October 2014
<<September October November>>
We are giving Alan Murray's solution (thanks, Alan):
Observe the following congruences:
2^(n+4*5^9) = 2^n mod 10^10 for n>=10
3^(n+8*5^8) = 3^n mod 4*5^9 for n>=0
4^(n+2*5^7) = 4^n mod 8*5^8 for n>=2
5^n = 5^7 mod 2*5^7 for n>=7
Now let n=6^(7^(8^9)), which is >=7.
So then 4^(5^n) = 390624 mod 8*5^8
And 3^(4^(5^n)) = 5765981 mod 4*5^9
And 2^(3^(4^(5^n))) = 8170340352 mod 10^10
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