20141002, 12:47  #1 
Aug 2002
209C_{16} Posts 
October 2014

20141002, 21:46  #2  
"Ed Hall"
Dec 2009
Adirondack Mtns
3·37^{2} Posts 
Quote:


20141002, 23:03  #3 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2×5×31^{2} 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(2^9911)'. But what [B]are[/B] those last ten digits, eh? 
20141002, 23:30  #4 
Aug 2002
2^{2}·2,087 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/numbertheo...exponent.html http://mymathforum.com/numbertheory...stdigits.html http://math.stackexchange.com/questi...digitsof999 
20141003, 00:18  #5 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
1000010101101_{2} 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))
Last fiddled with by MiniGeek on 20141003 at 01:13 
20141003, 02:55  #6 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2·5·31^{2} 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. 
20141003, 12:53  #7 
Aug 2006
3×1,993 Posts 
Don't the moduli change at each step?

20141003, 13:23  #8 
"Forget I exist"
Jul 2009
Dumbassville
20300_{8} 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 20141003 at 13:23 
20141102, 19:03  #9 
Aug 2002
2^{2}·2,087 Posts 

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