Question

Remainder when 23^101 is divided by 9 is?

Answer 1

@experimentX @Arnab09 @yash2651995

Answer 2

weird...my calculator cant solve 23^101

Answer 3

Some tricks to solve it!!

Answer 4

2

Answer 5

@Arnab09 u r correct can u show it

Answer 6

u know the method or just testing? :)

Answer 7

no i dont know!

Answer 8

okay

Answer 9

23=5(mod 9)

Answer 10

so, 23^3=125(mod 9)=-1 (mod 9)
so, 23^101=23^99* 529= (-1)^3*529( mod 9)= -529(mod 9)=2( mod 9)

Answer 11

sorry, thats (-1)^33 ^^^up there

Answer 12

got it?

Answer 13

No @Arnab09 i did nt understand!

Answer 14

find some cyclic pattern... try dividing few powers of 23 by 9 until it gets repeated

Answer 15

23=5(mod 9) ?????????????

Answer 16

u know the divisibility formulae? @Yahoo! ?

Answer 17

No....CAn u show that

Answer 18

that means 23 leaves remainder 5 when divided by 9

Answer 19

Oh .. yes that's right!!
it's equivalent to dividing
5^101 by 9

Answer 20

@Arnab09 ok then this so, 23^3=125(mod 9)=-1 (mod 9)
so, 23^101=23^99* 529= (-1)^3*529( mod 9)= -529(mod 9)=2( mod 9)????????

Answer 21

thats (-1)^33, i told before

Answer 22

it is bit confusing can u write it lol!!

Answer 23

@experimentX wat is that pattern

Answer 24

23^99=(23^3)^33=(-1)^33 (mod 9)=-1(mod 9)
now?

Answer 25

5^1 mod 9 = 5
5^2 mod 9 = 7
5^3 mod 9 = 8
5^4 mod 9 = 4
5^5 mod 9 = 2
5^6 mod 9 = 1
5^7 mod 9 = 5 <<<--- it will just repeat after that ... 7,8,4,2,1,5 ... so on and on
..
so on and on ..

Answer 26

as 23^3=(-1) (mod 9) as stated before..

Answer 27

when we divide 23^3 by 9 we get remainder as 8 not 1

Answer 28

-1 means 8 :)

Answer 29

wat?

Answer 30

difficult to explain..... number theory.. divisibility rule

Answer 31

@experimentX if the pattern is 7,8,4,2,1,5 . then....

Answer 32

find the 101 th term

Answer 33

it's 2 i solve with calc

Answer 34

how @Arnab09 it is not a AP so??

Answer 35

the remainder is repeated every 6th time

Answer 36

so, to get 101th term, it will complete 16 cycles+the 5th term.. that is 2

Answer 37

ok got it thxx

Answer 38

better if u apply divisibility rule.. get acquainted with it in number theory!!

Answer 39

but can u give some basics abt that

Answer 40

it repeats after 6th term, so
101 can be written as 16*6+5
so it reduces to
5^(16*6+5) mod 9 = 5^5 mod 9
http://www.wolframalpha.com/input/?i=5^%286*16%2B5%29+mod+9
http://www.wolframalpha.com/input/?i=23^101+mod+9
http://www.wolframalpha.com/input/?i=5^5+mod+9

Answer 41

http://www.wolframalpha.com/input/?i=quotient+remainder+%2823^101%2C+9%29

Answer 42

http://en.wikipedia.org/wiki/Divisibility

Answer 43

u can get something here^^ @Yahoo!

Answer 44

@Arnab09 look my new quest

Answer 45

oh .. new trick on wolfram!!

Answer 46

what is it?? LOL
http://www.wolframalpha.com/input/?i=quotient+remainder+%285%5E101%2C+9%29