Mongolian Mathematical Olympiad

Consider $f(x) = (1+x)(1+x^2)\dots(1+x^{2013}) = \sum_n a_n x^n$. Then
$$|T_r| = \sum_k [x^{7k+r}] f(x) = \sum_k a_{7k+r}.$$
Let $\epsilon = e^{\frac{2\pi i}{7}}$, i.e. $\epsilon$ is a 7th root of unity. We use the following well known facts.
$$1 + \epsilon + \epsilon^2 + \dots + \epsilon^6 = 0, \quad (1)$$
$$\sum_{k=1}^{6} \epsilon^{kr} = 6, \text{ if } 7 \mid r, \quad \sum_{k=1}^{6} \epsilon^{kr} = -1, \text{ if } 7 \nmid r. \quad (2)$$
Hence $\sum_{i=0}^{6} f(\epsilon^i) = \sum_{i=0}^{6} \sum_{n} a_n \epsilon^{ni} = \sum_{n} \sum_{i=0}^{6} \epsilon^{ni} = \sum_{7|n} 7a_n = 7|T_0|$. Analogously,
$$|T_r| = \frac{1}{7} \sum_{i=0}^{6} \epsilon^{-ri} f(\epsilon^i) = \frac{1}{7} (2^{2013} + \sum_{i=1}^{6} \epsilon^{-ri} f(\epsilon^i)).$$
Let $g(x) = x^7 - 1 = (x - \epsilon)(x - \epsilon^2)\dots(x - \epsilon^7)$. Then
$$g(-1) = -2 = (-1 - \epsilon)(-1 - \epsilon^2)\dots(-1 - \epsilon^7),$$
and
$$(1 + \epsilon)(1 + \epsilon^2)\dots(1 + \epsilon^7) = 2.$$
Since $\epsilon^7 = 1$ and $2013 = 7 \cdot 287 + 4$,
$$f(\epsilon) = (1+\epsilon)(1+\epsilon^2)\dots(1+\epsilon^{2013}) = [(1+\epsilon)(1+\epsilon^2)\dots(1+\epsilon^7)]^{287}(1+\epsilon)(1+\epsilon^2)(1+\epsilon^3)(1+\epsilon^4) = \\ = 2^{287}[(1+\epsilon)(1+\epsilon^2)(1+\epsilon^4)](1+\epsilon^3) = 2^{287}[1+\epsilon+\dots+\epsilon^7](1+\epsilon^3) = 2^{287}(1+\epsilon^3).$$
Thus $f(\epsilon^i) = 2^{287}(1 + \epsilon^{3i})$, for $1 \le i \le 6$.
$$|T_r| = \frac{1}{7} [2^{2013} + 2^{287} \cdot \sum_{i=1}^{6} (\epsilon^{-ri} + \epsilon^{(3-r)i})].$$
From (1) and (2)
$$\sum_{i=1}^{6} (\epsilon^{-ri} + \epsilon^{(3-r)i}) = \begin{cases} 6-1=5, & r=0,3 \\ -1-1= -2, & r=1,2,4,5,6 \end{cases}$$
Hence
$$|T_r| = \begin{cases} \frac{2^{2013} + 5 \cdot 2^{287}}{7}, & r = 0, 3 \\ \frac{2^{2013} - 2^{288}}{7}, & r = 1, 2, 4, 5, 6 \end{cases}$$