Graphics Calculator can't work out...

This a very simple question. What is 2^2006, is the last digit 8? See I am a very stupid and dumb person so please if you do insult me, it just won't work. Okay I have already come to that reality, everybody has told me.

2^2006 =
734803644975522895509013248753716469774283329336764928305691351568490407291117
000868262071242564012747805417368951976175006156134952355412112596067946658038
650431930529998592004408186548736287177145132736406311966602338967336885529262
828966658984665413791977972239352377245629475989194557031451020610129160602681
017500534435855228702770649976965982569602745617386922405315340852046312200267
272114453230122233570537076012939391093785869823060372809861692376973388291551
729814973901050624524526186435729634434723735846499381432233084365148857057765
0168488011227873843252661937332349789040779830473537880064

All I wanted to know if it the number ended with 8 and it didn't thank you

Originally posted by Knight Square
All I wanted to know if it the number ended with 8 and it didn't thank you

Do you mean I calculated all that for nothing? 😉

How can you calculate the number?

Originally posted by Knight Square
All I wanted to know if it the number ended with 8 and it didn't thank you

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.

[(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)

Originally posted by Knight Square
How can you calculate the number?

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.

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.

Then you don't get an 8 till you reach 2^2011 I think.
2011-3 = 2008, evenly divisible by 4. Right?
K has restrictions also, right? Only positve natural numbers.

Originally posted by sonhouse

2011-3 = 2008, evenly divisible by 4. Right?
K has restrictions also, right? Only positve natural numbers.

I agree (though zero is not positive). BTW, 2^2007 ends in 8.

Why did you want to know, Knight Square? A private bet?

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?

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?

No it was in a test I did at school called the Australian Maths Competion

Originally posted by sonhouse
Hey, can you tell me the URL of the other puzzle site we were on, what, a year ago?

That would be http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi#riddles_cat

Originally posted by ThudanBlunder
I used a program called Mathematica.

Although Mathematica is good, it's also quite expensive.

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.

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).
😉

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).
😉

But 2^x (where x is a positive integer) can only end with the digits 2,4,8 or 6.