You can write 2311 and 3912 in the form [tex]20q+r[/tex]:
[tex]2311=115\cdot20+11[/tex]
[tex]3912=125\cdot20+12[/tex]
Then
[tex]2311\cdot3912=(115\cdot20+11)(125\cdot20+12)[/tex]
[tex]2311\cdot3912=115\cdot125\cdot20^2+(11\cdot125+12\cdot115)\cdot20+11\cdot12[/tex]
Taken modulo 20, the terms containing powers of 20 vanish and you're left with
[tex]2311\cdot3912\equiv11\cdot12\equiv132\pmod{20}[/tex]
We further have
[tex]132=6\cdot20+12[/tex]
so we end up with
[tex]2311\cdot3912\equiv12\pmod{20}[/tex]
and so [tex]n=12[/tex].
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If instead you're trying to find [tex]2311^{3912}\pmod{20}[/tex], you can apply Euler's theorem. We can show that [tex]\mathrm{gcd}(2311,20)=1[/tex] using the Euclidean algorithm. Then since [tex]\varphi(20)=8[/tex], and 8 divides 3912, we have
[tex]2311^{3912}\equiv2311^{489\cdot8}\equiv(2311^{489})^8\equiv1\pmod{20}[/tex]
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
