8. 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?
8 . Note Quality.
First, consider whether the combination number $\mathrm{C}_{2012}^{k}$ is a multiple of $p=503$.
If $p \nmid k$, then
$\mathrm{C}_{2012}^{k}=\mathrm{C}_{4 p}^{k}=\frac{(4 p)!}{k!\cdot(4 p-k)!}=\frac{4 p}{k} \mathrm{C}_{4 p-1}^{k-1}$
$\Rightarrow p\left|k \mathrm{C}_{4 p}^{k} \Rightarrow p\right| \mathrm{C}_{t_{p}}^{k}$
$\Rightarrow p \mid \mathrm{C}_{2012}^{k}$
If $p \mid k$, then there are only the following five cases:
$\mathrm{C}_{4 p}^{0}=\mathrm{C}_{s_{p}}^{4_{p}}=1$,
$\mathrm{C}_{4 p}^{p}=\mathrm{C}_{4 p}^{3 p}=\frac{4 p \cdot(4 p-1) \cdots \cdot(3 p+1)}{p \cdot(p-1) \cdots \cdot 1}$
$\equiv 4(\bmod p)$,
$\mathrm{C}_{4 p}^{2}=\frac{4 p \cdot(4 p-1) \cdots \cdot(3 p+1) \cdot 3 p \cdot(3 p-1) \cdots \cdot(2 p+1)}{2 p \cdot(2 p-1) \cdots \cdot(p+1) \cdot p \cdot(p-1) \cdots \cdot 1}$
$\equiv 6(\bmod p)$.
None of these five cases are multiples of $p$.
Second, consider the parity of the combination number.
First, examine the power of 2 in $n$ factorial.
Assume the binary representation of $n$ is
$n=\left(a, a_{r-1} \cdots a_{1} a_{0}\right)_{2}$
$\qquad$
The power of 2 in $n$ factorial is
$\left[\frac{n}{2}\right]+\left[\frac{n}{4}\right]+\cdots+\left[\frac{n}{2^{m}}\right]+\cdots$
$=\left(a_{r} a_{t-1} \cdots a_{2} a_{1}\right)_{2}+\left(a a_{1} a_{t-1} \cdots a_{3} a_{2}\right)_{2}+\cdots+a_{t}$
$\begin{aligned}= & a_{r}\left(2^{\prime}-1\right)+a_{r-1}\left(2^{2-1}-1\right) \\ & a_{1}\left(2^{2}-1\right)+a_{0}\left(2^{0}-1\right)\end{aligned}$
$=n-s(n)$,
where $s(n)=a_{r}+a_{t-1}+\cdots+a_{0}$ represents the sum of binary digits, which is the number of 1s.
If the combination number
$\mathrm{C}_{2012}^{k}=\frac{2012!}{k!\cdot(2012-k)!}$
$=\frac{(k+m)!}{k!\cdot m!}(m=2012-k)$

is odd, i.e., $\square$
the power of 2 in the numerator and denominator is the same
$\Rightarrow k+m-s(k+m)=k-s(k)+m-s(m)$
$\Rightarrow s(k+m)=s(k)+s(m)$,

which means the binary addition of $k+m=2012$ does not carry (each carry means changing a $(10)_{2}$ in a certain position to a $(1)_{2}$ in the previous position, reducing the digit sum by 1).

Notice that $2012=(11111011100)_{2}$, consisting of eight 1s and three 0s.

If $k+m=(11111011100)_{2}$ does not carry, then in the 0 positions, $k$ and $m$ must both be 0 (0=0+0), with only one choice; in the 1 positions, $k$ and $m$ are one 0 and one 1 (1=1+0 and 1=0+1), with two choices. Therefore, there are $2^{8}=256$ cases where the binary addition of $k+m=2012$ does not carry, meaning there are 256 combination numbers that are odd. The remaining $2013-256=1757$ are even.
Third, consider the combination number
$\mathrm{C}_{2012}^{k}=\frac{2012!}{k!\cdot(2012-k)!}=\frac{(k+m)!}{k!\cdot m!}$

is even but not a multiple of 4.
Then the power of 2 in the numerator is exactly one more than in the denominator
$\Rightarrow k+m-s(k+m)=k-s(k)+m-s(m)+1$
$\Rightarrow s(k+m)=s(k)+s(m)-1$,

which means the binary addition of $k+m=2012$ has exactly one carry, and this carry is from a lower position to an adjacent higher position. Since the lower position does not carry to the next higher position, the binary addition at the lower position must be $(1)_{2}+(1)_{2}=(10)_{2}$ (otherwise, it cannot carry to the next position), meaning $k$ and $m$ are both 1 at the lower position. The higher position, receiving the carry from the lower position, does not carry further, which can only be $0+0=0$, becoming 1 after receiving the carry, meaning $k$ and $m$ are both 0 at the higher position.
The binary addition at the two positions is
$(01)_{2}+(01)_{2}=(10)_{2}$.
From the binary representation of the sum
$k+m=2012=(11111011100)$

the carry can only be from the fifth and sixth positions or the ninth and tenth positions from the front. In any case, the two positions of the carry are fixed as $(01)_{2}+(01)_{2}=(10)_{2}$. The remaining positions do not carry, 0 positions can only be $0+0=0$, 1 positions can be $1+0=1$ or $0+1=1$, with two choices. There are $2^{7}=128$ cases for each position, totaling 256 cases. That is, there are 256 combination numbers that are even but not multiples of 4.
Thus, the combination numbers that are multiples of 4 are
$2013-256-256=1501$ (cases).
Finally, consider the cases that are not multiples of $p=503$.
$\mathrm{C}_{4 p}^{0}=\mathrm{C}_{4 p}^{4 p}=1$ is not a multiple of 4.
$\mathrm{C}_{s_{p}}^{s}=\mathrm{C}_{4 p}^{3 p}$ has a power of 2 of
$s(p)+s(3 p)-s(4 p)=s(3 p) \geqslant 2$.
$\mathrm{C}_{s p}^{2}$ has a power of 2 of
$s(2 p)+s(2 p)-s(4 p)$
$=s(p)=s(2012)=8$.
These three combination numbers are multiples of 4 but not multiples of $p=503$.

Therefore, the combination numbers that are multiples of 4 and also multiples of $p=503$ are $1501-3=1498$.
In summary, the combination numbers that are multiples of 2012 are