Difference between revisions of "2017 AMC 10A Problems/Problem 13"
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==Solution== | ==Solution== | ||
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| + | A pattern starts to emerge as the function is continued. The repeating pattern is <math>0,1,1,2,0,2,2,1\ldots</math> The problem asks for the sum of eight consecutive terms in the sequence. Because there are eight numbers in the repeating sequence, we just need to find the sum of the numbers in the sequence, which is <math>\boxed{\qquad\textbf{(D)}\ 9}</math> | ||
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==See Also== | ==See Also== | ||
{{AMC10 box|year=2017|ab=A|num-b=12|num-a=14}} | {{AMC10 box|year=2017|ab=A|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:59, 8 February 2017
Problem
Define a sequence recursively by 
and 
the remainder when 
is divided by 
for all 
Thus the sequence starts 
What is 
Solution
A pattern starts to emerge as the function is continued. The repeating pattern is 
The problem asks for the sum of eight consecutive terms in the sequence. Because there are eight numbers in the repeating sequence, we just need to find the sum of the numbers in the sequence, which is 
See Also
| 2017 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
