What is the solution for 27 mod 4 in this math problem?

[Math9999]

Homework Statement

I'll upload the picture of the math problem.

Homework Equations

None.

The Attempt at a Solution

Here's the work:
[2, 0, 3, 2]*[6, 4, 3, 3]=12+9+6=27
Since [3, 1, 1, 2]+[3, 3, 2, 1]=[6, 4, 3, 3].
But how do I figure out 27 mod 4?

 

[fresh_42]

Here's the explanation of what modulo means:
https://en.wikipedia.org/wiki/Modular_arithmetic

It's the remainder by a division with a certain number, here $4$ and $5$. What you also used here, is that the modulo operation is a ring homomorphism, i.e. you don't have to bother whether you first apply the modulo operations and multiply in $\mathbb{Z}_4$ resp. $\mathbb{Z}_5$, or first calculate in $\mathbb{Z}$ and apply the modulo operation on the result.

 

[Math9999]

So for the first part of the problem, it's 27 mod 4=3. Since the remainder of 27-4*6=3, right? But what about the second one? ZZ 4 and 5?

 

[fresh_42]

If you can do it in $\mathbb{Z}_4$ what prevents you from doing the same with $5$? The upper $4$ in the notation only means, that we have four dimensional vectors here.

 

[Math9999]

So 27 mod 5=2 since the remainder of 27-5*5=2, right? So what you're saying is to ignore the upper part since that's the number of dimensional vectors?

 

[fresh_42]

Right. The notation means $\mathbb{Z}_5^4 = \mathbb{Z}_5 \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_5$. So the initial vectors, e.g. $(2,0,3,2)$ are meant to have their components, which are four of them, within $\mathbb{Z}_5$, that is $2 \in \mathbb{Z}_5\; , \;0 \in \mathbb{Z}_5\; , \; 3 \in \mathbb{Z}_5\; , \;2 \in \mathbb{Z}_5$. Together it is $(2,0,3,2) \in \mathbb{Z}_5^4$. So the upper four is a count for the number of components, whereas the lower $4$ or $5$ in my example tells us where those components belong to. $\mathbb{Z}_4 = \{0,1,2,3\}\; , \;\mathbb{Z}_5 = \{0,1,2,3,4\}$ the possible remainders of a division by $4$, resp. $5$.

 

[Math9999]

Thank you so much for the help!