Problem 3. Let $a, b, c, d, x, y, z, t$ be digits such that $0<a<b \leq$ $c<d$ and

$$
\overline{d c b a}=\overline{a b c d}+\overline{x y z t}
$$

Determine all possible values of the sum

$$
S=\overline{x y z t}+\overline{t z y x}
$$
Solution. Since $\overline{x y z t}=\overline{d c b a}-\overline{a b c d}$ it follows that $t=10+a-d \ldots \ldots \ldots \ldots 1 \mathbf{p}$ Case I.

If $b=c$, then $z=y=9$ and $x=d-a-1 . \ldots \ldots \ldots \ldots \ldots \ldots . \ldots . \ldots$ p Case II.

If $b<c$, then $z=9+b-c, y=c-b-1$ and $x=d-a \ldots \ldots . .1 \mathbf{p}$ We have $\overline{x y z t}+\overline{t z y x}=1001(x+t)+110(y+z) \ldots \ldots \ldots \ldots \ldots \ldots . \ldots \mathbf{p}$ In Case I

$x+t=9$ and $y+z=18$, from which $S=9009+1980=10989 \ldots \ldots . . .1 \mathbf{p}$ In Case II

$x+t=10$ and $y+z=8$, from which $S=10010+880=10890 \ldots \ldots . .1 \mathbf{p}$