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

## 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 take some time to draw a table of values of $f(i,j)$ for the first few values of $i$:

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

Now we claim that for $i \ge 5$, $f(i,j) = 1$ for all values $0 \le j \le 4$. We will prove this by induction on $i$ and $j$. The base cases for $i = 5$, have already been proven.

For our inductive step, we must show that for all valid values of $j$, $f(i, j) = 1$ if for all valid values of $j$, $f(i - 1, j) = 1$.

We prove this itself by induction on $j$. For the base case, $j=0$, $f(i, 0) = f(i-1, 1) = 1$. For the inductive step, we need $f(i, j) = 1$ if $f(i, j-1) = 1$. Then, $f(i, j) = f(i-1, f(i, j-1)).$ $f(i, j-1) = 1$ by our inductive hypothesis from our inner induction and $f(i-1, 1) = 1$ from our outer inductive hypothesis. Thus, $f(i, j) = 1$, completing the proof.

It is now clear that for $i \ge 5$, $f(i,j) = 1$ for all values $0 \le j \le 4$.

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