<s>[INST] 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 $(d,a)\in\{(5,1),(6,5),(8,3),(9,7)\}$
 [/INST] \begin{array}{r}
abcd\\
bcd\\
cd\\
d\\
1\\
\hline
dcba
\end{array}
The leftmost $a$ in the top row cannot recieve more than $1$ as a carry. The worst cases are $(b,c)=(9,8)$ and $(b,c) = (8,9)$ as $(b,c) = (9,9)$ leads quickly to no solution. Hence $d = a + 1$ or $d = a$.
Case: $d = a + 1$
Summing the rightmost column, we get
\begin{align}
   4d+1 &= a \pmod{10}\\
   4a+5 &= a \pmod{10}\\
   3a &= 5 \pmod{10}\\
   a &= 5\\
   d &= 6
\end{align}
Carrying the $2$ and summing the third column, we get
$3c + 2 = 10k + b$ where $k \in \{0, 1, 2\}$.
Finally, from the second column, we get
$2b + k = 10 + c$.
So $2b = 10 + c - k = 6c + 4 - 20k$
$5c = 19k + 6$
The only solution is $k = 1, c = 5$, and $ b = 7$. 
So $\overline{abcd} = 5756$
Case: $d=a$.
Summing the rightmost column, we get
\begin{align}
   4d+1 &= a \pmod{10}\\
   4a+1 &= a \pmod{10}\\
   3a &= 9 \pmod{10}\\
   a &= 3\\
   d &= 3
\end{align}
Carrying the $1$ and summing the third column, we get
$3c + 1 = 10k + b$ where $k \in \{0, 1, 2\}$.
Finally, from the second column, we get
$2b + k = c$.
So $2b = c - k = 6c + 2 - 20k$
$5c = 19k - 2$
which has no solution.
</s>