for (e) and (f).... I dont know how to solve such congruences, I only know how to get to an answer when I check for different values of x....

(g) I can do.

(h) The answer that was given users Euler's function... can some1 plz show me how to use it for this question....

any help greatly appreciated..

2. Originally Posted by Dreamer78692

for (e) and (f).... I dont know how to solve such congruences, I only know how to get to an answer when I check for different values of x....

(g) I can do.

(h) The answer that was given users Euler's function... can some1 plz show me how to use it for this question....

any help greatly appreciated..

You'll likely need FLT = Fermat's Last Theorem (google it if you don't know it):

(e) as $2^5=-1\!\!\pmod {11}\,,\,2$ is a generator of the multiplicative group $\left(\mathbb{Z}/11\mathbb{Z}\right)^*$ , and thus there does exist some

unique power of it that equals 7 modulo 11. Try and error (since discrete logarithms seem to be out of our depth for

the time being) gives us $x=7$

(f) Do a list of quadratic residues modulo 11, and check which one, if any, equals $9^x=(-2)^x\!\!\pmod {11}$ ...

(h) It's easy to check that $5^6=1\!\!\pmod 7\,,\,\,5^5=1\!\!\pmod {11}\Longrightarrow 5^{60}=1\!\!\pmod {77}$ (why?)

In fact, try to show that already $5^{30}=1\!\!\pmod{77}$ ...

Tonio