14. Find the four-digit number $\overline{a b c d}$ satisfying
$$
2(\overline{a b c d})+1000=\overline{d c b a} .
$$
(For example, if $a=1, b=2, c=3$ and $d=4$, then $\overline{a b c d}=1234$.)
14. Answer: 2996 .
Rewrite the equation as the following:
$$
\begin{array}{r}
a b c c \\
a b c \quad d \\
+\quad 1 \quad 0 \quad 0 \quad 0 \\
\hline d \quad c \quad b \quad a
\end{array}
$$

Since $a$ is the last digit of $2 d, a$ is even; since $2 a+1 \leq d \leq 9, a \leq 4$. So $a=2$ or $a=4$.
If $a=4$, then $d \geq 2 a+1=9$ and thus $d=9$; but then the last digit of $2 d$ would be $8 \neq a$, a contradiction.

If $a=2$, then $d \geq 2 a+1=5$ and the last digit of $2 d$ is 2 so $d=6$. The equation reduces to
$$
\begin{array}{r}
b c \\
b \quad c \\
+\quad 1 \\
\hline 1 \quad c \quad b
\end{array}
$$

There are 2 cases: either $2 c+1=b$ and $2 b=10+c$, which has no integer solution; or $2 c+1=10+b$ and $2 b+1=10+c$, which gives $b=c=9$.