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

 
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{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #4]] and [[2000 AMC 10 Problems|2000 AMC 10 #6]]}}
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== Problem ==
 
== Problem ==
The Fibonacci sequence <math>1,1,2,3,5,8,13,21,\ldots </math> 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?
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The [[Fibonacci sequence]] <math>1,1,2,3,5,8,13,21,\ldots </math> starts with two 1s, and each term afterwards is the sum of its two predecessors. Which one of the ten [[digit]]s is the last to appear in the units position of a number in the Fibonacci sequence?
  
<math> \mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 4 } \qquad \mathrm{(C) \ 6 } \qquad \mathrm{(D) \ 7 } \qquad \mathrm{(E) \ 9 </math>
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<math> \textbf{(A)} \ 0 \qquad \textbf{(B)} \ 4 \qquad \textbf{(C)} \ 6 \qquad \textbf{(D)} \ 7 \qquad \textbf{(E)} \ 9   </math>
  
 
== Solution ==
 
== Solution ==
Looking at the Fibonacci Sequence in <math>\bmod{10}</math>:
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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 <math>\bmod{10}</math>:
  
 
<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 \boxed{\mathrm{C}} </math>.
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==Video Solution by Daily Dose of Math==
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https://youtu.be/4dWDOG31tQM?si=Ne3F969T10goxIjC
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~Thesmartgreekmathdude
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== 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/Problem 3|Previous Problem]]
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{{AMC10 box|year=2000|num-b=5|num-a=7}}
*[[2000 AMC 12/Problem 5|Next problem]]
 
  
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]
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{{MAA Notice}}

Latest revision as of 23:39, 14 July 2024

The following problem is from both the 2000 AMC 12 #4 and 2000 AMC 10 #6, so both problems redirect to this page.

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?

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

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 $6 \Longrightarrow \boxed{\mathrm{C}}$.

Video Solution by Daily Dose of Math

https://youtu.be/4dWDOG31tQM?si=Ne3F969T10goxIjC

~Thesmartgreekmathdude


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
2000 AMC 10 (ProblemsAnswer KeyResources)
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
All AMC 10 Problems and Solutions

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