1981 IMO Problems/Problem 6: Difference between revisions

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Revision as of 20:45, 8 April 2024

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 $f(1, y+1) = f(0, f(1,y)) = f(1,y) + 1$ , 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) • Resources
Preceded by Problem 5	1 • 2 • 3 • 4 • 5 • 6	Followed by Last question
All IMO Problems and Solutions

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== Solution 2 ==https://beastacademy.com/

\begin{center} \begin{tabular}{||c c c c||}

\hline
Col1 & Col2 & Col2 & Col3 \\ [0.5ex] 
\hline\hline
1 & 6 & 87837 & 787 \\ 
\hline
2 & 7 & 78 & 5415 \\
\hline
3 & 545 & 778 & 7507 \\
\hline
4 & 545 & 18744 & 7560 \\
\hline
5 & 88 & 788 & 6344 \\ [1ex] 
\hline

\end{tabular} \end{center}

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1981 IMO Problems/Problem 6: Difference between revisions

Revision as of 20:45, 8 April 2024

Problem

Solution

@@ Line 22: / Line 22: @@
 {{IMO box|num-b=5|after=Last question|year=1981}}
+https://beastacademy.com/
+https://beastacademy.com/
 == Solution 2 ==https://beastacademy.com/