G1.3 A six-digit number 1234xy is divisible by both 8 and 9 . Given that $x+y=c$, find the value of $c$.
The number formed by last 3 digits must be divisible by 8 and the sum of digits must be divisible by 9 . i.e. $400+10 x+y$ is divisible by 8 and $1+2+3+4+x+y=9 m$
$$
\begin{array}{l}
10 x+y=8 n \ldots \ldots(1) ; x+y=9 m-10 \text {. } \\
\text { (1) }- \text { (2): } 9 x=8 n-9 m+9+1 \\
\Rightarrow n=1 \text { or } 10
\end{array}
$$
$\qquad$
When $n=1$, (1) has no solution; when $n=10, x=8, y=0 ; c=x+y=8$