Join the USAMO Guide Discord Server. Regular updates are posted there!

Let me Join!
Dismiss

HMMT 2014 #0jjd

HMMT 2014Hard

From module Functional Equations II

Community difficultyNo ratings yetBe the first to rate this problem.
Understand the scale
Scale 0.5: Easiest math competition problems, often solvable even for early middle school students without experience. 1: Problems strictly for beginners, often at a moderate middle school level. 1.5: Problems for stronger beginner students, on the level of the middling problems in most middle school contests. 2: For motivated beginners, harder questions from the previous categories. 2.5: More advanced beginner problems, hardest questions from previous categories. 3: Early intermediate problems that require more creative thinking. 3.5: Tougher early intermediate problems that consistently stretch into higher-level creative thinking or conceptual knowledge. 4: Intermediate-level problems. 4.5: Upper intermediate problems approaching the upper bound for non-invitational sprint math competitions. 5: More difficult AIME problems or simple proof-based Olympiad-style problems. 6: High-level AIME-styled questions or introductory Olympiad-level questions. 7: Tougher Olympiad-level questions, may require more technical knowledge. 8: High-level Olympiad-level questions. 9: Expert Olympiad-level questions. 9.5: The hardest problems appearing on Olympiads which the strongest students could reasonably solve. 10: Historically notorious problems, generally unsuitable for even very hard competitions because they are exceedingly tedious, long, and difficult.
Functional EquationsRecursion, bijection

Problem

Problem:

Let f:N→Nf: \mathbb{N} \rightarrow \mathbb{N} be a function satisfying the following conditions: (a) f(1)=1f(1)=1. (b) f(a)≤f(b)f(a) \leq f(b) whenever aa and bb are positive integers with a≤ba \leq b. (c) f(2a)=f(a)+1f(2 a)=f(a)+1 for all positive integers aa.

How many possible values can the 2014-tuple (f(1),f(2),…,f(2014))(f(1), f(2), \ldots, f(2014)) take?

← Back to all problems