Question

modulos math,

Which of the following gives the 3rd root of 92 modulo 187? (Note that 187=11⋅17.)

Answer 1

okay

Answer 2

3rd root ?
or
\(92^3\)
?

Answer 3

not making sense

Answer 4

92 mod 187 = 92 , the remainder is simply 92 in this case

Answer 5

i am sure, she knows that...
lets assume its
\(\Large 92^3 mod ~187\)

Answer 6

ok

Answer 7

can you post the screenshot of the question ?

Answer 8

ok hold on one sec

Answer 9

https://www.dropbox.com/s/kmwny6v0eziwuqy/Screenshot%202014-12-30%2021.01.37.jpg?dl=0

Answer 10

what is the question telling me the prime factors for 187 ( 11 * 17) ? and what does it mean by 3rd root ?

Answer 11

is it like x^3 = 92 modulo 187? and I have to find x ?

Answer 12

I am not sure,
i'll wait for other experts to look into it...sorry

Answer 13

its ok np

Answer 14

in that case i think you would just take 3rd root of 92

Answer 15

I would like to see "the following", please.

Answer 16

i thought mod arithmetic works with integers only

Answer 17

sumbody temme what she wants
!!

Answer 18

https://www.dropbox.com/s/8h1iub9daga1qvr/Screenshot%202014-12-30%2021.05.21.jpg?dl=0

Answer 19

should very well be the one hartnn assumed , but I am more like looking for the reasoning

Answer 20

all i can say is :
its not 2nd option :P

Answer 21

xD

Answer 22

is it asking for 3rd primitive root ... :O

Answer 23

hartnn seems to know

Answer 24

most likely

Answer 25

but wats dat

Answer 26

I need to use the prime factors of 187 somehow

Answer 27

@ganeshie8 @mathmath333

Answer 28

you want to solve
\[x^3 \equiv 92 \pmod {187}\]

Answer 29

what does it mean by
'give 3rd root of 92 modulo 187'
?

Answer 30

find some integer whose cube is 92

Answer 31

it doesnt exist

Answer 32

it might exist in modulo 187

Answer 33

looking at your options, it does exists

Answer 34

how can you pin point so fast though

Answer 35

looking at the options i mean

Answer 36

what we use (decimal) is modulo 10 ?

Answer 37

we need to work the root
was just saying you dont have an option like "root does not exist"

Answer 38

I don't @hartnn im new to modular arithmetic , might very well be

Answer 39

lets use the hint maube..

Answer 40

hmm

Answer 41

by the way right answer is: https://www.dropbox.com/s/fx0z4egq4ket670/Screenshot%202014-12-30%2021.22.15.jpg?dl=0

Answer 42

really got to brush up my modulos

Answer 43

just in case it helps with anything regarding the process

Answer 44

thats the right answer but idk any easy method to explain you

Answer 45

got it, thanks a lot

Answer 46

you will need to know euler phi function and few other things to make sense of above ^
let me see if i can figure out a solution using basic mod properties

Answer 47

nah I know phi function and the others, I understand perfectly now

Answer 48

fixed typo :
\[\phi(187) = \phi(11\cdot 17) = 10\cdot 16 = 160\]
so we have \[92^{160} \equiv \color{Red}{1} \pmod {187}\]

\[x^3\equiv 92\times 92^{2\times 160} \pmod {187}\]
\[x^3\equiv 92^{2\times 321} \pmod {187}\]
\[x\equiv 92^{ 321/3} \pmod {187}\]

Answer 49

thanks a bunch

Answer 50

you're good then :)

Answer 51

*sigh*
some day :P

Answer 52

xD

Answer 53

phi function is just an extension to fermat little theorem

Answer 54

Fermat little theorem : whenp is prime and gcd(a,p)=1 below holds

\[a^{p-1}\equiv 1 \pmod {p}\]

Answer 55

euler generalization of fermat little theorem using phi function :

\[a^{\phi(n)}\equiv 1 \pmod {n}\]

this is true for all integers n, it need not be prime anymore
however gcd(a,n)=1 is still required

Answer 56

\(\phi(n)\) gives the number of integers less than \(n\) that are coprime to \(n\)

Answer 57

\(\phi (n) =n-1\)
is tru only if n is prime

Answer 58

yes when n is prime, euler generalization becomes fermat little thm

Answer 59

ok thanks :)
i'll dig more :)