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2015 AMC 12B Problem 20

Problem 20 of 25HarderAlgebraNumber Theory

For every positive integer n,n, let mod5(n)\operatorname{mod}_5(n) be the remainder obtained when nn is divided by 5.5. Define a function f:{0,1,2,3,}f : \{0, 1, 2, 3, \ldots\} ×{0,1,2,3,4}\times \{0, 1, 2, 3, 4\} {0,1,2,3,4}\to \{0, 1, 2, 3, 4\} recursively as follows: f(i,j)={mod5(j+1)if i=0 and 0j4,f(i1,1)if i1 and j=0, andf(i1,f(i,j1))if i1 and 1j4. \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)?f(2015, 2)?

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Solution

Applying the recursion from left to right in each row gives i\j01234012340123401230241303410431313511111 \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 11s, the recursion makes the next row entirely 11s as well. Hence, by induction, f(i,2)=1f(i,2)=1 for every i5.i\ge5. Since 20155,2015 \ge 5, we get f(2015,2)=1.f(2015, 2) = 1. Thus, the correct answer is B.

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Concepts: recursion · pattern recognition · modular arithmetic

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.