Question

Modular equation

Answer 1

how do you solve this?

Answer 2

i give up
i would probably start with
\[m^7\equiv 26(33)\] and see if i could get a pattern of some kind

Answer 3

you have a method you are supposed to use?
btw the answer is 20

Answer 4

must have something to do with \(33=3\times 7\) and primitive roots perhaps

Answer 5

and perhaps the fact that \(\phi(33)=2\times 10=20\) so there are 20 congruence classes

Answer 6

i dont understand T^T

Answer 7

@FibonacciChick666

Answer 8

I can do this!!!!!

Answer 9

give me a minute I need my book

Answer 10

\(\large m^7 \equiv 26 \pmod {3}\)
\(\large m^7 \equiv 26 \pmod {11}\)

same as

\(\large m^7 \equiv 2 \pmod {3}\)
\(\large m^7 \equiv 4 \pmod {11}\)

Answer 11

I think that is legal

Answer 12

i see "-1" satisfies first congruence

Answer 13

should i check m= -5->5 for second congruence ?

Answer 14

well, I was thinking we should expect multiple answers

Answer 15

but hmm...

Answer 16

ok yea, so we can use chinese remainder theorem

Answer 17

\[m^7≡26(mod3)~~~~~~~~~
m^7≡26(mod11)\]
\[m^7≡2(mod3)~~~~~~~~~~~~
m^7≡4(mod11)\]

Answer 18

-1 satisfies mod 3

-2 satisfies mod 11

next we apply chinese remainder thm, thats it ? xD thought its more complicated for some reason haha!

Answer 19

I think so.... That's all I'd do

Answer 20

you might be able to use primitive roots though as suggested before, there is \(x^m-g\equiv 0 mod p\) where g is a prim root

Answer 21

actually, 33 cannot have a primitive root.... soo

Answer 22

chinese remainder thm wont work is it ?

Answer 23

the solution to system :
x^7 = 2 mod 3
x^7 = 4 mod 11

is not same as
x = 2 mod 3
x = 9 mod 11

?

Answer 24

gotta ask, where did you get the 9?

Answer 25

ah ok nvm

Answer 26

just tried all the numbers from 0->10 for x^7 = 4 mod 11
only 9 satisfies

Answer 27

ok I buy that

Answer 28

I'd say try it

Answer 29

why are you taking inverses again?

Answer 30

one sec, let me review chinese remainder thm quick

Answer 31

yeah there was a mistake, il delete

Answer 32

although I guess you could just take the numbers from (2-32)^7 and see if they are congruent to 26?

Answer 33

oh wolfram says it is 20, i must applying CRT incorrectly
http://www.wolframalpha.com/input/?i=solve+x+%3D+2+mod+3%2C+x+%3D+9+mod+11

Answer 34

then again http://www.di-mgt.com.au/rsa_alg.html

Answer 35

guess you could use this for inspiration http://www.math.upenn.edu/~ssneha/soln2.pdf