mersenneforum.org October 2014
 Register FAQ Search Today's Posts Mark Forums Read

 2014-10-02, 12:47 #1 Xyzzy     "Mike" Aug 2002 2×13×313 Posts October 2014
2014-10-02, 21:46   #2
EdH

"Ed Hall"
Dec 2009

72018 Posts

Quote:
 Originally Posted by Xyzzy http://domino.research.ibm.com/Comm/...tober2014.html
I guess yafu '2^(3^(4^(5^(5^(6^(7^(8^9))))))' probably wouldn't be considered an "elegant solution."

 2014-10-02, 23:03 #3 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 3·43·73 Posts 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?
 2014-10-02, 23:30 #4 Xyzzy     "Mike" Aug 2002 177128 Posts 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
 2014-10-03, 00:18 #5 Mini-Geek Account Deleted     "Tim Sorbera" Aug 2006 San Antonio, TX USA 17×251 Posts 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
 2014-10-03, 02:55 #6 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 3·43·73 Posts 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.
 2014-10-03, 12:53 #7 CRGreathouse     Aug 2006 32·5·7·19 Posts Don't the moduli change at each step?
 2014-10-03, 13:23 #8 science_man_88     "Forget I exist" Jul 2009 Dumbassville 20C016 Posts 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
 2014-11-02, 19:03 #9 Xyzzy     "Mike" Aug 2002 2·13·313 Posts

 Similar Threads Thread Thread Starter Forum Replies Last Post Xyzzy Puzzles 9 2017-11-07 15:18 R. Gerbicz Puzzles 10 2016-11-01 13:35 LaurV Puzzles 3 2015-11-02 15:22 Joe O Prime Sierpinski Project 1 2010-10-09 06:12 lukethomas Science & Technology 18 2006-11-13 03:51

All times are UTC. The time now is 08:11.

Fri May 7 08:11:32 UTC 2021 up 29 days, 2:52, 0 users, load averages: 1.57, 1.68, 1.65