Walk me through some basic modular arithmetic? The problem is as follows:
Positive integers $a$, $b$, and $c$ are randomly and independently selected with replacement from the set $\{1, 2, 3,\dots, 2010\}$. What is the probability that $abc + ab + a$ is divisible by $3$?
The solution is as follows:
First we factor $abc + ab + a$ as $a(bc + b + 1)$, so in order for the number to be divisible by 3, either $a$ is divisible by $3$, or $bc + b + 1$ is divisible by $3$.
We see that $a$ is divisible by $3$ with probability $\frac{1}{3}$. We only need to calculate the probability that $bc + b + 1$ is divisible by $3$.
We need $bc + b + 1 \equiv 0\pmod 3$ or $b(c + 1) \equiv 2\pmod 3$. (I get it up to this point because of my lacking modular arithmetic knowledge). Using some modular arithmetic, $b \equiv 2\pmod 3$ and $c \equiv 0\pmod 3$ or $b \equiv 1\pmod 3$ and $c \equiv 1\pmod 3$. The both cases happen with probability $\frac{1}{3} * \frac{1}{3} = \frac{1}{9}$ so the total probability is $\frac{2}{9}$.
Then the answer is $\frac{1}{3} + \frac{2}{3}\cdot\frac{2}{9} = \frac{13}{27}$
Is there a rule relating that two things being multiplied (b and (c+1)), if you mod them and add the result, would equal the other side? As in, I don't know why  $b \equiv 2\pmod 3$ and $c \equiv 0\pmod 3$ or $b \equiv 1\pmod 3$ and $c \equiv 1\pmod 3$. Trying to find a pattern (2+0 = 2 and 1+1 = 2) lead me to that question. Also, where did the 2/3 come from?
I would appreciate someone walking me through the solution as I am not experienced with mod (and my brain is having a hard time functioning without food). 

In this case you can think mod as an abreviate way for write the conditions. Concretely, you need $3|bc+b+1$. Now, as $3|3$, then $3|bc+b+1-3=bc+b-2$ (it means $b(c+1)\equiv 2(mod3)$). Now, $b$ only can be of the form $b=3n$ or $b=3n+1$ or $b=3n+2$. For the first one, non $c$ exists in order to get $3|bc+b-2$. If $b=3n+1$, then $c$ must be of the form $c=3m+1$ in order to get $3|bc+b-2$ (this means that $b\equiv 1$ and $c\equiv 1$). Analogously the other case.
