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Difference between revisions of "2016 AMC 8 Problems/Problem 17"
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17. 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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== Problem ==
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<math>(A)\mbox{ }30\mbox{          }(B)\mbox{ }7290\mbox{          }(C)\mbox{ }9000\mbox{          }(D)\mbox{ }9990\mbox{          }(E)\mbox{ }9999\mbox{          }</math>
 
  
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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?
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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>
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==Solutions==
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===Solution 1===
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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>.
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~CHECKMATE2021
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===Solution 2===
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Counting the prohibited cases, we find that there are 10 of them. This is because, when we start with 9,1, and 1, 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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~CHECKMATE2021
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==Video Solution 1 (HOW TO THINK CREATIVELY!!!)==
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https://youtu.be/bP2698xypP0
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~Education, the Study of Everything
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==Video Solution==
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https://youtu.be/slvWHYXz-20
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~savannahsolver
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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}}

Latest revision as of 21:44, 21 January 2024

Problem

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?

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

Solutions

Solution 1

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. $999 \cdot 10 = 9990 \rightarrow \boxed{\textbf{(D)}\ 9990}$.

~CHECKMATE2021

Solution 2

Counting the prohibited cases, we find that there are 10 of them. This is because, when we start with 9,1, and 1, we can have any of the 10 digits for the last digit. So, our answer is $10^4-10=\boxed{\textbf{(D)}\ 9990}.$

~CHECKMATE2021

Video Solution 1 (HOW TO THINK CREATIVELY!!!)

https://youtu.be/bP2698xypP0

~Education, the Study of Everything


Video Solution

https://youtu.be/slvWHYXz-20

~savannahsolver

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

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