Example 6 How many elements $k$ are there in the set $\{0,1, \cdots, 2012\}$ such that the binomial coefficient $\mathrm{C}_{2012}^{k}$ is a multiple of 2012?
【Analysis】Perform prime factorization $2012=2^{2} \times 503$.
First consider 503 I $\mathrm{C}_{2012}^{k}$.
By Corollary 3, we know that when and only when $k$ and $2012-k$ do not produce a carry when added in base 503, $\left(503, \mathrm{C}_{2012}^{k}\right)=1$.
Writing 2012 in base 503 gives $2012=(40)_{503}$.
Let $k=\left(a_{1} a_{2}\right)_{503}$.
No carry must have $a_{2}=0, a_{1} \in\{0,1,2,3,4\}$.

Converting to decimal, we know that when and only when
$k \in\{0,503,1006,1509,2012\}$

then $503 \nmid \mathrm{C}_{2012}^{k}$.
Next, consider the opposite.
Determine $k$ such that $v_{2}\left(\mathrm{C}_{2012}^{k}\right) \leqslant 1$.
There are two cases.
(1) $v_{2}\left(\mathrm{C}_{2012}^{k}\right)=0$.
By Corollary 3, we know that $k$ and $2012-k$ do not produce a carry when added in binary, and
$$
2012=(11111011100)_{2},
$$

From this, it is easy to see that there are $2^{8}=256$ such $k$.
(2) $v_{2}\left(\mathrm{C}_{2012}^{k}\right)=1$.
By Corollary 2, we know that $k$ and $2012-k$ produce exactly one carry when added in binary.
First, if this carry occurs at a position where 2012's binary representation is 1, then the position to the right of this 1 must also produce a carry during the addition, resulting in at least two carries, which is a contradiction.
Thus, this single carry must occur at a position where 2012's binary representation is 0.
We will discuss this in binary.
(i) If the carry occurs at the first position from the right, note that $(11111011100)_{2}$ has a 0 at the second position from the right, so a carry must also occur at the second position, which is a contradiction.
(ii) If the carry occurs at the sixth position from the right, note that $(11111011100)_{2}$ has a 1 at the seventh position from the right, and combining this with the fact that only one carry occurs, we deduce that the seventh position of $k$ must be 0.
Thus, there are $2^{7}=128$ such $k$.
(iii) If the carry occurs at the second position from the right, similarly, there are $2^{7}=128$ such $k$.
Combining (1) and (2), there are 512 $k$ such that $v_{2}\left(\mathrm{C}_{2012}^{k}\right) \leqslant 1$.
It is easy to verify that $\mathrm{C}_{2012}^{0}=\mathrm{C}_{2012}^{2012} \equiv 1(\bmod 2)$,
$$
\begin{array}{l}
\mathrm{C}_{2012}^{503}=\mathrm{C}_{2012}^{1509} \equiv 0(\bmod 4), \\
\mathrm{C}_{2012}^{1006} \equiv 0(\bmod 4) .
\end{array}
$$

Thus, the number of non-negative integers $k$ in the set that satisfy $2012 \mid \mathrm{C}_{2012}^{k}$ is
$$
2013-5-512+2=1498
$$

That is, there are 1498 elements $k$ in the set $\{0,1, \cdots, 2012\}$ such that the binomial coefficient $\mathrm{C}_{2012}^{k}$ is a multiple of 2012.