Using each of the eight digits $1,3,4,5,6,7,8$ and 9 exactly once, a three-digit number $A$, two twodigit numbers $B$ and $C, B<C$, and a one-digit number $D$ are formed. The numbers are such that $A+D=B+C=143$. In how many ways can this be done?
From $A=143-D$ and $1 \leq D \leq 9$, it follows that $134 \leq A \leq 142$. The hundreds digit of $A$ is therefore 1 , and the tens digit is either 3 or 4 . If the tens digit of $A$ is 4 , then the sum of the units digits of
$A$ and $D$ must be 3 , which is impossible, as the digits 0 and 2 are not among the eight digits given. Hence the first two digits of $A$ are uniquely determined as 1 and 3 . The sum of the units digits of $A$ and $D$ must be 13. This can be achieved in six different ways as $13=4+9=5+8=6+7=7+6=8+5=9+4$.

The sum of the units digits of $B$ and $C$ must again be 13, and as $B+C=143$, this must also be true for the tens digits. For each choice of the numbers $A$ and $D$, the remaining four digits form two pairs, both with the sum 13. The units digits of $B$ and $C$ may then be chosen in four ways. The tens digits are then uniquely determined by the remaining pair and the relation $B<C$. The total number of possibilities is therefore $6 \cdot 4=24$.