Difference between revisions of "2000 AMC 12 Problems/Problem 4"

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<math>1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,7,7,4,1,5,6,....</math>
 
<math>1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,7,7,4,1,5,6,....</math>
  
The last digit to appear in the units position of a number in the Fibonacci sequence is <math> 6 \Rightarrow C </math>.
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The last digit to appear in the units position of a number in the Fibonacci sequence is <math> 6 \Longrightarrow \mathrm{C} </math>.
  
 
== See also ==
 
== See also ==
* [[2000 AMC 12 Problems]]
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{{AMC12 box|year=2000|num-b=3|num-a=5}}
*[[2000 AMC 12 Problems/Problem 3|Previous Problem]]
 
*[[2000 AMC 12 Problems/Problem 5|Next problem]]
 
  
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]

Revision as of 18:16, 4 March 2007

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