- 28 Jul '06 10:07 / 3 edits2^2006 =

734803644975522895509013248753716469774283329336764928305691351568490407291117

000868262071242564012747805417368951976175006156134952355412112596067946658038

650431930529998592004408186548736287177145132736406311966602338967336885529262

828966658984665413791977972239352377245629475989194557031451020610129160602681

017500534435855228702770649976965982569602745617386922405315340852046312200267

272114453230122233570537076012939391093785869823060372809861692376973388291551

729814973901050624524526186435729634434723735846499381432233084365148857057765

0168488011227873843252661937332349789040779830473537880064 - 28 Jul '06 11:28

You can save a bit of a time by using the fact that 6*6=36, or any number ending with a 6 when multiplied by a number ending in 6 will give you a number ending in 6.*Originally posted by Knight Square***All I wanted to know if it the number ended with 8 and it didn't thank you**

[(10a+6)(10b+6)=100ab+60a+60b+36=10(10ab+6a+6b+3)+6]

2^2=4,

2^2 * 2^2 = 2^4=16 (last digit 6)

2^2006 = (2^4)^501 * 2^2

(2^4)^501 ends with a 6, 2^2 ends with a 4

2^2006 ends with a 4 (6*4=24) - 28 Jul '06 16:46

Then you don't get an 8 till you reach 2^2011 I think.*Originally posted by ThudanBlunder***I used a program called Mathematica.**

2^n ends in 8 only when n is of the form 4k+3

eg,

k = 0 gives n = 3 and 2^n = 2^3 = 8

k = 1 gives n = 7 and 2^n = 2^7 = 128

k = 2 gives n = 11 and 2^n = 2^11 = 2048

etc.

2011-3 = 2008, evenly divisible by 4. Right?

K has restrictions also, right? Only positve natural numbers. - 29 Jul '06 05:34

Hehe, I knew you were going to say that about zero! Hey, can you tell me the URL of the other puzzle site we were on, what, a year ago?*Originally posted by ThudanBlunder***I agree (though zero is not positive). BTW, 2^2007 ends in 8.**

Why did you want to know, Knight Square? A private bet? - 29 Jul '06 08:43

Although Mathematica is good, it's also quite expensive.*Originally posted by ThudanBlunder***I used a program called Mathematica.**

If you want an excellent free piece of software for arbitrary precision arithmetic you should download gp/PARI.

Just use Google with gp PARI in the search field.

It's a calculator for number theory, but typing in stuff like 2^1000000 is not a problem, and it can factor(N) numbers N etc. etc.

It's very good and easy to use. - 31 Jul '06 13:58

But 2^x (where x is a positive integer) can only end with the digits 2,4,8 or 6.*Originally posted by Alcra***if the question was of the form "**

is the last digit of the answer to 2^2006 = 8" and you had no idea whatsoever as to the answer, then a betting man would say "no" - since there are 10 different digits the last number could be (0-9).