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

Revision as of 05:54, 17 January 2023

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

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 $10^4-10=\boxed{\textbf{(D)}\ 9990}.$

Video Solution

https://youtu.be/slvWHYXz-20

~savannahsolver

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.

@@ Line 10: / Line 10: @@
 ===Solution 2===
-Counting the prohibited cases, we find that there are 10 of them. This is because we start with 9,1, and 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, 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>
 ==Video Solution==

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