5. If a non-negative integer $m$ and the sum of its digits are both multiples of 6, then $m$ is called a "Lucky Six Number". Find the number of Lucky Six Numbers among the non-negative integers less than 2012.
5. Solution 1 It is easy to know that a non-negative integer is a hexagonal number if and only if its last digit is even and the sum of its digits is a multiple of 6.
For convenience, let
$$
M=\{0,1, \cdots, 2011\}
$$

write each number in $M$ as a four-digit number $\overline{a b c d}$ (when it is less than four digits, add several "0"s in front of the highest digit to make it exactly four digits), and use $f(k)$ to denote the number of hexagonal numbers in $M$ whose last digit is $k$, where $k \in\{0,2,4,6,8\}$.
For $n \in \mathbf{N}$, the number of pairs $(x, y)$ that satisfy $x+y=n$, and
$$
x, y \in\{0,1, \cdots, 9\}
$$

is denoted as $p_{n}$.
Obviously, $p_{n}=\left\{\begin{array}{ll}n+1, & n=0,1, \cdots, 9 ; \\ 19-n, & n=10,11, \cdots, 18 ; \\ 0, & n \geqslant 19 .\end{array}\right.$
First, consider all hexagonal numbers $\overline{a b c k}$ less than 2000.
If $k=0$, then when $a=0$, $b+c=0,6,12,18$; when $a=1$, $b+c=5,11,17$.
$$
\begin{array}{l}
\text { Hence } f(0)=\left(p_{0}+p_{6}+p_{12}+p_{18}\right)+\left(p_{5}+p_{11}+p_{17}\right) \\
=16+16=32 .
\end{array}
$$

If $k=2$, then when $a=0$, $b+c=4,10,16$; when $a=1$, $b+c=3,9,15$.
$$
\begin{array}{l}
\text { Hence } f(2)=\left(p_{4}+p_{10}+p_{16}\right)+\left(p_{3}+p_{9}+p_{15}\right) \\
=17+18=35 .
\end{array}
$$

If $k=4$, then when $a=0$, $b+c=2,8,14$; when $a=1$, $b+c=1,7,13$.
Hence $f(4)=\left(p_{2}+p_{8}+p_{14}\right)+\left(p_{1}+p_{7}+p_{13}\right)$
$$
=17+16=33 \text {. }
$$

When $k=6,8$, similar to the cases of $k=0,2$, we have
$$
f(6)=f(0)=32, f(8)=f(2)=35 \text {. }
$$

Therefore, the number of hexagonal numbers less than 2000 is
$$
f(0)+f(2)+f(4)+f(6)+f(8)=167 \text { (numbers). }
$$

Notice that, from 2000 to 2011, there is exactly one hexagonal number 2004. Thus, the total number of hexagonal numbers is
$$
167+1=168 \text {. }
$$

Solution 2 For a non-negative integer $n$, let $S(n)$ be the sum of its digits.

First, pair all multiples of 6 (a total of 334) among the non-negative integers less than 2000 into the following 167 pairs:
$$
\begin{array}{l}
(0,1998),(6,1992),(12,1986), \\
\cdots,(996,1002) . \\
\text { For each pair of numbers }(x, y), \text { let } \\
x=\overline{a_{1} a_{2} a_{3} a_{4}, y=b_{1} b_{2} b_{3} b_{4}}
\end{array}
$$
(Agree that when $x$ or $y$ is less than four digits, add several "0"s in front of the highest digit to make it exactly four digits). Then
$$
\begin{array}{l}
1000\left(a_{1}+b_{1}\right)+100\left(a_{2}+b_{2}\right)+ \\
10\left(a_{3}+b_{3}\right)+\left(a_{4}+b_{4}\right) \\
=x+y=1998 .
\end{array}
$$

Since $x, y$ are even, $a_{4}, b_{4} \leqslant 8$.
Therefore, $a_{4}+b_{4} \leqslant 16<18$, so $a_{4}+b_{4}=8$.
Also, since $a_{3}+b_{3} \leqslant 18<19$, we have $a_{3}+b_{3}=9$.
Similarly, $a_{2}+b_{2}=9$.
Finally, we must have $a_{1}+b_{1}=1$.
Thus, $S(x)+S(y)$
$$
\begin{array}{l}
=\left(a_{1}+b_{1}\right)+\left(a_{2}+b_{2}\right)+\left(a_{3}+b_{3}\right)+\left(a_{4}+b_{4}\right) \\
=1+9+9+8=27 .
\end{array}
$$

Therefore, one and only one of $S(x)$ and $S(y)$ is a multiple of 6 (because $x, y$ are both divisible by 3, so $S(x)$ and $S(y)$ are both divisible by 3). Hence, one and only one of $x, y$ is a hexagonal number.
So, the number of hexagonal numbers less than 2000 is 167.
Also, from 2000 to 2011, there is exactly one hexagonal number 2004, so the total number of hexagonal numbers is $167+1=168$.