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View Poll Results: The last three digits for the number 9^387420489
089 2 9.09%
289 16 72.73%
489 2 9.09%
689 1 4.55%
889 1 4.55%
Voters: 22. You may not vote on this poll

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Old 2012-05-23, 04:48   #34
axn
 
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Quote:
Originally Posted by LaurV View Post
So much talking for nothing...
So much typing for nothing...
Code:
? #
   timer = 1 (on)
? Mod(9,1000)^9^9
time = 0 ms.
%1 = Mod(289, 1000)
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Old 2012-05-23, 04:56   #35
retina
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Quote:
Originally Posted by LaurV View Post
So much talking for nothing...
Code:
(11:39:36) gp > lift((x=Mod(9,1000))^lift(Mod(9,znorder(x))^9))
%8 = 289
(11:39:36) gp > ##
  ***   last result computed in 0 ms.
(11:39:37) gp >
So with your method can it also say what are the last three digits of 99[sup]9[sup]9[sup]9[sup]9[/sup][/sup][/sup][/sup]?

[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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Old 2012-05-23, 04:58   #36
LaurV
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@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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Old 2012-05-23, 05:02   #37
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Quote:
Originally Posted by retina View Post
So with your method can it also say what are the last three digits of 99[sup]9[sup]9[sup]9[sup]9[/sup][/sup][/sup][/sup]?

[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]

Nice text obfuscation technique, by the way
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Old 2012-05-23, 05:31   #38
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Quote:
Originally Posted by retina View Post
... What if the program is written wrongly? How would you know?
I totally agree. At Project Euler, the problems that are solved with 10% googling research, 89% pencil-on-paper and only 1% computation are the most dear in my heart. #385 is great in that regard, and #148 is another example.
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Old 2012-05-24, 14:39   #39
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Quote:
Originally Posted by LaurV View Post
@retina: yes, with a bit of... more typing (?)
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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Old 2012-05-24, 14:42   #40
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Quote:
Originally Posted by Batalov View Post
I totally agree. At Project Euler, the problems that are solved with 10% googling research, 89% pencil-on-paper and only 1% computation are the most dear in my heart. #385 is great in that regard, and #148 is another example.
Thanks for the tip. I'll try to find out what these are when I get some time.
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Old 2012-05-24, 16:19   #41
Raman
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Quote:
Originally Posted by retina View Post
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.
You can want to try out the Project Euler problems #188, #282

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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Old 2012-05-24, 19:25   #42
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Quote:
Originally Posted by retina View Post
Thanks for the tip. I'll try to find out what these are when I get some time.
This discussion has become silly. Could an admin move it to misc. math?
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Old 2012-05-25, 02:59   #43
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Quote:
Originally Posted by R.D. Silverman View Post
This discussion has become silly. Could an admin move it to misc. math?
I vote to make RDS a mod in _this_ forum.
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