Let n=391=17*23. Show that 2n-11 (mod n). Find an exponent j>0 such that
2j1 (mod n)

2. If those are exponents you might want to be a little more clear.

Are you saying :

Let $\displaystyle n=391=17\times23$. Show that $\displaystyle 2^{n-1} \not\equiv 1 \mod n$. Find an exponent $\displaystyle j>0$ such that
$\displaystyle 2^j\equiv 1 \mod n$ .
?

If so, then by no means do you need a computer to solve this problem.

3. Yes, this question need to solve by Matlab, but I am not good for matlab. Do you have any ideas??

thanks for help!!

4. Use Euler's theorem, which states that for any integer $\displaystyle a$, relatively prime to $\displaystyle n$,

$\displaystyle a^{\phi(n)} \equiv 1 \mod n$

where $\displaystyle \phi$ is Euler's phi function.

So, by Euler's theorem, with $\displaystyle \phi(391)=(17-1)(23-1)=352$, you have $\displaystyle 2^{352} \equiv 1 \mod 391$.

For the first part you can use the Chinese Remainder Theorem. You have :

$\displaystyle 2^{391} \equiv (2^{17})^{23} \equiv 2^{23} \equiv 2^{17}2^6 \equiv 2^7 \equiv 128 \equiv 9 \mod 17$

so it's impossible that $\displaystyle 2^{391} \equiv 2 \mod 391$.

Matlab is for engineers.