i was bored so i attempted to find patterns with 9^whatnot...
col 10=1,9,1,9 etc
col 9=8,2,6,4,4,6,2,8,0,0 then repeats
so if it's 9^an odd number, it'll end in 9.
9^a even number, it'll end in 1.
divide the (power-1) by 10, and if the remainders 1 then the second last digit of 9^ that number shall be a 8, if it's 2 it'll be a 2 etc (there are 10 numbers that repeat over)
so, 9^(9^(9^9)).
9^(9^9) is odd, so the last digit shall be 9. and 9^9 is 387420489, which is odd, so the second last digit of 9^(9^9) shall be 2. it'll end in 29. thus 9^(9^(9^9)) shall end in 89.
i am bored. if you see a flaw, tell me. but from what i can see, that is correct...so, who's willing to find out the next 8 digits? 😉
coloum 8 goes (if we sub 1 for odd and 2 for even) 1122112222 and repeats...
also-this is just using 9^1 to 25...
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Originally posted by cosmic voice3111098189
FIND THE LAST 10 DIGITS OF 9^(9^(9^9))
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 = 0,1,2,3,4,5,6,7,8,9
9^9 is 387420489
9^9^9 has the same last 10 digits as 200000001.20000001.8000001.1.400001.1.601.21.9 which has the same last 10 digits as 2835673589 (obviously when you multiply you can just disgard any digits after the last 10).
Repeat the process and you get the answer.
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 ?
THUDandBLUNDER's number is just 9^(9^8), which has relatively modest number of digits (fewer than 100 million).