@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 $$10A+A-1-16 \pmod{7} \equiv \overline{[A-1][A-1]} \text{ } (\textcolor{red}{6 \mod 7})$$   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 $$[6-1][6-1] = 55 \equiv 49 + 6 \pmod{7} \equiv 6 \pmod{7}.$$