Difference between revisions of "2009 AMC 8 Problems/Problem 5"
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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 == | ||
| + | https://youtu.be/USVVURBLaAc?t=221 | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2009|num-b=4|num-a=6}} | {{AMC8 box|year=2009|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 23:56, 28 January 2021
Contents
Problem
A sequence of numbers starts with 
, 
, and 
. The fourth number of the sequence is the sum of the previous three numbers in the sequence: 
. 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?
Solution
List them out, adding the three previous numbers to get the next number,
![\[1,2,3,6,11,20,37,\boxed{\textbf{(D)}\ 68}\]](http://latex.artofproblemsolving.com/3/f/b/3fba5a20cd1e94a9ffbc9900192c13205d0fcc86.png)
Video Solution
https://youtu.be/USVVURBLaAc?t=221
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. 
