Hello,
My question is as stated in the title.
Does anyone know an easy way to solve this.
It doesn't show the answer on a calculator and I don't want to always rely on Wolfram Alpha
Any help would be appreciated.
we want to find 2008^{2008} (mod 100)
note that this will be the same as 8^{2008} (mod 100).
(since 2008 = 8 (mod 100)).
now 8^{21} = (8^{3})^{7} = 12^{7} = (12^{2})^{3}(12) (mod 100)
= (44^{3})(12) = (44^{2})(44)(12) = (36)(44)(12) = (84)(12) = 8 (mod 100).
since 2008 = (21)(95) + 13:
8^{2008} = (8^{21})^{95}(8^{13}) = (8^{95})(8^{13}) (mod 100)
and 95 = (21)(4) + 11:
(8^{95})(8^{13}) = (8^{21})^{4}(8^{11})(8^{13}) (mod 100)
= 8^{4+11+13} = 8^{28} = (8^{21})(8^{7}) = (8)(8^{7}) (mod 100)
= 8^{8} = (8^{3})^{2}(8^{2}) = (12)^{2}(64) = (44)(64) = 16 (mod 100).
so the last two digits are 16.
Just in case (like me) you are not comfortable with all the modular arithmetic,
Now except the last term, the rest of the terms are going to have atleast 3 Zeroes in the end. So the last two digit of the expression will be decided by the last expression, namely alone.
Now we just have to see the pattern in powers of 2. So it comes out as 2,4,8,16,32,64,28,56,12,24,48,96,92,84,68,36,72,44 ,88,76,52,04,08,16 ..
What we see is the the last digit is 6 for all multiples of 4 and that the last 2 digit follows a pattern namely - 16,56,96,36,76,16 ... (for powers in multiple of 4) , so for etc, last digit is 16. Now we have and
Hence the last two digits are 16.