(In reply to
re: Solution by Charlie)
Thanks, I didn' know that notation.
I've seen on the web that 7^5 have the same three last as (7^5)^5 and in general 7^(5^n). Last digits are in 807.
So we have 9^something finishing in 807
This will be 9^(a number of hundreds) * 9^7
The first factor ends in 001 (my firs post) the second in 969.
So it's 969.
Edited on July 28, 2016, 4:32 pm
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Posted by armando
on 2016-07-28 16:29:15 |
re(2): Solution
