Difference between revisions of "2016 AMC 8 Problems/Problem 17"
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<math>\textbf{(A)}\mbox{ }30\qquad\textbf{(B)}\mbox{ }7290\qquad\textbf{(C)}\mbox{ }9000\qquad\textbf{(D)}\mbox{ }9990\qquad\textbf{(E)}\mbox{ }9999</math> | <math>\textbf{(A)}\mbox{ }30\qquad\textbf{(B)}\mbox{ }7290\qquad\textbf{(C)}\mbox{ }9000\qquad\textbf{(D)}\mbox{ }9990\qquad\textbf{(E)}\mbox{ }9999</math> | ||
==Solution 1== | ==Solution 1== | ||
| − | For the first three digits, there are <math>10^3-1=999</math> combinations since <math>911</math> is not allowed. For the final digit, any of the <math>10</math> numbers are allowed. <math>999 \cdot 10 = 9990 \rightarrow \boxed{D}</math> | + | For the first three digits, there are <math>10^3-1=999</math> combinations since <math>911</math> is not allowed. For the final digit, any of the <math>10</math> numbers are allowed. <math>999 \cdot 10 = 9990 \rightarrow \boxed{\textbf{(D)}\ 9990}</math> |
==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=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> |
{{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 23:43, 27 November 2016
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?
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 
| 2016 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
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| All AJHSME/AMC 8 Problems and Solutions | ||
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