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1981 IMO Problems/Problem 6

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Problem

The function f(x,y) satisfies

(1) f(0,y)=y+1,

(2) f(x+1,0)=f(x,1),

(3) f(x+1,y+1)=f(x,f(x+1,y)),

for all non-negative integers x,y. Determine f(4,1981).

Solution

We observe that f(1,0) = f(0,1) = 2 and that f(1, y+1) = f(1, f(1,y)) = f(1,y) + 1, so by induction, f(1,y) = y+2. Similarly, f(2,0) = f(1,1) = 3 and f(2, y+1) = f(2,y) + 2, yielding f(2,y) = 2y + 3.

We continue with f(3,0) + 3 = 8; f(3, y+1) + 3 = 2(f(3,y) + 3); f(3,y) + 3 = 2^{y+3}; and f(4,0) + 3 = 2^{2^2}; f(4,y) + 3 = 2^{f(4,y) + 3}.

It follows that f(4,1981) = 2^{2\cdot ^{ . \cdot 2}} - 3 when there are 1984 2s, Q.E.D.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

1981 IMO (Problems)
Preceded by
Problem 5
1 2 3 4 5 6 Followed by
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Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's Intermediate Counting & Probability by David Patrick.
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