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2000 AMC 10 Problems/Problem 6

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Problem

The Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, 21, \ldots starts with two 1s, 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?

\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 4 \qquad\mathrm{(C)}\ 6 \qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 9

Solution

The pattern of the units digits are

1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,7,7,4,1,5,6

In order of appearance:

1,2,3,5,8,4,9,7,0,6.

6 is the last.

\boxed{\text{C}}

See Also

2000 AMC 10 (ProblemsResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!
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