IMC 2003 · Day 1 · P2

Let $S = a_1 + a_2 + \dots + a_{51}$. Then $b_1 + b_2 + \dots + b_{51} = 50 S$. Since $b_1, b_2, \dots, b_{51}$ is a permutation of $a_1, a_2, \dots, a_{51}$, we get $50 S = S$, so $49 S = 0$. Assume that the characteristic of the field is not equal to 7. Then $49 S = 0$ implies that $S = 0$. Therefore $b_i = -a_i$ for $i = 1, 2, \dots, 51$. On the other hand, $b_i = a_{\varphi(i)}$, where $\varphi \in S_{51}$. Therefore, if the characteristic is not 2, the sequence $a_1, a_2, \dots, a_{51}$ can be partitioned into pairs $\{a_i, a_{\varphi(i)}\}$ of additive inverses. But this is impossible, since 51 is an odd number. It follows that the characteristic of the field is 7 or 2.

The characteristic can be either 2 or 7. For the case of 7, $x_1 = \dots = x_{51} = 1$ is a possible choice. For the case of 2, any elements can be chosen such that $S = 0$, since then $b_i = -a_i = a_i$.