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

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==Solution==
 
==Solution==
Simply take 30 minutes to draw a table of values of <math>f(i,j)</math> for the first few values of <math>i</math>:
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Simply take some time to draw a table of values of <math>f(i,j)</math> for the first few values of <math>i</math>:
  
 
<cmath>\begin{array}{|c || c | c | c | c | c |}
 
<cmath>\begin{array}{|c || c | c | c | c | c |}

Revision as of 17:08, 6 December 2019

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

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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