2000 AMC 12 Problem 4

Attempt Problem 4 of the 2000 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2000 AMC 12 solutions, or check the answer key.

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4.

The Fibonacci sequence 1,1,2,3,5,8,13,21,1, 1, 2, 3, 5, 8, 13, 21, \ldots starts with two 11s, and each term afterwards is the sum of its two predecessors. Which one of the ten digits is the last to appear in the units position of a number in the Fibonacci sequence?

00

44

66

77

99

Answer: C
Concepts:Fibonacciunits digitsystematic listing
Difficulty rating: 1240
Solution:

The sequence of units digits begins 1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,7,7,4,1,5,6, \begin{gathered} 1, 1, 2, 3, 5, 8, 3, 1, 4, 5, 9, \\ 4, 3, 7, 0, 7, 7, 4, 1, 5, 6, \ldots \end{gathered}

Scanning this list, the digit 66 is the last of the ten digits to appear.

Thus, the correct answer is C.

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