Number of ordered pairs $(x, y)$ such that $0 \leq x, y\leq 18$ and $3x+4y+5$ is divisible by $19$ The problem would have been much simpler if there was no constant term, (like $3x+4y$ divisible by 19) because then all the solutions could have been generated from just the solution to $3x+4y=19$. 
I tried many things, one of which was painful enumeration.
The following are the solutions to $3x+4y+5=19k$ as ordered triplets $(x, y, k)$
( 0, 13, 3  )
( 1, 17, 4  )
( 2, 2, 1  )
( 3, 6, 2  )
( 4, 10, 3  )
( 5, 14, 4  )
( 6, 18, 5  )
( 7, 3, 2  )
( 8, 7, 3  )
( 9, 11, 4  )
( 10, 15, 5  )
( 11, 0, 2  )
( 12, 4, 3  )
( 13, 8, 4  )
( 14, 12, 5  )
( 15, 16, 6  )
( 16, 1, 3  )
( 17, 5, 4  )
( 18, 9, 5  )
Clearly there are 19 ordered pairs, one for each $x$ from $0$ to $18$. Why is this so? I mean, why is it that there is one and only one ordered pair for each $x$ ? I think there is one ordered pair for a particular $x$ in every $19$ consecutive numbers for $y$, for example $1$ in $0$ to $18$ another in $2$ to $20$ ...
Moreover, how can we conclude that there will always be some ordered pair for every $x$ ?
Also I saw that this is not special to $19$, it happens for all numbers. I feel like I'm missing something big here. 
Any help would be appreciated.
 A: Hint: We are looking for solutions of $3x+4y+5\equiv 0\pmod{19}$. Equivalently (multiply by $5$) we are solving $15x+y=\equiv -6\pmod{19}$, or equivalently $y\equiv 4x-6\pmod{19}$.  Now it is clear that for every $x$ there is a unique $y$ in the interval $0$ to $18$ for which the congruence holds. It is the remainder when $4x-6$ is divided by $19$.
A: Note that $3x+4y+5$ is divisible by $19$ if and only if $5(3x+4y+5)=15x+20y+25$ is divisible by $19$, if and only if $15x+y+6$ is divisible by $19$. So given $x$, you can compute $y$ by computing $4x+13$ modulo $19$ to get $y$.
You can do this because $4$ is relatively prime to $19$. In general, given $a,b,n$, if $b$ and $n$ are relatively prime then the number of pairs $0\leq x,y<n$ such that $ax+by+c$ is divisible by $n$ is $n$. 
That's because there is a $b'$ so that $bb'=1+nk$ for some $k$ and $ax+by+c$ is divisible by $n$ if and only if $ab'x+bb'y+cb'$ is divisible by $n$ if and oly if $y+(ab'x+cb')$ is divisible by $n$. So there is exactly one $y$ for each $x$.
