1984 AIME Problem 7

Attempt Problem 7 of the 1984 AIME below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1984 AIME solutions, or check the answer key.

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7.

The function ff is defined on the set of integers and satisfies f(n)=n3f(n)=n-3 if n1000,n\geq1000, and f(n)=f(f(n+5)) f(n)=f(f(n+5)) if n<1000.n<1000. Find f(84).f(84).

Answer: 997
Concepts:recursioninductionparity
Difficulty rating: 2440
Small Hint:

Start by evaluating f(995),f(995), f(996),f(996), ,\ldots, f(999)f(999)

Big Hint:

Use downward induction to find a parity pattern for every integer below 10001000

Solution:

Directly from the definition, f(999)=f(f(1004))=f(1001)=998,f(998)=f(f(1003))=f(1000)=997. \begin{aligned} f(999)&=f(f(1004))\\ &=f(1001)=998,\\ f(998)&=f(f(1003))\\ &=f(1000)=997. \end{aligned} Continuing gives f(997)=998,f(997)=998, f(996)=997,f(996)=997, and f(995)=998.f(995)=998.

We now use downward induction. If n<995n<995 is even, then n+5n+5 is odd, so the established pattern above nn gives f(n)=f(f(n+5)).f(n)=f(f(n+5)). Thus f(n)=f(998)=997.f(n)=f(998)=997. If nn is odd, the same argument gives f(n)=f(997)=998.f(n)=f(997)=998. Thus every even n<1000n<1000 has value 997.997. Since 8484 is even, f(84)=997.f(84)=997.

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