by collegebookworm » Fri Apr 06, 2007 10:07 pm
**edit: misread
**re-edited: corrected
first, we must find the pattern for 8^8^8^8....
notice that the pattern for 8^n, where n is any integer, starting with 1, follows this 20-cycle pattern for the last 2 digits:
08,64,12,96,68,44,52,16,28,24,92,36,88,04,32,56,48,84,72,76
to find 8^n for any n, therefore, we must find the value of n mod 20; we can thus conclude that we need the last two digits of n to find the last two digits of 8^n.
We can now find the last two digits of 8^8. Since 8 mod 20 is 8, the last two digits are 16. Now we can find the last two digits of 8^8^8 by the same logic (16th value in the pattern: 56). We can do this again (56 mod 20 = 16; 16th value in the pattern is 56) and we find that we have reached a final resting place for the last two digits: 56.
Please inform me of any errors I have made.
(NS/CA/MA/CS/SC) "**" indicates TMSCA State
08 HI(LO): 142?-**(88-D2)/224-D2(17x-**)/320-D2(202-**)/248?-D2(206-**)/210-D2(17x-NovA+?)
08 D2: 88/224/320/210/248
Next meet: Texas Tech (Region, 4/11-12)