2016 AMC 8 Problem 17

Attempt Problem 17 of the 2016 AMC 8 below, then check your answer against the video solution and professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2016 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

17.

An ATM password at Fred’s Bank is composed of four digits from 00 to 9,9, with repeated digits allowable. If no password may begin with the sequence 9,9, 1,1, 1,1, then how many passwords are possible?
30 30
7290 7290
9000 9000
9990 9990
9999 9999
Answer: D
Concepts:multiplication principlecomplementary counting
Small hint:
First count all four-digit passwords when leading zeroes are allowed.
Big hint:
Then subtract the passwords whose first three digits are forced to be 9,9, 1,1, 11.
Video solution:
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Written solution:
The total number of passwords with no conditions is 104.10^4. The condition removes 1010 possible passwords since the first 33 are determined, and the last one can be anything. Therefore, the number of acceptable passwords is 10,000−10=9990.10{,}000 - 10 = 9990.
Thus, D is the correct answer.
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