Modular arithmetic and probability $a$, $b$ and $c$ are chosen from $1$-$999$ with order and replacement, what is the probability that $a^2 + bc$ is divisible by $3$? (each choice is equally likely)
I split it up into 2 cases
$a^2$ has a remainder of $1$ while $bc$ has a remainder of $2$
$a^2$ has a remainder of $1$, $2/3$ of the time, $bc$ has a remainder of $2$ if $b$ and $c$ have $2$ and $1$. $(2/3 \times 1/3 \times 1/3)2$ (because $b$ and $c$ can be interchanged) = $4/27$
$a^2$ has a remainder of 0 while $bc$ has a remainder of $0$
$a^2$ has a remainder of $0$ $1/3$ of the time while $bc$ has a remainder of $0$ if $b$ or $c$ have remainders of $0$. probability $1/3 \times 1/3 \times 2 = 2/9$
$4/27 + 2/9 = 10/27$
Please tell me what i did wrong.

The error is here:

$a^2$ has a remainder of $0$ 1/3 of the time while $bc$ has a remainder of $0$ if $b$ or $c$ have remainders of $0$. probability $1/3 \times 1/3 \times 2 = 2/9$

You multiplied these numbers, which would be appropriate if you had "...if $b$ and $c$ have remainders of $0$." Since it's or instead, this changes. We can use the inclusion-exclusion principle here:
\begin{align*}
P(b\text{ or } c \text{ has remainder }0 ) &= P(b \text{ has remainder } 0) + P(c \text{ has remainder }0) \\
&- P(b \text{ and } c \text{ have remainder }0) \\
&=1/3 + 1/3 - 1/9 = 5/9.
\end{align*}
