting the prohibited cases, we find that there are 10 of them. This is because we start with 9,1,1 and we can have any of the 10 digits for the last digit. So our answer is 
See Also
| 2016 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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. 
