I can't give a complete lesson on elementary number theory, but I can explain what the mod business is about.
We have 
) 
Add the 1 to the 10 and divide by 11.
^{k}(mod \;\ 11)) 
Let k=2:
) 
 
Say we have 
^{3}(mod \;\ 11)) 
 
and so on.
It's about remainders. It is sometimes called clock arithmetic and is very useful when dealing with huge numbers.
Arbitrary example:
Solve 
) 
We want to find all values of x that satisfy the congruence.
We have gcd(42,90)=6. Since 6 is a factor of 12, there is a solution.
Solving the congruence is the same as solving

 for integers x and q.
This reduces to 

 or 
) 
The numbers congruent to 1 modulo 15 are 16,31,46,61,. 
and also -14,-29,-44,. 
See, they're all off by 1 from being a multiple of 15.
7 is a factor of -14, so we multiply both sides by -2 since 
) 
Thus, we have 
\rightarrow x\equiv 11(mod \;\ 15)) 
The solution is 
) 
Check: 

. 90 divides into 450.
I know this is a lot to digest, but a lot of proofs come from this area of number theory. It is a fascinating field and can be used to solve many problems.