Maths Olympiad Prep

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Algebra Difficulty 5.0 AIME, harder Prove it United States

Problem:

The real function ff has the property that, whenever aa, bb, nn 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)?

Solution

Solution:

We know f(a)=n2f(2na)f(a) = n^{2} - f\left(2^{n} - a\right) for any aa, nn with 2n>a2^{n} > a; repeated application gives
f(2002)=112f(46)=112(62f(18))=112(62(52f(14)))=112(62(52(42f(2)))). \begin{gathered} 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) . \end{gathered}
But f(2)=22f(2)f(2) = 2^{2} - f(2), giving f(2)=2f(2) = 2, so the above simplifies to 112(62(52(422)))=9611^{2} - \left(6^{2} - \left(5^{2} - \left(4^{2} - 2\right)\right)\right) = 96.

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