Question

question..

Answer 1

\[\large 2^{1505}\] is divided by 9, the remainder is ?

Answer 2

Any idea?

Answer 3

are you supposed to show working??

Answer 4

it is one of the following
\[2,4,8,7,5,1\]

Answer 5

Nope

Answer 6

they come in that order

Answer 7

How @satellite73 ?

Answer 8

use bionomial theorm

Answer 9

well its an objective question with only one correct answer. Options I have are

a)8 b)7 c)5 d)6 e)1

Answer 10

once you get to 1, you start over

Answer 11

damn i made a mistake!! hold on

Answer 12

Sorry I didnt understand ^^

Answer 13

*historic moment :satellite makes a mistake ! :)

Answer 14

there is probably a snappier way to do this, but i like to think simple
take the remainder for powers of 2,i did wrong, let me try again

Answer 15

no i make lots

Answer 16

what about binomial theorem ?

Answer 17

oh ok :)

Answer 18

\(2^1\equiv 2\)
\(2^2\equiv 4\)
\(2^3\equiv 8\)
\(2^4\equiv 7\)
\(2^5\equiv 5\)
\(2^6\equiv 1\)

Answer 19

actually i didn't make a mistake! what i had was correct
once you get to 1, you start over
so if the exponent is divisible by 6, it is 1
for example \(2^{24}\equiv 1\)

Answer 20

@Hoa i am doing it the simple way

Answer 21

hmm...still having problem understanding the logic ^

Answer 22

it should be pretty clear that if you divide 1505 by 6, the remainder is 5, so you answer is the same as \(2^5\)

Answer 23

okay here is what i did
i started finding remainders for successive powers of two

Answer 24

ok

Answer 25

is that much clear?

Answer 26

yep

Answer 27

i listed the remainders
in order they are \(\{2,4,8,7,5,1\}\)

Answer 28

once you get to 1, you start over again

Answer 29

I see now !

Answer 30

that is, since \(2^6\equiv 1\) you know
\[2^7\equiv 2, 2^8\equiv 4,2^9\equiv 8,2^{10}\equiv 7,2^{11}\equiv 5, 2^{12}\equiv 1\]

Answer 31

they just repeat in that pattern
so your job in solving this is to find the integer remainder when you divide 1505 by 6

Answer 32

since the remainder is 5, it will be the same as \(2^5\)

Answer 33

Ok I get it now ...Thanks a lot !