Question

For every positive integer $n,$ let $\operatorname{mod}_5(n)$ be the remainder obtained when $n$ is divided by $5.$ Define a function $f : \{0, 1, 2, 3, \ldots\}$ $\times \{0, 1, 2, 3, 4\}$ $\to \{0, 1, 2, 3, 4\}$ recursively as follows:

$\tiny f(i, j) = \begin{cases} \operatorname{mod}_5(j + 1) & \text{if } i = 0 \text{ and } 0 \le j \le 4, \\ f(i - 1, 1) & \text{if } i \ge 1 \text{ and } j = 0, \text{ and} \\ f(i - 1, f(i, j - 1)) & \text{if } i \ge 1 \text{ and } 1 \le j \le 4. \end{cases}$

What is $f(2015, 2)?$

$0$

$1$

$2$

$3$

$4$

Answer: B

Concepts:recursion pattern recognition modular arithmetic

Difficulty rating: 2100

Solution:

Applying the recursion from left to right in each row gives $\begin{array}{c|ccccc} i\backslash j&0&1&2&3&4\\ \hline 0&1&2&3&4&0\\ 1&2&3&4&0&1\\ 2&3&0&2&4&1\\ 3&0&3&4&1&0\\ 4&3&1&3&1&3\\ 5&1&1&1&1&1 \end{array}$ If a row consists entirely of $1$ s, the recursion makes the next row entirely $1$ s as well. Hence, by induction, $f(i,2)=1$ for every $i\ge5.$

Since $2015 \ge 5,$ we get $f(2015, 2) = 1.$

Thus, the correct answer is B.

2015 AMC 12B Problem 20

Answer: B

Difficulty rating: 2100

Solution:

Problem 20 in Other Years