Question

Determine n between 0 and 19 such that (2311)(3912) ≡ n mod 20.

Answer 1

You can write 2311 and 3912 in the form $20q+r$:

$2311=115\cdot20+11$

$3912=125\cdot20+12$

Then

$2311\cdot3912=(115\cdot20+11)(125\cdot20+12)$

$2311\cdot3912=115\cdot125\cdot20^2+(11\cdot125+12\cdot115)\cdot20+11\cdot12$

Taken modulo 20, the terms containing powers of 20 vanish and you're left with

$2311\cdot3912\equiv11\cdot12\equiv132\pmod{20}$

We further have

$132=6\cdot20+12$

so we end up with

$2311\cdot3912\equiv12\pmod{20}$

and so $n=12$.

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If instead you're trying to find $2311^{3912}\pmod{20}$, you can apply Euler's theorem. We can show that $\mathrm{gcd}(2311,20)=1$ using the Euclidean algorithm. Then since $\varphi(20)=8$, and 8 divides 3912, we have

$2311^{3912}\equiv2311^{489\cdot8}\equiv(2311^{489})^8\equiv1\pmod{20}$

To show 2311 and 20 are coprime:

2311 = 115*20 + 11

20 = 1*11 + 9

11 = 1*9 + 2

9 = 4*2 + 1   =>  gcd(2311, 20) = 1