66. Let the 6-digit number $n=\overline{a b c d e f}$, $n^{\prime}=\overline{c d e f a b}$, $n^{\prime \prime}=\overline{\text { efabcd. }}$.
(1) Find $n$ such that $2 n-111111=n^{\prime}$;
(2) Find $n$ such that $3 n-222222=n^{\prime \prime}$.
Solution: (1) Let $\overline{a d}=x, \overline{c d e f}=y$. Then $2(10000 x+y)-111111=x+100 y$. Simplifying, we get $2875 x=14 y+15873$. Therefore, (1) $x$ is an odd number;
(2) $2875 x \leqslant 14 \times 9999+15873 \Rightarrow x \leqslant 54$;
(3) $7(408 x-2 y-2267)=4-x \Rightarrow 7 \mid 4-x$.

Thus, $x \in\{11,25,39,53\}$.
Upon inspection, the solutions to the equation are $n=111111,253968$, 396825, or 539682.
(2) Let $\overline{a b c d}=u, \overline{e f}=v$. Then $3(100 u+v)-222222=u+10000 v$.
Simplifying, we get $23 u=769 v+17094$, which implies $23(u-33 v-743)=10 v+5$, $23|10 v+5 \Rightarrow 23| 2 v+1, v \in\{11,34,57,80\}$.
Upon inspection, the solutions to the equation are
$n=111111,188034,264957$ or 341880.
Note: Let the repeating decimal of $\frac{1}{7}$ have a repeating cycle $A$, then the 4 solutions in (1) are $111111+i A, i=0,1,2,3$; Let the repeating cycle of $\frac{1}{13}$ be $B$, then the 4 solutions in (2) are $111111+i B, i=0,1,2,3$.
(Wu Kang, Department of Mathematics, South China Normal University, 510631)