How many distinct four-digit numbers are divisible by $3$ and have $23$ as their last two digits?
$\textbf{(A) } 27 \qquad\textbf{(B) } 30 \qquad\textbf{(C) } 33 \qquad\textbf{(D) } 81 \qquad\textbf{(E) } 90$
Solution 1 (Slow)
To test if a number is divisible by three, you add up the digits of the number. If the sum is divisible by three, then the original number is a multiple of three. If the sum is too large, you can repeat the process until you can tell whether it is a multiple of three. This is the basis for the solution. Since the last two digits are $23$, the sum of the digits is $2+3 = 5$ (therefore it is not divisible by three). However certain numbers can be added to make the sum of the digits a multiple of three.
$5+1 = 6$,
$5+4 = 9$, and so on. 
However since the largest four-digit number ending with $23$ is $9923$, the maximum sum is
$5+18 = 23$.
Using that process we can fairly quickly compile a list of the sum of the first two digits of the number. 
\[\{1, 4, 7, 10, 13, 16\}\]
Now we find all the two-digit numbers that have any of the sums shown above. We can do this by listing all the two digit numbers $xy$ in separate cases.
\[I. x+y = 1, \{10\} = 1\]
\[II. x+y = 4, \{13, 22, 31, 40\} = 4\]
\[III. x+y = 7, \{16, 25, 34, 43, 52, 61, 70\} = 7\]
\[IV. x+y = 10, \{19, 28, 37, 46, 55, 64, 73, 82, 91\} = 9\]
\[V. x+y = 13, \{49, 58, 67, 76, 85, 94\} = 6\]
\[VI. x+y = 16, \{79, 88, 97\} = 3\]
And finally, we add the number of elements in each set.
\[1+4+7+9+6+3 = \boxed{\textbf{(B)}\ 30}\]

Solution 2 (Medium)
A number divisible by $3$ has all its digits add to a multiple of $3.$ The last two digits are $2$ and $3$ and add up to $5 \equiv 2\ (\text{mod}\ 3).$ Therefore the first two digits must add up to $1\ (\text{mod}\ 3).$ $4$ digits (including $0$) are $0\ (\text{mod}\ 3),$ $3$ are $1\ (\text{mod}\ 3),$ and $3$ are $2\ (\text{mod}\ 3).$ The following combinations are equivalent to $1\ (\text{mod}\ 3)$:
\[0\ (\text{mod}\ 3)+1\ (\text{mod}\ 3) \equiv 1\ (\text{mod}\ 3)+0\ (\text{mod}\ 3) \equiv 2\ (\text{mod}\ 3) +2\ (\text{mod}\ 3)\]
Let the first term in each combination represent the thousands digit and the second term represent the hundreds digit. We can use this to find the total amount of four-digit numbers.
\[3\cdot3 + 3\cdot4 + 3\cdot3 = 9 + 12 + 9 = \boxed{\textbf{(B)}\ 30}\]

Solution 3 (Fast)
We have the following: $n \equiv 0 \pmod{3}$ and $n \equiv 23\pmod{100}$. Then $n = 3a = 100b+23$ for some integers $a$ and $b$.  Taking mod 3 gives:
$0 \equiv b+2 \pmod 3 \implies b\equiv 1\pmod 3$ so $b=3c+1$ for some integer $c$.  But $N = 100b+23 = 100(3c+1)+23$ so $n = 300c+123$.  Bounding this gives us:
$999 < 300c+123 < 10000$ so $876 < 300c < 9877$.  Dividing by $300$ gives $2.92 < c < 32.9222$ so $3\le c \le 32$.  This gives $32-3+1 = \boxed{\textbf{(B)} \ 30 }$

Solution 4 (Extremely Fast)
The number is in the form $xy23$
Notice that the number is divisible by $3$ if the sum of the digits of $xy23$ is divisible by $3$
Our first number that is divisible by $3$ is $1023$, next is $1323$..
Notice $xy$ goes from $10\implies 97$
Hence, there are $\frac{97-10}{3}+1=30$ distinct four digit numbers.

Solution 5 (Even Faster than Extremely Fast)
Following the form $xy23$ in Solution 4, notice that $x+y \equiv 1 \pmod 3$ to satisfy our condition. Choose a value of $x$ from 1 to 9. For $x=1, 4, 7$, there are exactly 4 values of $y: 0, 3, 6, 9$. For the remaining 6 digits, there are 3 choices for y. So our answer is $(3)(4)+(6)(3)=30$.

Solution 6 (Even Faster than Even Faster than Extremely Fast)
There are $9 \cdot 10 = 90$ possible values for the first two digits. $\frac{1}{3}$ of them yield a multiple of $3$, so the answer is $\frac{90}{3} = \boxed{\textbf{(B)}\ 30}$

Solution 7 (Easy to understand)
Denote the four digit number by $ab23_{10}$. Then it is well known the sum of the digits of an integer $N$ determine whether or not they are divisible by $3$. Thus we want $a + b + 2 + 3 \equiv 0 \pmod 3$.
$\underline{a = 1}: \implies$ $b \equiv 0 \pmod 3 \implies b = 0, 3, 6, 9$.
$\underline{a = 2}: \implies$ $b \equiv 2 \pmod 3 \implies b = 2, 5, 8$.
$\underline{a = 3}: \implies$ $b \equiv 1 \pmod 3 \implies b = 1, 4, 7$.

Now we have that when $a = 4$ this is the same case as $a = 1$ as $4 \equiv 1 \pmod 3$. Hence if we count the above number of solutions and multiply by $3$ (as $1 \leq a \leq 9$), we are home free. There are $4 + 3 + 3 = 10$ solutions. Multiplying this by $3$ yields $\boxed{30}$. $\blacksquare$

Solution 8 (Very very inefficient and slow)
Don't do this on a test unless you have lots of time left.
Pounding out all of the possible numbers and counting them, we get $\boxed{\textbf{(B)}\ 30}.$