2000 AMC 12 Problems/Problem 4

Revision as of 18:16, 4 March 2007 by Azjps (talk | contribs) (box)

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

Note that any digits other than the units digit will not affect the answer. So to make computation quicker, we can just look at the Fibonacci sequence in $\bmod{10}$:

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

The last digit to appear in the units position of a number in the Fibonacci sequence is $6 \Longrightarrow \mathrm{C}$.

See also

2000 AMC 12 (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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
All AMC 12 Problems and Solutions