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 1

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

Solution 2

We are given that $f(0,n) \equiv n+1\pmod{5} .$ Then, $f(1,n) = f(0, f(1, n-1)) \equiv f(1,n-1) + 1\pmod{5}$ . Thus $f(1,n) \equiv n+f(1,0)\pmod{5}$ . Since $f(1,0)=f(0,1)=2$ , we get $f(1,n) \equiv n+2\pmod{5} .$ Then, $f(2,n) = f(1, f(2, n-1)) \equiv f(2,n-1) + 2\pmod{5}$ . Thus $f(2,n) \equiv 2n+f(2,0)\pmod{5}$ . Since $f(2,0)=f(1,1)=3$ , we get $f(2,n) \equiv 2n+3\pmod{5} .$ Now $f(3,n) = f(2, f(3, n-1)) \equiv 2f(3,n-1) + 3\pmod{5}$ . Thus $\begin{align*} f(3,n) &\equiv 2f(3,n-1) + 3 &\pmod{5} \\ 2f(3,n-1) &\equiv 2^2f(3,n-2) + 3\cdot 2 &\pmod{5} \\ \vdots \qquad &\quad\vdots \quad\qquad \vdots \qquad\qquad \vdots &\\ 2^{n-1}f(3,1) &\equiv 2^nf(3,0) + 3\cdot 2^{n-1} &\pmod{5} \end{align*}$ Adding them all up we get $f(3,n) \equiv 3(2^n-1)\pmod{5} .$ This means that $f(3,0)=0$ , $f(3,1)=3$ , $f(3,2)=4$ , $f(3,3)=1$ , and $f(3,4)=0$ . Thus $f(3,n)$ never takes the value 2.

Since $f(4,n)=f(3,f(4,n-1))$ , this implies that $f(4,n) \neq 2$ for any $n$ . By induction, $f(3,n) \neq f(3,2) = 4$ for any $n$ . It follows that $f(3,n) \neq f(3,4) = 0$ for any $n$ . Thus $f(4,n)$ only takes values in $\{1,3\}$ . In fact, it alternates between 1 and 3: $f(4,0)=f(3,1)=3$ , then $f(4,1)=f(3,f(4,0))=f(3,3)=1$ , then $f(4,2)=f(3,f(4,1))=f(3,1)=3$ , and so on.

Repeating the argument above, we see that $f(5,n) = f(4, f(5,n-1))$ can only take values in $\{1,3\}$ . However, $f(5,n-1)\neq 0$ for any $n$ implies that $f(5,n)\neq f(4,0)=3$ for any $n$ . Thus $f(5,n)=1$ for all $n$ . We can easily verify this: $f(5,0)=f(4,1)=1$ , then $f(5,1)=f(4,f(5,0))=f(4,1)=1$ , then $f(5,2)=f(4,f(5,1))=f(4,1)=1$ , and so on.

Then $f(6,0)=f(5,1)=1$ . Moreover, $f(6,n) = f(5,f(6,n-1)) = 1$ for all $n$ . Continuing in this manner we see that $f(m,n)=1$ for all $m\ge 5$ .

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

2015 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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2015 AMC 12B Problems/Problem 20

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Problem

Solution 1

Solution 2

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