What is the relation between the linear combination and modular arithmetic? What is the relation between the linear combination and modular arithmetic? The linear combination is in a field and there must be some fundamental relation between them. What is it?
 A: When first learning modular arithmetic, you learn how to add, subtract, and multiply congruence classes of integers mod $n$ --- just as you learned basic arithmetic on numbers when you were in elementary school. There is no deep relationship between linear combinations here.
A bit later, you begin to look at division. This takes a bit a care, since in general it is not true that $ac \equiv bc \pmod n \implies a \equiv b \pmod n$, or rather one cannot simply divide or cancel on both sides in a congruence. We ask: by which elements does it make sense to "cancel" from a congruence?
This boils down to finding modular inverses. The inverse of $a \pmod n$ is the integer $x$ such that $ax \equiv 1 \pmod n$. Multiplying by such an inverse is a lot like dividing each side by $a$, except that one can say that $a \equiv b \pmod n \implies ac \equiv bc \pmod n$.
So when does an element $a$ have an inverse mod $n$? This is deeply connected to linear combinations of $a$ and $n$. Namely, the congruence
$$ ax \equiv 1 \pmod n$$
is equivalent to the linear combination
$$ ax + ny = 1$$
for integers $x,y$.
Since division (or rather modular inverses) is so fundamental, one can think of this as a deep relationship between modular arithmetic and linear combinations.
A: There is no canonical or natural "connection" between modular arithmetic and linear combinations. The only thing that I can think of mentioning is Bezout's Identity, which states that if in a PID $\gcd(a,b) = d$, then there are $p,q$ such that $ap+bq=d$; in $\mathbb{Z}_n$, then, there is a linear combination of $a,b$ that equals 0 if and only if $p$ divides the gcd of $a$ and $b$. 
A: Modular arithmetic uses the operator modulus to create a system which repeats or is cyclical. This is because the modulus or c in the example repeats it's self. Euclidean division is a = bx + c where b is the number of times x goes into a (output of division or input of multiplication) and c is the remainder (output of modulus) (there is no input because there are an infinite number of cases that satisfy this result but you could limit it to the first ones to make it). It is cyclical because if you consider x the base or basis which is trying to reach a then there exists a number which is not a linear combination of only x as the basis (b is just a scalar) so you have to choose  a c from another bases to make it work.
