Difference between revisions of "2015 AMC 12B Problems/Problem 20"
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==Solution== | ==Solution== | ||
+ | Simply draw a table of values of <math>f(i,j)</math> for the first few values of <math>i</math>: | ||
+ | <cmath>\begin{array}{|c || c | c | c | c | c |} | ||
+ | \hline | ||
+ | i \text{\ \textbackslash\ } j & 0 & 1 & 2 & 3 & 4\\ \hline\hline | ||
+ | 0 & 1 & 2 & 3 & 4 & 0\\ \hline | ||
+ | 1 & 2 & 3 & 4 & 0 & 1\\ \hline | ||
+ | 2 & 3 & 0 & 2 & 4 & 1\\ \hline | ||
+ | 3 & 0 & 3 & 4 & 1 & 0\\ \hline | ||
+ | 4 & 3 & 1 & 3 & 1 & 3\\ \hline | ||
+ | 5 & 1 & 1 & 1 & 1 & 1\\ \hline | ||
+ | 6 & 1 & 1 & 1 & 1 & 1\\ \hline | ||
+ | \end{array}</cmath> | ||
+ | |||
+ | It becomes quickly obvious that for <math>i \ge 5</math>, <math>f(i,j) = 1</math> for all values <math>0 \le j \le 4</math>. | ||
+ | |||
+ | Thus, <math>f(2015,2) = \boxed{\textbf{(B)} \; 1}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2015|ab=B|num-a=21|num-b=19}} | {{AMC12 box|year=2015|ab=B|num-a=21|num-b=19}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 03:07, 7 March 2015
Problem
For every positive integer , let be the remainder obtained when is divided by 5. Define a function recursively as follows:
What is ?
Solution
Simply draw a table of values of for the first few values of :
It becomes quickly obvious that for , for all values .
Thus, .
See Also
2015 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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 |
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