2016 AMC 8

Complete problem set with solutions and individual problem pages

Problem 17 Hard

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?

  • A.

    30

  • B.

    7290

  • C.

    9000

  • D.

    9990

  • E.

    9999

Answer:D

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}.

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