# What would the last 2 digits of 2008^2008 be?

• July 23rd 2012, 08:48 PM
Atrey_G
What 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.
• July 23rd 2012, 09:28 PM
earboth
Re: What would the last 2 digits of 2008^2008 be?
Quote:

Originally Posted by Atrey_G
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.

Examine only the last digit:

(2008)^1 = xxxx...xxxx8
(2008)^2 = xxxx...xxxx4
(2008)^3 = xxxx...xxxx2
(2008)^4 = xxxx...xxxx6
----------------------------
(2008)^5 = xxxx...xxxx8
...

Do you see the pattern?
• July 24th 2012, 12:49 AM
Deveno
Re: What would the last 2 digits of 2008^2008 be?
we want to find 20082008 (mod 100)

note that this will be the same as 82008 (mod 100).

(since 2008 = 8 (mod 100)).

now 821 = (83)7 = 127 = (122)3(12) (mod 100)

= (443)(12) = (442)(44)(12) = (36)(44)(12) = (84)(12) = 8 (mod 100).

since 2008 = (21)(95) + 13:

82008 = (821)95(813) = (895)(813) (mod 100)

and 95 = (21)(4) + 11:

(895)(813) = (821)4(811)(813) (mod 100)

= 84+11+13 = 828 = (821)(87) = (8)(87) (mod 100)

= 88 = (83)2(82) = (12)2(64) = (44)(64) = 16 (mod 100).

so the last two digits are 16.
• July 24th 2012, 08:29 AM
pratique21
Re: What would the last 2 digits of 2008^2008 be?
Just in case (like me) you are not comfortable with all the modular arithmetic,
$2008^{2008}$
$= (2000+8)^{2008}$
$= 2000C0 + 2000^{2008} + 2008C1 (2000^{2007})(8^1)+ ..........+2008C2008 (8^{2008})$
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 $8^{2008}$ alone.
$8^{2008} = 2^{6024}$
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 $2^4, 2^{24}$ etc, last digit is 16. Now we have $2^{6024}$ and $6024=24*251$
Hence the last two digits are 16.