1. What is the last digit of 7⁷⁷⁷⁷⁷⁷ = 7^(7^(7^(7^(7^(7^7))))) ?

2. More generally, if we define the hyper power ⁿa by

¹a:=a, ⁿ⁺¹a:=a^(na) for n>=1

does the series of the last digits of ¹a, ²a, ³a,... always become periodic at some point?

If so, can you provide a sharp upper limit for the period length?

Note: The hyper power ⁿa is also often denoted as a^^n.

I've been playing around with Mathematica and used this definition of hyper powers to quickly compute the last digit of each hyper power for various a.  Note that I keep all my computations modulo 10 to keep the numbers from getting too big.

HypPow[a_,n_]:=(
    If[n==1, Return[Mod[a,10]],
      Return[Mod[a^HypPow[a,n-1],10]]
      ];
);

Now I next computed the last digit of the first 10 hyper powers for a=2 to 30  and I got the following

a=1    {1,1,1,1,1,1,1,1,1,1}
a=2    {2,4,6,4,6,4,6,4,6,4}
a=3    {3,7,7,7,7,7,7,7,7,7}
a=4    {4,6,6,6,6,6,6,6,6,6}
a=5    {5,5,5,5,5,5,5,5,5,5}
a=6    {6,6,6,6,6,6,6,6,6,6}
a=7    {7,3,3,3,3,3,3,3,3,3}
a=8    {8,6,4,6,4,6,4,6,4,6}
a=9    {9,9,9,9,9,9,9,9,9,9}
a=10  {0,1,0,1,0,1,0,1,0,1}

a=11  {1,1,1,1,1,1,1,1,1,1}
a=12  {2,4,6,4,6,4,6,4,6,4}
a=13  {3,7,7,7,7,7,7,7,7,7}
a=14  {4,6,6,6,6,6,6,6,6,6}
a=15  {5,5,5,5,5,5,5,5,5,5}
a=16  {6,6,6,6,6,6,6,6,6,6}
a=17  {7,3,3,3,3,3,3,3,3,3}
a=18  {8,6,4,6,4,6,4,6,4,6}
a=19  {9,9,9,9,9,9,9,9,9,9}
a=20  {0,1,0,1,0,1,0,1,0,1}

a=21  {1,1,1,1,1,1,1,1,1,1}
a=22  {2,4,6,4,6,4,6,4,6,4}
a=23  {3,7,7,7,7,7,7,7,7,7}
a=24  {4,6,6,6,6,6,6,6,6,6}
a=25  {5,5,5,5,5,5,5,5,5,5}
a=26  {6,6,6,6,6,6,6,6,6,6}
a=27  {7,3,3,3,3,3,3,3,3,3}
a=28  {8,6,4,6,4,6,4,6,4,6}
a=29  {9,9,9,9,9,9,9,9,9,9}
a=30  {0,1,0,1,0,1,0,1,0,1}

now as you can see the cycles have a cycle of their own of 10.  So now for the hard part (which I'm currently stuck on) is showing that this cycle continues for all a, if that can be shown then it is easy to show that the max cycle is 3 when a has a last digit of 8.

Posted by Daniel on 2006-09-08 18:11:09