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Difference between revisions of "2009 AMC 8 Problems/Problem 5"
 
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==Problem==
 
==Problem==
  
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A sequence of numbers starts with <math> 1</math>, <math> 2</math>, and <math> 3</math>. The fourth number of the sequence is the sum of the previous three numbers in the sequence: <math> 1\plus{}2\plus{}3\equal{}6</math>. In the same way, every number after the fourth is the sum of the previous three numbers. What is the eighth number in the sequence?  
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A sequence of numbers starts with <math> 1</math>, <math> 2</math>, and <math> 3</math>. The fourth number of the sequence is the sum of the previous three numbers in the sequence: <math> 1+2+3=6</math>. In the same way, every number after the fourth is the sum of the previous three numbers. What is the eighth number in the sequence?  
  
 
<math> \textbf{(A)}\  11  \qquad
 
<math> \textbf{(A)}\  11  \qquad
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<cmath>1,2,3,6,11,20,37,\boxed{\textbf{(D)}\ 68}</cmath>
 
<cmath>1,2,3,6,11,20,37,\boxed{\textbf{(D)}\ 68}</cmath>
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 +
==Video Solution ==
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https://youtu.be/USVVURBLaAc?t=221
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 +
==Video Solution 2==
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https://youtu.be/bsqdmV1x9aM
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 +
~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2009|num-b=4|num-a=6}}
 
{{AMC8 box|year=2009|num-b=4|num-a=6}}
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{{MAA Notice}}

Latest revision as of 07:22, 30 May 2023

Problem

A sequence of numbers starts with $1$, $2$, and $3$. The fourth number of the sequence is the sum of the previous three numbers in the sequence: $1+2+3=6$. In the same way, every number after the fourth is the sum of the previous three numbers. What is the eighth number in the sequence?

$\textbf{(A)}\ 11 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 37 \qquad \textbf{(D)}\ 68 \qquad \textbf{(E)}\ 99$

Solution

List them out, adding the three previous numbers to get the next number,

\[1,2,3,6,11,20,37,\boxed{\textbf{(D)}\ 68}\]

Video Solution

https://youtu.be/USVVURBLaAc?t=221

Video Solution 2

https://youtu.be/bsqdmV1x9aM

~savannahsolver

See Also

2009 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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