I need to show that for any given ISBN-10 code $\sum_{i=1}^{10} ix_i$ holds
$x_{10} = \sum_{i=1}^{9} ix_i \equiv 0 \ \ mod \ 11$
I understand that $x_{10} \equiv -\sum_{i=1}^{9} ix_i \ \ mod \ 11$ and therefore holds $x_{10} = \sum_{i=1}^{9} ix_i $.
But I don't get why $\sum_{i=1}^{9} ix_i \equiv 0 \ \ mod \ 11$ should hold. I mean if this were true we wouldn't need the check digit in the first place wouldn't we?
Do I misunderstand something or is there a mistake in the task?
ISBN-10 by example from wiki;
ISBN-10 of 0-306-40615-2:
$\begin{align} s &= (0\times 10) + (3\times 9) + (0\times 8) + (6\times 7) + (4\times 6) + (0\times 5) + (6\times 4) + (1\times 3) + (5\times 2) + (2\times 1) \\ &= 0 + 27 + 0 + 42 + 24 + 0 + 24 + 3 + 10 + 2\\ &= 132 = 12\times 11 \end{align} $
if we consider the code as $x_i$'s $x_1-x_2 x_3 x_4- x_5 x_6 x_7 x_8 x_9-x_{10}$ is the indexing. So the summations and modulus as follows.
$(10x_1+9x_2+8x_3+7x_4+6x_5+5x_6+4x_7+3x_8+2x_9+x_{10})\equiv 0 \pmod{11}$
$$\sum_{i=1}^{10} i \cdot x_{10-i+1} \equiv 0 \pmod{11}$$