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#341696 A mean, mean minimization problem
Posted by Rainman on 25 February 2015  05:45 PM
#341584 Hello guys i have a puzzle/riddle i can not work out
Posted by Rainman on 16 February 2015  02:17 AM
 1
#337280 Inequalities for large integers
Posted by Rainman on 29 January 2014  12:49 PM
These problems use Knuth's uparrow notation (http://en.wikipedia....arrow_notation), with ^ being the uparrow.
1. Find the smallest value for n, such that 1000^^n > 12^^(n+1).
2. We define the sequence g, by assigning g(1) = 3^^^^3, and g(n+1) = 3^^^...^3, where the number of uparrows is g(n). A famously large number is Graham's number = g(64). Which is larger, Graham's number or 2^^^...^2, where the number of uparrows is Graham's number?
 1
#336956 Can you decript this?
Posted by Rainman on 09 December 2013  11:26 PM
I don't have friends
That sucks, I don't have many friends myself. Aaryan's point, though, is that asking us to solve it for you is cheating. But since you got the hint of using a 9x6 matrix, I'll at least do you the favor of putting the message in a 9x6 matrix.
WH3TEE
HERWDN
A2OO7T
TSO6AY
0QTHN9
IU4UDF
SAON8I
1RFDTV
TE5RWE
Hopefully you can see it now. Good luck with the competition!
 1
#334853 Chase on an endless road, continued
Posted by Rainman on 30 July 2013  04:46 PM
Caught again, your friend concedes that you are the supreme runner. Maybe it's not a coincidence that your profile pic is an ostrich
 1
#327168 piece of cake
Posted by Rainman on 10 January 2013  04:44 PM
 
   
    _{<cut along this line}
 
 

 1
#326785 Blood runs thick in Transylvania
Posted by Rainman on 19 December 2012  05:04 PM
 1
#326671 Amoeba evacuation puzzle
Posted by Rainman on 08 December 2012  01:18 AM
Yeah I think that is correct, here's why:
Spoiler for
Divide the board into diagonal lines like so:
Now let's define c{i} to be the number of amoebas at diagonal i, let's give weights to the diagonals w{i}=0.5^{i1}, so the weight of the first diagonal is 1 and the weight of every diagonal after that is half the weight of diagonal before it...
Now define board_sum to be the sum of the number of amoebas on each diagonal times the weight of that diagonal, boardsum_ = sum(c{i}w{i})...
At the beginning we have board_sum=1, whenever we make a move we split an amoaba at diagonal i into two amoebas at diagonal i+1, since the weight of diagonal i+1 is half that of diagonal i we can see that the board_sum has not changed...
So the board_sum is always constantly 1.
Now imagine the case where there is an infinite number of amoebas one in every cell, the board_sum in that case is:
(the convergence was calculated using the formula for the mean of a geometric distribution with parameter 0.5)
Now take away the amoebas at the cells we want to clear, their weight is:
1 + 0.5 + 0.5 + 0.25 + 0.25 + 0.25 + 0.125 + 0.125 = 3.
So the board_sum of the board when there is an infinite number of amoebas filling every cell except the ones that we need to clear is 1, which means that if the number of amoebas was finite then the board_sum would be less than 1 which is impossible...
That's the solution I had in mind for the problem, well done. I also enjoyed your javascript implementation of the problem.
 1
#326621 Amoeba evacuation puzzle
Posted by Rainman on 29 November 2012  11:01 PM
...........
oooooooo...
oooooooo...
xxoooooo...
xxxooooo...
xxxooooo...
 1
#325187 What is the missing term of the sequence?
Posted by Rainman on 08 September 2012  06:58 PM
0 (switch zeros and ones to get 1, add this at the end of the string to get the next number)
01 (switch zeros and ones to get 10, add this at the end to get...)
0110 (switch to get 1001, add to get...)
01101001 (switch to get 10010110, add to get...)
0110100110010110 (switch to get 1001011001101001, add to get...)
01101001100101101001011001101001 (this is the missing string, bring it back to decimal base to get 1,771,476,589)
 1
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