Difference between revisions of "1981 IMO Problems/Problem 6"
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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 == | |
− | + | 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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− | == | + | 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>f(4,1981) = 2^{2\cdot ^{ . \cdot 2}} - 3 </math> with 1984 2s, 2 tetration 1983. | ||
+ | -Multpi12 | ||
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Latest revision as of 18:15, 12 April 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 |