I am trying to solve the following question:
Determine $n$ between $0$ and $19$ such that $(2311)(3912) = n \mod 20$.
Where I am now:
(2311) = 11(mod 20)
(3912) = 12(mod 20)
Multiply the 11X12 to get 132;
So: (2311)(3912) = 132(mod 20)
My question is, do I need to further simplify this result or is my solution above in a generally acceptable form. Sorry if this is a really lame question, I am still learning this subject.
Hint:
$$\begin{cases}2,311=11\pmod{20}\\{}\\ 3,912=12\pmod{20}\end{cases}\;\;\;\implies 2,311\cdot3,912=11\cdot12\pmod{20}=\ldots$$