Macedonian Mathematical Competitions

Let the first student received $x_1$ candies, the second $x_2$ candies, ..., and the twentieth received $x_{20}$ candies. So $x_1 + x_2 + \dots + x_{20} = 800$. Let $\frac{x_k}{x_k + 2k-1} = M$ from where $x_k = (2k-1)\frac{M}{1-M}$. For the sum $x_1 + x_2 + \dots + x_{20}$ we obtain
$$x_1 + x_2 + \dots + x_{20} = \frac{M}{1-M}(1+3+5+\dots+39) = \frac{M}{1-M}\frac{40 \cdot 20}{2} = 400\frac{M}{1-M}$$

Because $x_1 + x_2 + \cdots + x_{20} = 800$ we get $400^{\frac{M}{1-M}} = 800$ or $\frac{M}{1-M} = 2$. Hence the twelfth student received $x_{12} = (2 \cdot 12 - 1) \cdot 2 = 46$.