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Question:

let $x_{i}=1$ or $-1$,$i=1,2,\cdots,1990$, show that $$x_{1}+2x_{2}+\cdots+1990x_{1990}\neq 0$$

this problem it seem is easy,But I think is not easy.

I think note $$1+2+3+\cdots+1990\equiv \pmod { 1990}?$$

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  • $\begingroup$ is there a low for $x_i$ that is, we can know the value of $x_8$ for exemple? or it is not said? $\endgroup$ – math_man Sep 23 '14 at 15:46
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suppose all the $x_i$ are $1$, then we have $$1+2+3+\cdots+1990 = \frac{1991\times 1990}{2}$$ is odd.

Then each time you change one of $x_i$ from $1$ to $-1$, you change the sum by an even number, so the sum is always odd

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Consider the equation modulo 2. Regardless of whether $x_i = 1,or -1$, $x_i\equiv 1\pmod{2}$.

Thus $\sum_{i=1}^{1990}{ix_i}\equiv 1+0+1+\cdots +0\pmod{2}$, where there are $\dfrac{1990}{2}=995$ $1$'s in the summation.

We conclude that $\sum_{i=1}^{1990}{ix_i}\equiv 1\pmod{2}$, so it definitely cannot be equal to $0$.

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You are summing up $\frac{1990\times 1991}{2}=1981045$ numbers each of which is odd. The result must be odd and in particularly cannot be $0$.

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