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 $(d,a)\in\{(5,1),(6,5),(8,3),(9,7)\}$

I'm late to the game here, but I find it interesting that no one used the mod 3 identity for digit sums in base 10, which lets you go from the inside digits out, instead of the outside in.
\begin{align}
\overline{abcd} + \overline{bcd} + \overline{cd} + d + 1 &= \overline{dcba}&(\textrm{mod 3}) \\
(a + b + c + d) + (b + c + d) + (c + d) + d + 1 &= (a + b + c + d) &\quad(\textrm{mod 3}) \\
a + 2b + 3c + 4d + 1 &= a + b + c + d &(\textrm{mod 3}) \\
a + 2b + d + 1 + 3(c+d) &= a + b + c + d &(\textrm{mod 3}) \\
a + 2b + d + 1 &= a + b + c + d &(\textrm{mod 3}) \\
b + 1 &= c &(\textrm{mod 3}) \\
\end{align}
Since $a \le d$, we know that the $4d+1$ in the ones place produces a carry of at least 1, so for some carry digits $k$ and $l$:
\begin{align}
l &\in \{1, 2, 3\} \\
3c + l &= \overline{kb} \\
b + 1 &= c &(\textrm{mod 3}) \\
2b + k &= c &(\textrm{mod 10})
\end{align}
You can go through each of the 9 possible values of $c$, and for any given $c$, you only need to pick one of the three possible values of $l$ to uniquely determine $k$ and $b$. You can usually tell pretty quickly which $l$ value (if any) will give a $b$ with the right value mod 3, so this is only a little tedious by hand. This leads to only one solution, $(b, c, k, l) = (7, 5, 1, 2)$.
From this you know that $5 \le d \le 7$ (to get the carry digit $l=2$), and $d = a+1$ (because $2b+l=15$, carrying a 1). Since $a$ must be odd, $d$ must be even, so the answer is $5756$.
