Find numbers $\overline{abcd}$ so that $\overline{abcd}+\overline{bcd}+\overline{cd}+d+1=\overline{dcba}$ Find the numbers $\overline{abcd}$, with digits not null that satisfy the equality
\begin{equation}\overline{abcd}+\overline{bcd}+\overline{cd}+d+1=\overline{dcba}\end{equation}
where 
\begin{equation}\overline{abcd}=1000a+100b+10c+d\end{equation}
I see that $4d+1=\overline{..a},\ d\neq0,\ a\neq 0,\ a<d$  but that gives too many pairs to start with $(d,a)\in\{(5,1),(6,5),(8,3),(9,7)\}$

As you observed, $$(d,a) \in \{(5,1), (8,3), (9,7), (6, 5)\}$$ but notice also that $d = a+k_3$, where $k_3$ is the carry resulting from $bcd+cd+d+1$.  This quantity is at most $999+99+9+1 = 1108$ so the carry is either $0$ or $1$ and thus $d = a$ or $d=a+1$. Combined with the foregoing, this means that $(d,a) = (6, 5)$.
Now we have $$\overline{5bc6} + \overline{bc6} + \overline{c6} + 6 + 1 = \overline {6cb5}$$ and at this point anything we do will solve the problem as there are only 9 possibilities to check. Remembering we need $10\le 2b+k_2 \le 19$ where $k_2$ is the carry resulting from $\overline{c6} + 7$, so $b\ge 5$, we do trial and error on $b\in \{5,6,7,8,9\}$ and immediately find $b=5, c=7$.
Computer search confirms the uniqueness of the answer.
