Difference between revisions of "2010 IMO Problems/Problem 5"
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, where <math>X, Y = 2^{2^{.^{.^{.^2}}}}</math> where X has <math>19</math> 2's and Y has <math>X</math> 2's, and <math>Y</math> is clearly bigger then <math>T/4</math> | , where <math>X, Y = 2^{2^{.^{.^{.^2}}}}</math> where X has <math>19</math> 2's and Y has <math>X</math> 2's, and <math>Y</math> is clearly bigger then <math>T/4</math> | ||
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+ | == See also == | ||
+ | {{IMO box|year=2010|num-b=4|num-a=6}} | ||
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+ | [[Category:Olympiad Geometry Problems]] |
Revision as of 00:36, 24 October 2010
Problem
Each of the six boxes , , , , , initially contains one coin. The following operations are allowed
Type 1) Choose a non-empty box , , remove one coin from and add two coins to ;
Type 2) Choose a non-empty box , , remove one coin from and swap the contents (maybe empty) of the boxes and .
Determine if there exists a finite sequence of operations of the allowed types, such that the five boxes , , , , become empty, while box contains exactly coins.
Author: Unknown currently
Solution
Let the notation be the configuration in which the -th box has coin,
Let .
Our starting configuration is
We need these compound moves:
Compound move 1: , this is just repeated type 1 move on all coins.
Compound move 2: , apply type 1 move on 1 of the coin to get , then apply compound move 1 to the 2 coins to get , apply type 2 move and get , and repeat compound move 1 and type 2 move until is achieve.
Compound move 3: , apply compound move 2 to obtain and use type 2 move to get
Compound move 4: with 's. Apply compound move 3 times.
Compound move 5: with , use type 2 move times.
Let's follow this move:
Using Compound move 1, 4 and 5, We can obtain:
, where where X has 2's and Y has 2's, and is clearly bigger then
See also
2010 IMO (Problems) • Resources | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
All IMO Problems and Solutions |