6. Find all four-digit numbers $\overline{a b c d}$ (in decimal notation), for which the equality

$$
\overline{a b c d}+1=(\overline{a c}+1)(\overline{b d}+1) .
$$

holds.
Solution. The equation $1000 a+100 b+10 c+d+1=(10 a+c+1)(10 b+d+1)$ can be rearranged to the form

$$
100 a(9-b)+10 a(9-d)+10 b(9-c)+c(9-d)=0 .
$$

Each of the four terms on the left side is a non-negative integer, so this equation will be satisfied if and only if each of them is equal to zero. Since $a>0$, it must be that $b=d=9$ and subsequently $c=9$. In this case, the equation is satisfied by any digit $a, a \in\{1,2, \ldots, 9\}$. Therefore, the solutions to the problem are precisely the following four-digit numbers: 1999, 2999, 3999, 4999, 5999, 6999, 7999, 8999, and 9999.