Difference between revisions of "2009 AMC 8 Problems/Problem 5"
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List them out, adding the three previous numbers to get the next number, | List them out, adding the three previous numbers to get the next number, | ||
| − | 1,2,3,6,11,20,37,68 | + | <cmath>1,2,3,6,11,20,37,\boxed{\textbf{(D)}\ 68}</cmath> |
==Video Solution == | ==Video Solution == | ||
https://youtu.be/USVVURBLaAc?t=221 | https://youtu.be/USVVURBLaAc?t=221 | ||
| + | |||
| + | ==Video Solution 2== | ||
| + | https://youtu.be/bsqdmV1x9aM | ||
| + | |||
| + | ~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}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 07:22, 30 May 2023
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
Video Solution 2
~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. 
