Intuition - Linear Congruence Theorem 
Let a and b be integers (not both 0) with greatest common divisor d.
  Then an
  integer $c = ax + by$ for some $x, y \in Z$  $\iff d|c$.
In particular, d is the least positive integer of the form ax +by.

Is there intuition? Or illustration? I keep forgetting which variable is supposed to go. I'm not querying proofs. 
Withal, if I write $ax + by = c$ as $ax \equiv c \; (mod \, b)$, then an error is even more likely! I can't remember if it's  $c|d$, $b|c$, $c|a$, or some other wrong combination...
Origin - Elementary Number Theory, Jones, p10, Theorem 1.8
 A: It's definitely something you can develop intuition for. Since $a$ and $b$ have common divison $d$, we have 
$$c = ax+by = dzx+dwy = d(zx+dw)$$
for some $z,w$. Hence $d | c$.  Think about it like simplifying fractions you learned in middle school or earlier. If everything in the numerator is a multiple of $d$, and the denominator is $d$, then the result is an integer. 
A: You have two integers $a$ and $b$ (not both $0$) with the greatest common divisor $d$. It means that $d|a$ and $d|b$. As we know, if $d|a$ and $d|b$ then $d|ax+by$ for all $x,y\in\mathbb{Z}$. Furthermore, one wonders what the set $$S=\{ax+by:x,y\in\mathbb{Z}\}.$$
Of course, $d$ divides each element in $S$ since is one of the common factor(s) between $a$ and $b$ which is equivalent to the necessary part of your statement.
The sufficient part of your statement is motivated by Bezout's_identity.
A: Given $d=\gcd(a,b)$
Thus $a=d.k_1$ and $b=d.k_2$
Necessary part
Suppose $c=ax+by$  where $x,y\in Z$.
Then $c=(dk_1)x+(dk_2)y$ i.e. $c=d(xk_1+yk_2)$
Thus $d|c$
Sufficient part
Suppose $d|c$. Let $c=d.k$
Since $gcd$ can be expressed as linear combination of the two numbers, there exist $x,y\in Z$ such that $d=ax+by$. Multiply with $k$ to get $c=dk=k(ax+by)=a(kx)+b(ky)=ax_0+by_0$
