Competition problem — Algebra — Maths Olympiad Prep

Algebra Difficulty 4.7 AIME Prove it Brazil

Let $Z$ be the set of integers. $f: Z \rightarrow Z$ is defined by $f(n) = n - 10$ for $n > 100$ and $f(n) = f(f(n+11))$ for $n = 100$ . Find the set of possible values of $f$ .

Solution

We show that $f(n) = 91$ for $n \le 100$ . For $n = 100, 99, \dots, 90$ , we have $f(n) = f(f(n+11)) = f(n+11-10) = f(n+1)$ . But $f(101) = 91$ , so $f(n) = 91$ for $n = 100, 99, \dots, 90$ . Now for $n = 89, 88, \dots, 1, 0, -1, \dots$ we use induction. We have $f(n) = f(f(n+11)) = f(91)$ which, by induction, is 91. So the range of $f$ is all the integers greater than or equal to 91.

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