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2000 AMC 10 Problem 6

Problem 6 of 25EasierAlgebraNumber Theory

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

Recording only the units digits gives the sequence 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 for the first appearance of each digit, the digit 66 is the last of the ten digits to show up. Thus, the correct answer is C.

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Concepts: Fibonacci · units digit · pattern recognition

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