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.

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- Jul 23rd 2012, 08:48 PMAtrey_GWhat would the last 2 digits of 2008^2008 be?
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. - Jul 23rd 2012, 09:28 PMearbothRe: What would the last 2 digits of 2008^2008 be?
- Jul 24th 2012, 12:49 AMDevenoRe: What would the last 2 digits of 2008^2008 be?
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. - Jul 24th 2012, 08:29 AMpratique21Re: What would the last 2 digits of 2008^2008 be?
Just in case (like me) you are not comfortable with all the modular arithmetic,

$\displaystyle 2008^{2008} $

$\displaystyle = (2000+8)^{2008} $

$\displaystyle = 2000C0 + 2000^{2008} + 2008C1 (2000^{2007})(8^1)+ ..........+2008C2008 (8^{2008}) $

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 $\displaystyle 8^{2008} $ alone.

$\displaystyle 8^{2008} = 2^{6024} $

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 $\displaystyle 2^4, 2^{24} $ etc, last digit is 16. Now we have$\displaystyle 2^{6024}$ and $\displaystyle 6024=24*251 $

Hence the last two digits are 16.