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Old 2013-02-13, 07:56   #3
Mr. P-1
 
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Jun 2003

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Quote:
Originally Posted by Nick View Post
Nothing deep here, just a light-hearted puzzle.
Find the next number (i.e. its first digit) in the following sequence:
5
25
625
0625
90625
890625
2890625
12890625
212890625
8212890625
18212890625
918212890625
...
2166509580863811000557423423230896109004106619977392256259918212890625
and then?
The last n+1 digits of 5^(2^n) starting with n=0

Now do the same for:
6
76
376

...
7833490419136188999442576576769103890995893380022607743740081787109376[/QUOTE]

The last n+1 digits of 376^(2^n) starting with n=0
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