Using Fermat's Little Theorem find the value for x.

x ≡ 9^(794) mod 73

I don't even know where to begin, I need step by step help majorly : )

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- Jun 1st 2008, 06:40 PMduggaboyCongruence Fermat's little theorem help
Using Fermat's Little Theorem find the value for x.

x ≡ 9^(794) mod 73

I don't even know where to begin, I need step by step help majorly : ) - Jun 1st 2008, 06:44 PMThePerfectHacker
Begin with 9^72 = 1 (mod 73)

Now keep on rasing exponents to get 794 - Jun 2nd 2008, 01:08 PMduggaboyfeeling dumb
This is totally stupid I know, but what is the base for using Mods? I've never used them by hand, only TI89 which now I have to show by hand, I'm totally lost..

- Jun 2nd 2008, 01:12 PMMoo
Hello,

Basis :

$\displaystyle a=b\mod n$ means that $\displaystyle a-b$ is a multiple of n.

In some cases, b represents the remainder of a in the Euclidian division by n.

a and b can be negative or positive.

Basic operations :

$\displaystyle a=b \mod n \Longleftrightarrow a+c=b+c \mod n$

Also, $\displaystyle a=b \mod n \Longleftrightarrow a=b+kn \mod n$, for any k in $\displaystyle \mathbb{Z}$

$\displaystyle a=b \mod n \implies a^c=b^c \mod n$ << for this one, always look for congruences to 1, because any power of 1 always yields 1.

$\displaystyle a=b \mod n \implies a*c=b*c \mod n$

Be careful :

$\displaystyle a=b \mod n \not \implies \frac ac=\frac bc \mod n$

There are several properties you should know...

Fermat's little theorem states :

$\displaystyle \text{If p is a prime number, then } a^{p-1}=1 \mod p$ (with a, not multiple of p) - Jun 2nd 2008, 01:15 PMMathstud28
- Jun 2nd 2008, 01:24 PMduggaboycongruence mods
Okay that makes a lot of sense thank you so much!!!

In solving x ≡ 9^(794) mod 73

I see where 9^72 *9^72 (until I get 794) but here I'm a little stuck cause 9^72 (11 times) gives me 9^792 so then at that point would I also include 9^2?? - Jun 2nd 2008, 01:27 PMMoo
Yep, nearly ^^

Actually, when dealing with such questions, always find the power to which the number (here, 9) will be congruent to 1 modulo n (794).

Then, do the Euclidian division :

$\displaystyle 794=72*11+2$

Therefore $\displaystyle 9^{794}=(9^{72})^{11}*9^2$ (these are powers rules)

-------> $\displaystyle 9^{794}=1^{11}*9^2 \mod 794$

Conclude ? - Jun 2nd 2008, 01:32 PMduggaboygetting so close
BTW your the BEST!!

now that I have the 1^11*9^2 (mod 794)

Would I use the equation a^p = 9 + 794k to find a? - Jun 2nd 2008, 01:38 PMMoo
Hmmmm strange oO

what is this formula ?

All you need is to "simplify" $\displaystyle 9^{794} \mod 73$, in order to get x.

$\displaystyle x=1^{11}*9^2 \mod 73$, not 794 :) Don't confuse with the numbers, it's quite common to see that kind of mistakes ^^

Now, just simplify 1^11 and calculate 9². It will give you x :p - Jun 2nd 2008, 01:43 PMduggaboyahha
not sure where i came across that formula ??(Thinking)

mods kill me!!

THANK YOU AGAIN! - Jun 2nd 2008, 01:44 PMThePerfectHacker
- Jun 2nd 2008, 01:49 PMduggaboychecking point
when solving for x here using the approach from the previous problem

x^86 = = 6 (mod 29)

I am getting

x = 36 (mod 29)

36^86 = = 6 (mod 29) or just

36 = = 6 (mod 29)