1981 IMO Problems/Problem 6: Difference between revisions

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Revision as of 20:57, 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
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Solution 2

We can start by creating a list consisting of certain x an y values and their outputs. $f(0,0)=1, f(0,1)=2, f(0,2)=3, f(0,3)=4, f(0,4)=5$

Revision as of 20:56, 8 April 2024 view source Multpi12 (talk \| contribs) editor 290 edits →Solution 2 ← Older edit		Revision as of 20:57, 8 April 2024 view source Multpi12 (talk \| contribs) editor 290 edits →Solution 2 Newer edit →
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	== Solution 2 ==		== Solution 2 ==

	We can start by creating a list consisting of certain x an y values and their outputs. <cmath>f(0,0)=1, f(0,1)=2, f(0,2)=3, f(0,3)=4, f(0,4)=5</cmath>		We can start by creating a list consisting of certain x an y values and their outputs. <cmath>f(0,0)=1, f(0,1)=2, f(0,2)=3, f(0,3)=4, f(0,4)=5</cmath>

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1981 IMO Problems/Problem 6: Difference between revisions

Revision as of 20:57, 8 April 2024

Problem

Solution

Solution 2