October 2014

[Xyzzy]

http://domino.research.ibm.com/Comm/...tober2014.html

[EdH]

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."

[Batalov]

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?

[Xyzzy]

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

[TimSorbet]

I can't say I fully grok the math behind it, but I found a solution by calculating the following in PARI/GP:

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.

[Batalov]

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.

[CRGreathouse]

Don't the moduli change at each step?

[science_man_88]

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.

[Xyzzy]

http://domino.research.ibm.com/Comm/...tober2014.html