I1.4 設 $d=\overline{x y z}$ 為一不能被 10 整除的三位數。若 $\overline{x y z}$ 與 $\overline{z y x}$ 之和可被 $c$ 整除, 求此整數的最大可能值 $d$ 。
Let $d=\overline{x y z}$ be a three-digit integer that is not divisible by 10 . If the sum of integers $\overline{x y z}$ and $\overline{z y x}$ is divisible by $c$, determine the greatest possible value of such an integer $d$.
$$
\begin{array}{l} 
100 x+10 y+z+100 z+10 y+x \\
= 100(x+z)+20 y+x+z \\
= 101(x+z)+20 y \\
= 99(x+z)+22 y+2(x+z-y), \text { which is divisible by } 11 \\
x+z-y \text { is a multiple of } 11 \\
x+z-y=0 \text { or } 11
\end{array}
$$

To maximize $d, x$ should be as large as possible
$$
\begin{array}{l}
x+z=11+y \\
x=9, z=9, y=7
\end{array}
$$

The greatest possible value of $d=979$
Check: $\overline{x y z}+\overline{z y x}=979+979=1958=11 \times 178$, which is divisible by 11 .