Maths Olympiad Prep

Library / HMMT February /149 of 1302

HMMT February

Algebra Difficulty 4.8 AIME Find the answer

The real function ff has the property that, whenever a,b,na, b, n are positive integers such that a+b=2na+b=2^{n}, the equation f(a)+f(b)=n2f(a)+f(b)=n^{2} holds. What is f(2002)f(2002)?

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

We know f(a)=n2−f(2n−a)f(a)=n^{2}-f\left(2^{n}-a\right) for any aa, nn with 2n>a2^{n}>a; repeated application gives f(2002)=112−f(46)=112−(62−f(18))=112−(62−(52−f(14)))=112−(62−(52−(42−f(2))))f(2002)=11^{2}-f(46)=11^{2}-\left(6^{2}-f(18)\right)=11^{2}-\left(6^{2}-\left(5^{2}-f(14)\right)\right) =11^{2}-\left(6^{2}-\left(5^{2}-\left(4^{2}-f(2)\right)\right)\right) But f(2)=22−f(2)f(2)=2^{2}-f(2), giving f(2)=2f(2)=2, so the above simplifies to 112−(62−(52−(42−11^{2}-\left(6^{2}-\left(5^{2}-\left(4^{2}-\right.\right.\right. 2)) =96=96.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.