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

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(Problem)
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==Problem==
 
==Problem==
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For every positive integer <math>n</math>, let <math>\text{mod}_5 (n)</math> be the remainder obtained when <math>n</math> is divided by 5. Define a function <math>f: \{0,1,2,3,\dots\} \times \{0,1,2,3,4\} \to \{0,1,2,3,4\}</math> recursively as follows:
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<cmath>f(i,j) = \begin{cases}\text{mod}_5 (j+1) & \text{ if } i = 0 \text{ and } 0 \le j \le 4 \text{,}\\
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f(i-1,1) & \text{ if } i \ge 1 \text{ and } j = 0 \text{, and} \\
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f(i-1, f(i,j-1)) & \text{ if } i \ge 1 \text{ and } 1 \le j \le 4.
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\end{cases}</cmath> 
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What is <math>f(2015,2)</math>?
  
 
<math>\textbf{(A)}\; 0 \qquad\textbf{(B)}\; 1 \qquad\textbf{(C)}\; 2 \qquad\textbf{(D)}\; 3 \qquad\textbf{(E)}\; 4</math>
 
<math>\textbf{(A)}\; 0 \qquad\textbf{(B)}\; 1 \qquad\textbf{(C)}\; 2 \qquad\textbf{(D)}\; 3 \qquad\textbf{(E)}\; 4</math>

Revision as of 01:20, 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

See Also

2015 AMC 12B (ProblemsAnswer KeyResources)
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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