If you want to find b such that $a \equiv b \pmod{11}$, you do (assuming a has 4 digits, for example):
$$a_4*10^4+a_3*10^3+a_2*10^2+a_1*10^1+a_0*1$$
Then you calculate mod 11 for each product then add everything and do mod 11 again, like you do with other numbers.
The problem is that my theory book states that $10^k\equiv (-1)^k\pmod{11}$. I don't get this. How isn't the remainder of the division of 10 by 11 10? I thought that to calculate the remainder when your divisor is greater than the dividend, you to d*0 and your remainder is your dividend. For example, 10/11 = 0*11+10. Have I been doing it wrong?
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