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Seven powers of seven (Posted on 2006-09-08) Difficulty: 2 of 5
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.

See The Solution Submitted by JLo    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Answer to 1. | Comment 2 of 3 |

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.


  Posted by Dej Mar on 2006-09-09 05:33:39
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