Difference between revisions of "2015 AMC 12B Problems/Problem 20"
Pi over two (talk | contribs) m (→Solution) |
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\end{array}</cmath> | \end{array}</cmath> | ||
− | It | + | It is now clear 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>. | Thus, <math>f(2015,2) = \boxed{\textbf{(B)} \; 1}</math>. |
Revision as of 08:17, 8 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 is now clear 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.