<s>[INST] Determine $2$ missing digits for modulo $11$ An account number verification system works as follows:
All digits in a 10-digit account number all multiplied by following weights:
$$(6 \ 3 \ 7 \ 9 \ 10 \ 5 \ 8 \ 4 \ 2 \ 1)$$
Resulting numbers are summed and divided by $11$. The result must be $0$.
Example:
Account number: $1111111111$
Sum: $1 \cdot 6 + 1 \cdot 3 + 1 \cdot 7 + 1 \cdot 9 + 1 \cdot 10 + 1 \cdot 5 + 1 \cdot 8 + 1 \cdot 4 + 1 \cdot 2 + 1 \cdot 1$
Result: $55$ 
$55 \bmod{11} = 0$: Check OK

Goal:
I need to determine $2$ out of $10$ digits ($8$ are known) on positions $6$ and $7$.
Example:
$21942xy925$

where $x$ can only be $8$ or $9$ and $y$ can be any integer.
How can I calculate the correct values for $x$, $y$ without resorting to brute-force methods?
 [/INST] It amounts to solving $\ \color{#c00}8y \equiv -k - 5x\!\pmod{11}\,$ for $\,x =8\,$ or $\,9.\,$ and $\,k=$ sum from other digits.
Notice that  $\,{\rm mod}\ 11\!:\,\ 1/\color{#c00}8 \equiv 12/{-}3 = -4,\,$ so $\, y \equiv 4k+20x\equiv 4k-2x \equiv \smash[b]{\underbrace{4k+6}_{\large x=8},\ \underbrace{4k+4}_{\large x=9}} $
The same technique works for other unknown digits.
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