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Difference between revisions of "2015 AMC 12B Problems/Problem 20"

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==Solution==
 
==Solution==
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Simply draw a table of values of <math>f(i,j)</math> for the first few values of <math>i</math>:
  
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<cmath>\begin{array}{|c || c | c | c | c | c |}
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\hline
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i \text{\ \textbackslash\ } j & 0 & 1 & 2 & 3 & 4\\ \hline\hline
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0 & 1 & 2 & 3 & 4 & 0\\ \hline
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1 & 2 & 3 & 4 & 0 & 1\\ \hline
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2 & 3 & 0 & 2 & 4 & 1\\ \hline
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3 & 0 & 3 & 4 & 1 & 0\\ \hline
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4 & 3 & 1 & 3 & 1 & 3\\ \hline
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5 & 1 & 1 & 1 & 1 & 1\\ \hline
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6 & 1 & 1 & 1 & 1 & 1\\ \hline
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\end{array}</cmath>
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 +
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>.
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 +
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 02:07, 7 March 2015

Problem

For every positive integer $n$, let $\text{mod}_5 (n)$ be the remainder obtained when $n$ is divided by 5. Define a function $f: \{0,1,2,3,\dots\} \times \{0,1,2,3,4\} \to \{0,1,2,3,4\}$ recursively as follows:

\[f(i,j) = \begin{cases}\text{mod}_5 (j+1) & \text{ if } i = 0 \text{ and } 0 \le j \le 4 \text{,}\\ f(i-1,1) & \text{ if } i \ge 1 \text{ and } j = 0 \text{, and} \\ f(i-1, f(i,j-1)) & \text{ if } i \ge 1 \text{ and } 1 \le j \le 4. \end{cases}\]

What is $f(2015,2)$?

$\textbf{(A)}\; 0 \qquad\textbf{(B)}\; 1 \qquad\textbf{(C)}\; 2 \qquad\textbf{(D)}\; 3 \qquad\textbf{(E)}\; 4$

Solution

Simply draw a table of values of $f(i,j)$ for the first few values of $i$:

\[\begin{array}{|c || c | c | c | c | c |} \hline i \text{\ \textbackslash\ } j & 0 & 1 & 2 & 3 & 4\\ \hline\hline 0 & 1 & 2 & 3 & 4 & 0\\ \hline 1 & 2 & 3 & 4 & 0 & 1\\ \hline 2 & 3 & 0 & 2 & 4 & 1\\ \hline 3 & 0 & 3 & 4 & 1 & 0\\ \hline 4 & 3 & 1 & 3 & 1 & 3\\ \hline 5 & 1 & 1 & 1 & 1 & 1\\ \hline 6 & 1 & 1 & 1 & 1 & 1\\ \hline \end{array}\]

It becomes quickly obvious that for $i \ge 5$, $f(i,j) = 1$ for all values $0 \le j \le 4$.

Thus, $f(2015,2) = \boxed{\textbf{(B)} \; 1}$.

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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