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 $(d,a)\in\{(5,1),(6,5),(8,3),(9,7)\}$
 A: \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.
A: 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.
A: $$\begin{equation}\overline{abcd}+\overline{bcd}+\overline{cd}+d+1=\overline{dcba}\end{equation}$$
Expanding the numbers you get
$$999a+190b-70c-996d+1=0$$
modulo $10$ you get
$$-a+4d+1=0$$
which leads to the possibilities you listed.
Moreover
$$ 999(a-d) =70c+3d-190b-1 \in (-1711, 657)$$
This yields $$a-d \in \{ -1, 0 \}$$
This reduces the pairs to $a=5, d=6$. Plugging in in the original equation we get
$$999*5+190b-70c-996*6+1=0$$
$$190b-70c-980=0$$
$$19b-7c=98$$
As $7|98$ and $7|7c$ you have $7|19b$ and hence $7|b$.
This yields $b=0$ or $b=7$. $b=0$ is not possible, while $b=7$ yields
$$19*7-7c=14*7 \Rightarrow 19-c=14$$
A: 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 $5756$.
