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Old 2014-10-02, 12:47   #1
Xyzzy
 
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Default October 2014

http://domino.research.ibm.com/Comm/...tober2014.html
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Old 2014-10-02, 21:46   #2
EdH
 
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"Ed Hall"
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Quote:
Originally Posted by Xyzzy View Post
I guess yafu '2^(3^(4^(5^(5^(6^(7^(8^9))))))' probably wouldn't be considered an "elegant solution."
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Old 2014-10-02, 23:03   #3
Batalov
 
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Phi(4,2^7658614+1)/2

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But it isn't!

It is as if for the last 10 digits of the penultimate factor of M991 you would submit, yafu 'factor(2^991-1)'. But what [B]are[/B] those last ten digits, eh?
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Old 2014-10-02, 23:30   #4
Xyzzy
 
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Here is an easy way to get the answer: http://www.wolframalpha.com/input/?i...29%29%29%29%29

But, it didn't show us how to get the answer. We spent a lot of time working on this today without much progress.

http://mathhelpforum.com/number-theo...-exponent.html
http://mymathforum.com/number-theory...st-digits.html
http://math.stackexchange.com/questi...-digits-of-999
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Old 2014-10-03, 00:18   #5
Mini-Geek
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"Tim Sorbera"
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I can't say I fully grok the math behind it, but I found a solution by calculating the following in PARI/GP:
Code:
Mod(8,10^10)^9
Mod(7,10^10)^lift(%)
Mod(6,10^10)^lift(%)
Mod(5,10^10)^lift(%)
Mod(4,10^10)^lift(%)
Mod(3,10^10)^lift(%)
Mod(2,10^10)^lift(%)
Results in Mod(8170340352, 10000000000), or ...8170340352

Or as C#
var n = BigInteger.Pow(8, 9);
var mod = BigInteger.Pow(10, 10);
for (var b = 7; b >= 2; b--)
    n = BigInteger.ModPow(b, n, mod);
Console.WriteLine(n.ToString("D10"));

And just for good measure, Python:
n = 8**9
for b in range(7, 1, -1):
    n = pow(b, n, 10**10)
print("{:010d}".format(n))
It matches the result from Wolfram Alpha, and I submitted my solution to the site...I'm not sure if only sufficiently unique/elegant solutions count, I'm sure this approach has been done before.

Last fiddled with by Mini-Geek on 2014-10-03 at 01:13
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Old 2014-10-03, 02:55   #6
Batalov
 
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Phi(4,2^7658614+1)/2

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It happens to be the correct answer, but with generally incorrect solution.
The period of power function (mod n) is not n but eulerphi(n).
All but the last step needs to be done mod eulerphi(10^10), not 10^10.
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Old 2014-10-03, 12:53   #7
CRGreathouse
 
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Don't the moduli change at each step?
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Old 2014-10-03, 13:23   #8
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I tried to follow the easiest route I could in my head but it didn't get me far enough even after seeing the answers here:

2 to odd exponent : ends in 2 or 8
exponent is 3 to a power that's 0 mod 4: the exponent on the 2 ends in 1; this along with 0 mod 3 leads to 21,51,81 etc being a list of possible exponents before other powers are taken into effect


and then I was going to go from there but I'm too slow at trying things like this.

Last fiddled with by science_man_88 on 2014-10-03 at 13:23
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Old 2014-11-02, 19:03   #9
Xyzzy
 
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http://domino.research.ibm.com/Comm/...tober2014.html
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