For $x$ to be such a number is equivalent to $x$ being an $k^{\text {th }}$ root of unity for some $k$ up to 2012. For each $k$, there are $\varphi(k)$$primitive$$\text{k}$ {th }}$roots of unity, so the total number of roots is \sum_{k=1}^{2012} \varphi(k)$. We will give a good approximation of this number using well known facts about the Möbius function, defined by $\mu(n)=$$$\left\{\begin{array}{ll}0 & \text { if } n \text { is not squarefree } \\ (-1)^{r} & \text { if } n \text { has } r \text { distinct prime factors. }\end{array}\right.$$$. It turns out that if$$f(n)=\sum_{d \mid}$ n} g(d)$, then$$g(n)=\sum_{d \mid}$ n} $\mu(d) f\left(\frac{n}{d}\right)$$. Using this fact, since$$n=\sum_{d \mid}$ n} $\varphi(d)$$, we have that \varphi(n)=\sum_{d \mid n} \mu(d) \frac{n}{d}$. Now we have reduced the problem to estimating $\sum_{k=1}^{2012} \sum_{d \mid}$ k} $\mu(d) \frac{k}{d}$$. Let$$a=\frac{k}{d}$$, so we obtain \sum_{k=1}^{2012} \sum_{d \mid k} a \mu(d)$. We can interchange the order of summation by writing $$\begin{aligned} \sum_{d=1}^{2012} \sum_{a=1}^{\left\lfloor\frac{2012}{d}\right\rfloor} a \mu(d) & \approx \sum_{d=1}^{2012} \mu(d) \frac{1}{2}\left(\left\lfloor\frac{2012}{d}\right\rfloor\right)^{2} \\ & \approx \sum_{d=1}^{2012} \mu(d) \frac{2012^{2}}{2 d^{2}} \\ & =\frac{2012^{2}}{2} \sum_{d=1}^{2012} \frac{\mu(d)}{d^{2}} \\ & \approx \frac{2012^{2}}{2} \sum_{d=1}^{\infty} \frac{\mu(d)}{d^{2}} \end{aligned}$$ The Möbius function also satisfies the property that $\sum_{d \mid}$ n} $\mu(d)=$$$\left\{\begin{array}{ll}1 & \text { if } n=1 \\ 0 & \text { otherwise }\end{array}\right.$$, which can be seen as a special case of the theorem above (letting f(n)=1, g(n)=$$\left\{\begin{array}{ll}1 & \text { if } n=1 \\ 0 & \text { otherwise }\end{array}\right.$$$). We can then see that \left(\sum_{d=1}^{\infty} \frac{\mu(d)}{d^{2}}\right)\left(\sum_{c=1}^{\infty} \frac{1}{c^{2}}\right)=\frac{1}{1^{2}}=1$, so $\sum_{d=1}^{\infty} \frac{\mu(d)}{d^{2}}=\frac{6}{\pi^{2}}$$. Therefore, we have \sum_{k=1}^{2012} \varphi(k) \approx \frac{3}{\pi^{2}} \cdot 2012^{2}=$ 1230488.266... 2012 is large enough that all of our approximations are pretty accurate and we should be comfortable perturbing this estimate by a small factor to give bounding values.