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

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(Adding a proof to the solution.)
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<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>
  
==Solution==
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==Solution #1==
 
Simply draw a table of values of <math>f(i,j)</math> for the first few values of <math>i</math>:
 
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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5 & 1 & 1 & 1 & 1 & 1\\ \hline
 
5 & 1 & 1 & 1 & 1 & 1\\ \hline
 
\end{array}</cmath>
 
\end{array}</cmath>
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Now we claim that for <math>i \ge 5</math>, <math>f(i,j) = 1</math> for all values <math>0 \le j \le 4</math>.  We will prove this by induction on <math>i</math> and <math>j</math>.  The base cases for <math>i = 5</math>, have already been proven.
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For our inductive step, we must show that for all valid values of <math>j</math>, <math>f(i, j) = 1</math> if for all valid values of <math>j</math>, <math>f(i - 1, j) = 1</math>.
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We prove this itself by induction on <math>j</math>.  For the base case, <math>j=0</math>, <math>f(i, 0) = f(i-1, 1) = 1</math>.  For the inductive step, we need <math>f(i, j) = 1</math> if <math>f(i, j-1) = 1</math>.  Then, <math>f(i, j) = f(i-1, f(i, j-1)).</math>  <math>f(i, j-1) = 1</math> by our inductive hypothesis from our inner induction and <math>f(i-1, 1) = 1</math> from our outer inductive hypothesis.  Thus, <math>f(i, j) = 1</math>, completing the proof.
  
 
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>.
 
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>.

Revision as of 20:54, 11 February 2018

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 #1

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

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

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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