@spaceblastxy1428 Thanks for writing and asking about this! It's so great that you're trying to really understand it!
🏆 🎖 🎊 Instead of glossing over the solution, you are taking the time to go through everything thoroughly!
The solution is actually correct, though it skips a few steps, which might be the source of the confusion. Actually, the solution isn't exactly claiming that
\(\overline{AA4321} \equiv \overline{A[A-1]8} \equiv A(A-1)-2 \times 8 \equiv 10A+A-1-16 \pmod{7}\) is the same as the number with two digits equal to A-1, which we can denote as \(\overline{[A-1][A-1]}.\)
It is really saying that
$$ 10A+A-1-16 \pmod{7} \equiv \overline{[A-1][A-1]} \text{ } (\textcolor{red}{6 \mod 7}) $$
temp-20200806.png
This is because
$$\begin{aligned}
10A+A-1-16 \pmod{7} &= 11A - 17 \pmod{7} \\
& \equiv \textcolor{red}{[A-1][A-1] + 11} - 17 \pmod{7} \\
\end{aligned}
$$
Why did we do this? It was a clever move, because \( [A][A] \) and \( [A-1][A-1]\) are both double-digit numbers. So it makes no difference to us if we are searching for a number of the form \([A][A]\) or \( [A-1][A-1] \) -- we just think to ourselves, "I want to consider numbers of the form \( 11, 22, 33, 44, 55, \ldots \)" but, conveniently, removing an \(11\) from the \([A][A]\) allows us to cancel out some of the \( -17.\)
$$\begin{aligned}
[A-1][A-1] \textcolor{red}{+ 11} - 17 \pmod{7} & \equiv [A-1][A-1] \textcolor{red}{- 6} \pmod{7} \\
\end{aligned}
$$
This is congruent to \( 0 \pmod{7} \) when \( [A-1][A-1]\) is congruent to \(6 \pmod{7},\) and we see that when \( \boxed{A=6},\) this works, since \([6-1][6-1] = 55 \equiv 49 + 6 \pmod{7} \equiv 6 \pmod{7}.\)
🙂
