1. What is the last digit of
7
777777 = 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.
1. The last digit will be 3.
7n, such that n is a positive integer ending in the digit 3, where the 3 is preceded by an even digit, will end with the three digits 343 [if preceded by an odd digit, the last three digits would be 407]. Because 77 = 823543, which ends in a 3 preceded by the digit 4 [an even digit], 7^(7^(7^(7^(7^(7^7))))) will end in the digits 343.
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Posted by Dej Mar
on 2006-09-09 05:33:39 |