12 Nov '04 07:58

FIND THE LAST 10 DIGITS OF 9^(9^(9^9))

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Your Blackened Sky12 Nov '04 11:084 editsi 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? ðŸ˜‰

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...- Joined
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12 Nov '04 16:40

3111098189*Originally posted by cosmic voice***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.- Joined
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Account suspended13 Nov '04 00:01

Agree*Originally posted by Glenton***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.- Joined
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Loughborough21 Nov '04 18:363 edits

cosmic 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!*Originally posted by CoolPlayer***Whaoh......? REALLY ?**

THUDandBLUNDER's number is just 9^(9^8), which has relatively modest number of digits (fewer than 100 million).