Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
(Adding a proof to the solution.)
Line 11: Line 11:
 
<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==
+
==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>:
  
Line 24: Line 24:
 
5 & 1 & 1 & 1 & 1 & 1\\ \hline
 
5 & 1 & 1 & 1 & 1 & 1\\ \hline
 
\end{array}</cmath>
 
\end{array}</cmath>
 +
 +
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.
 +
 +
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>.
 +
 +
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 19:54, 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 (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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12B_Problems/Problem_20&oldid=91075"