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2000 AMC 12 Problem 4

Problem 4 of 25EasierAlgebraNumber TheoryCounting & Probability

The Fibonacci sequence 1,1, 1,1, 2,2, 3,3, 5,5, 8,8, 13,13, 21,21, \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?

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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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Concepts: Fibonacci · units digit · systematic listing

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.