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#34 |
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Jun 2003
117328 Posts |
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#35 | |
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Undefined
"The unspeakable one"
Jun 2006
My evil lair
24·389 Posts |
Quote:
[sup][sup][sup][sup][sup][sup][sup]I think my method is clearer than using a program. What if the program is written wrongly? How would you know?[/sup][/sup][/sup][/sup][/sup][/sup][/sup] |
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#36 |
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Romulan Interpreter
Jun 2011
Thailand
32·29·37 Posts |
@axn:
![]() ![]() ![]() @retina: yes, with a bit of... more typing (?)
Last fiddled with by LaurV on 2012-05-23 at 05:01 |
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#37 | |
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"Ben"
Feb 2007
3,517 Posts |
Quote:
Nice text obfuscation technique, by the way
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#38 |
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"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
251616 Posts |
I totally agree. At Project Euler, the problems that are solved with 10%
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#39 |
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Undefined
"The unspeakable one"
Jun 2006
My evil lair
24×389 Posts |
I am curious to know if your program can easily compute something like:
98[sup]7[sup]6[sup]5[sup]4[sup]3[sup]2[sup]1[sup]0[/sup][/sup][/sup][/sup][/sup][/sup][/sup][/sup] mod 100000000 ;(it is just a power tower going from 9 to 0) I already know the answer is 79806721 but I just want to see how the program can handle such a computation. |
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#40 |
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Undefined
"The unspeakable one"
Jun 2006
My evil lair
24·389 Posts |
Thanks for the tip. I'll try to find out what these are when I get some time.
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#41 | |
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Noodles
"Mr. Tuch"
Dec 2007
Chennai, India
3×419 Posts |
Quote:
But that the last few digits for the hyperexponentiation/tetration operation converges off very rapidly For example, try out with Ackermann(5,5), Ackermann(6,6), Ackermann(7,7) the last dozens of digits are being essentially going to be the same a↑↑b mod N { if b=1 return a if N=1 return 0 else return aa↑↑(b-1) mod φ(N) mod N } The project euler problem #176 was being essentially a classical pencil-paper exercise How many problems should I have solved at the project euler before I can start contributing into the new problems myself? I have many classical problems to contribute, for this example case 1. How many powers of 2 below 210[sup]20[/sup] begin with the digit 9? (For the powers of 3, this becomes easier, but) 2. If we write the powers of 5 in line, for example 1 5 25 125 625 3125 15625 78125 390625... the position at which twelve consecutive same (identical) digit (7) appears out 3. Number of positions from within the Peg Solitaire board, different boards, different starting positions that can be reached from the initial starting position, that can be reduced to the one peg ending position I have solved out 232 off counting furthermore |
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#42 |
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Nov 2003
22×5×373 Posts |
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#43 |
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Jun 2003
10011110110102 Posts |
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