Difference between revisions of "2015 AMC 12B Problems/Problem 20"
Pi over two (talk | contribs) m (→Solution) |
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:
![\[f(i,j) = \begin{cases}\text{mod}_5 (j+1) & \text{ if } i = 0 \text{ and } 0 \le j \le 4 \text{,}\\ f(i-1,1) & \text{ if } i \ge 1 \text{ and } j = 0 \text{, and} \\ f(i-1, f(i,j-1)) & \text{ if } i \ge 1 \text{ and } 1 \le j \le 4. \end{cases}\]](http://latex.artofproblemsolving.com/1/e/f/1efa0d34cf7ae66e852b1687a71b18515cda058b.png)
What is 
?
Solution
Simply draw a table of values of 
for the first few values of 
:
![\[\begin{array}{|c || c | c | c | c | c |} \hline i \text{\ \textbackslash\ } j & 0 & 1 & 2 & 3 & 4\\ \hline\hline 0 & 1 & 2 & 3 & 4 & 0\\ \hline 1 & 2 & 3 & 4 & 0 & 1\\ \hline 2 & 3 & 0 & 2 & 4 & 1\\ \hline 3 & 0 & 3 & 4 & 1 & 0\\ \hline 4 & 3 & 1 & 3 & 1 & 3\\ \hline 5 & 1 & 1 & 1 & 1 & 1\\ \hline \end{array}\]](http://latex.artofproblemsolving.com/8/6/a/86a0e49932036b4f8ac2202543492e3c25dc10c6.png)
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. 
