Example 3 How many elements $k$ are there in the set $\{0,1, \cdots, 2012\}$ such that $\mathrm{C}_{2012}^{k}$ is a multiple of 2012? ${ }^{[3}$
(2012, Girls' Mathematical Olympiad)
Notice that, $2012=4 \times 503$, where $p=503$ is a prime number, and
$$
\mathrm{C}_{2012}^{k}=\mathrm{C}_{4 p}^{k}=\frac{(4 p)!}{k!\cdot(4 p-k)!}=\frac{4 p}{k} \mathrm{C}_{4 p-1}^{k-1} \text {. }
$$

If $p \nmid k$, then $p \mid \mathrm{C}_{2012}^{k}$;
If $p \mid k$, then $k \in\{0, p, 2 p, 3 p, 4 p\}$.
Notice that, $\mathrm{C}_{4 p}^{0}=\mathrm{C}_{4 p}^{4 p}=1$,
$$
\begin{array}{l}
\mathrm{C}_{4 p}^{p}=\mathrm{C}_{4 p}^{3 p}=4 \mathrm{C}_{4 p-1}^{p-1} \\
=4 \times \frac{(4 p-1)(4 p-2) \cdots(3 p+1)}{(p-1)(p-2) \cdots 1} \\
\equiv 4(\bmod p), \\
\mathrm{C}_{4 p}^{2 p}=2 \mathrm{C}_{4 p-1}^{2 p-1} \\
=2 \times \frac{(4 p-1)(4 p-2) \cdots(3 p+1)}{(2 p-1)(2 p-2) \cdots(p+1)} \times 3 \times \\
\frac{(3 p-1)(3 p-2) \cdots(2 p+1)}{(p-1)(p-2) \cdots 1} \\
\equiv 6(\bmod p),
\end{array}
$$

neither of which is a multiple of $p$.
If $n$'s binary representation is
$$
n=\left(a_{r} a_{r-1} \cdots a_{0}\right)_{2}=\sum_{t=0}^{r} a_{t} 2^{t},
$$

then the power of 2 in $n!$ is
$$
\begin{array}{l}
\sum_{s=1}^{\infty}\left[\frac{n}{2^{s}}\right]=\sum_{s=1}^{\infty} \sum_{t=s}^{r} a_{t} t^{t-s}=\sum_{t=1}^{r} \sum_{s=1}^{t} a_{t} 2^{t-s} \\
=\sum_{t=1}^{r} a_{t}\left(2^{t}-1\right)=n-\sum_{t=0}^{r} a_{t}=n-S(n),
\end{array}
$$

where $[x]$ denotes the greatest integer not exceeding the real number $x$, and $S(n)$ denotes the sum of the binary digits of $n$ (i.e., the number of 1s).
$$
\begin{array}{l}
\text { Hence } 4 \nmid \mathrm{C}_{2012}^{k} \Leftrightarrow 4 \nmid \frac{2012!}{k!\cdot(2012-k)!} \\
\Leftrightarrow 4 \nmid \frac{(k+m)!}{k!\cdot m!}(m=2012-k) \\
\Leftrightarrow k+m-S(k+m) \\
\quad=k-S(k)+m-S(m)+\varepsilon(\varepsilon \in\{0,1\}) \\
\Leftrightarrow S(k+m)=S(k)+S(m)-\varepsilon(\varepsilon \in\{0,1\}) . \\
\Leftrightarrow k+m=2012=(11111011100)_{2} \text { at most }
\end{array}
$$

has one carry.
(1) If $k+m=2012$ has no carry, then the three 0s correspond to 0s in the digits of $k$ and $m$, and the eight 1s correspond to one 0 and one 1 in the digits of $k$ and $m$. Hence, there are $2^{8}$ cases, in which $\mathrm{C}_{2012}^{k}$ is odd.
(2) If $k+m=2012$ has exactly one carry, it must occur at one of the two 10s (i.e., from the 1st and 5th positions to the 2nd and 6th positions), i.e., $01+01=10$, and the other digits follow the conclusion of (1). Hence, there are $2 \times 2^{7}=2^{8}$ cases, in which $\mathrm{C}_{2012}^{k}$ is even but not a multiple of 4.
Thus, among $\mathrm{C}_{2012}^{0}, \mathrm{C}_{2012}^{1}, \cdots, \mathrm{C}_{2012}^{2012}$, there are
$$
2013-2^{8}-2^{8}=1501
$$

multiples of 4.
Additionally, $\mathrm{C}_{4 p}^{0}=\mathrm{C}_{4 p}^{4 p}=1$ is not a multiple of 4,
$\mathrm{C}_{4 p}^{p}=\mathrm{C}_{4 p}^{3 p}$ has a power of 2 of
$$
\begin{array}{l}
S(p)+S(3 p)-S(4 p) \\
=S(3 p)=S(1509)>2,
\end{array}
$$
$\mathrm{C}_{4 p}^{2 p}$ has a power of 2 of
$$
\begin{array}{l}
S(2 p)+S(2 p)-S(4 p)=S(p) \\
=S(503)=S(2012)=8,
\end{array}
$$

among which, there are three multiples of 4.
Therefore, among $\mathrm{C}_{2012}^{0}, \mathrm{C}_{2012}^{1}, \cdots, \mathrm{C}_{2012}^{2012}$, there are
$$
1501-3=1498
$$

multiples of 2012.