Difference between revisions of "1981 IMO Problems/Problem 6"
(→Solution 2) |
(→Solution) |
||
(6 intermediate revisions by 2 users not shown) | |||
Line 15: | Line 15: | ||
We observe that <math>f(1,0) = f(0,1) = 2 </math> and that <math>f(1, y+1) = f(0, f(1,y)) = f(1,y) + 1</math>, so by induction, <math>f(1,y) = y+2 </math>. Similarly, <math>f(2,0) = f(1,1) = 3</math> and <math>f(2, y+1) = f(2,y) + 2</math>, yielding <math>f(2,y) = 2y + 3</math>. | We observe that <math>f(1,0) = f(0,1) = 2 </math> and that <math>f(1, y+1) = f(0, f(1,y)) = f(1,y) + 1</math>, so by induction, <math>f(1,y) = y+2 </math>. Similarly, <math>f(2,0) = f(1,1) = 3</math> and <math>f(2, y+1) = f(2,y) + 2</math>, yielding <math>f(2,y) = 2y + 3</math>. | ||
− | We continue with <math>f(3,0) + 3 = 8 </math>; <math>f(3, y+1) + 3 = 2(f(3,y) + 3)</math>; <math>f(3,y) + 3 = 2^{y+3}</math>; and <math>f(4,0) + 3 = 2^{2^2}</math>; <math>f(4,y) + 3 = 2^{f(4,y) + 3}</math>. | + | We continue with <math>f(3,0) + 3 = 8 </math>; <math>f(3, y+1) + 3 = 2(f(3,y) + 3)</math>; <math>f(3,y) + 3 = 2^{y+3}</math>; and <math>f(4,0) + 3 = 2^{2^2}</math>; <math>f(4,y) + 3 = 2^{f(4,y-1) + 3}</math>. |
It follows that <math>f(4,1981) = 2^{2\cdot ^{ . \cdot 2}} - 3 </math> when there are 1984 2s, Q.E.D. | It follows that <math>f(4,1981) = 2^{2\cdot ^{ . \cdot 2}} - 3 </math> when there are 1984 2s, Q.E.D. | ||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
== Solution 2 == | == Solution 2 == | ||
Line 30: | Line 24: | ||
This pattern can be proved using induction. After proving, we continue to setting a list when <math>x=2</math>. <cmath>f(1,0)=2, f(1,1)=3, f(1,2)=4, f(1,3)=5, f(1,4)=6</cmath> This pattern can also be proved using induction. The pattern seems d up of a common difference of 1. Moving on to <math>x=3</math> <cmath>f(3,0)=5, f(3,1)=13, f(3,2)=29, f(3,3)=61, f(3,4)=125</cmath> All of the numbers are being expressed in the form of <math>3^a -3 | This pattern can be proved using induction. After proving, we continue to setting a list when <math>x=2</math>. <cmath>f(1,0)=2, f(1,1)=3, f(1,2)=4, f(1,3)=5, f(1,4)=6</cmath> This pattern can also be proved using induction. The pattern seems d up of a common difference of 1. Moving on to <math>x=3</math> <cmath>f(3,0)=5, f(3,1)=13, f(3,2)=29, f(3,3)=61, f(3,4)=125</cmath> All of the numbers are being expressed in the form of <math>3^a -3 | ||
− | </math> where <math>a=y+3</math>. Lastly where x=4 we have <cmath>f(4,0)=13, f(4,1)=65533, f(4,2)=^5 2, f(3,3)=^6 2, f(4,4)=^7 2</cmath> where each term can be represented as <math>^a 2</math> when <math>a=y+2</math>. In <math>^a 2</math> represents tetration or 2 to the power 2 to the power 2 to the power 2 ... where <math>a</math> amount of 2s. So therefore the answer is < | + | </math> where <math>a=y+3</math>. Lastly where x=4 we have <cmath>f(4,0)=13, f(4,1)=65533, f(4,2)=^5 2, f(3,3)=^6 2, f(4,4)=^7 2</cmath> where each term can be represented as <math>^a 2</math> when <math>a=y+2</math>. In <math>^a 2</math> represents tetration or 2 to the power 2 to the power 2 to the power 2 ... where <math>a</math> amount of 2s. So therefore the answer is <math>f(4,1981) = 2^{2\cdot ^{ . \cdot 2}} - 3 </math> with 1984 2s, 2 tetration 1983. |
+ | -Multpi12 | ||
+ | |||
+ | {{alternate solutions}} | ||
+ | |||
+ | {{IMO box|num-b=5|after=Last question|year=1981}} |
Latest revision as of 01:57, 12 December 2024
Problem
The function satisfies
(1)
(2)
(3)
for all non-negative integers . Determine
.
Solution
We observe that and that
, so by induction,
. Similarly,
and
, yielding
.
We continue with ;
;
; and
;
.
It follows that when there are 1984 2s, Q.E.D.
Solution 2
We can start by creating a list consisting of certain x an y values and their outputs.
This pattern can be proved using induction. After proving, we continue to setting a list when .
This pattern can also be proved using induction. The pattern seems d up of a common difference of 1. Moving on to
All of the numbers are being expressed in the form of
where
. Lastly where x=4 we have
where each term can be represented as
when
. In
represents tetration or 2 to the power 2 to the power 2 to the power 2 ... where
amount of 2s. So therefore the answer is
with 1984 2s, 2 tetration 1983.
-Multpi12
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 |