12 Nov '04 07:58>
FIND THE LAST 10 DIGITS OF 9^(9^(9^9))
Originally posted by cosmic voice3111098189
FIND THE LAST 10 DIGITS OF 9^(9^(9^9))
Originally posted by GlentonAgree
3111098189
raise 9 to the power of a number of form kx10^n:
- (n=0): The units digit cycles 1,9 and repeats
- (n=1): The tens digit cycles every 10 (01,01,81,21,61,41,41,61,21,81)
- (n>2): Subsequent digits cycle according to their digit.
kx10^n (k=0,1,2,...,9; n=2,3,...,9) has last (n+1) digits of
jx10^n+1 where j=0,4,8,2,6,0,4,8,2,6 for k = 0,1, ...[text shortened]... ou can just disgard any digits after the last 10).
Repeat the process and you get the answer.
Originally posted by CoolPlayercosmic voice's number has more than (9^(9^9))/2 digits... and THAT has more than (9^9)/2 digits, so even writing down the number of digits of the number of digits of 9^(9^(9^9) in full would require you to write down millions of digits!
Whaoh......? REALLY ?