In the set {1, 2, 3, ..., 2019}, we pick a, b, and c randomly without any conditions. What is the probability for abc + bc + c to be divisible by 3? 3 integers a, b, and c are randomly taken from the set {1, 2, 3, ..., 2019} without any conditions. What is the probability for abc + bc + c is divisible by 3?
Correct me if I'm wrong please.
abc + bc + c = c(ab + b + 1).
We can conclude that c has to be a multiple of 3, so we have 1/3 probability for c. a and b are any integers from the set.
AND, ab + b + 1 has to be  a multiple of 3 in the case that c is not a multiple of 3.
ab + b + 1 = b(a + 1) + 1.
b(a + 1) has to be in the form 3n - 1 for b(a + 1) + 1 to be a multiple of 3.
And this is where I get stuck. New solutions, hints, or suggestions are welcome.

For a solution to $b(a+1)+1=3n$ where $n$ is arbitrary, we need to have $$(b(a+1)+1) \mod 3=0$$ which means we need $(b(a+1)) \mod 3=2$. That can happen if $a\mod 3=0$ and $b\mod 3=2$ or $a\mod 3=1$ and $b\mod 3=1$.
For example $a=3$, $b=8$ has $a\mod3=0$ and $b\mod3=2$ and $8(3+1)+1=33=3\times11$, and $a=4$, $b=4$ has $a\mod3=b\mod3=1$ and $4(4+1)+1=21=3\times7$.
Since these two sets don't overlap each other, you can just add them, so $\frac292019$ of pairs of $a$ and $b$ match the criterion, in addition to the $c$s which match.
