Difference between revisions of "2016 AMC 8 Problems/Problem 17"

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An ATM password at Fred's Bank is composed of four digits from <math>0</math> to <math>9</math>, with repeated digits allowable. If no password may begin with the sequence <math>9,1,1,</math> then how many passwords are possible?
 
An ATM password at Fred's Bank is composed of four digits from <math>0</math> to <math>9</math>, with repeated digits allowable. If no password may begin with the sequence <math>9,1,1,</math> then how many passwords are possible?
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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==
 
==Solution==

Revision as of 12:39, 23 November 2016

An ATM password at Fred's Bank is composed of four digits from $0$ to $9$, with repeated digits allowable. If no password may begin with the sequence $9,1,1,$ then how many passwords are possible?

$\textbf{(A)}\mbox{ }30\qquad\textbf{(B)}\mbox{ }7290\qquad\textbf{(C)}\mbox{ }9000\qquad\textbf{(D)}\mbox{ }9990\qquad\textbf{(E)}\mbox{ }9999$

Solution

For the first three digits, there are $10^3-1=999$ combinations since $911$ is not allowed. For the final digit, any of the $10$ numbers are allowed. $999 \cdot 10 = 9990 \rightarrow \boxed{D}$

2016 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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