～～質數整除問題～～

易知111...111 ( p-1 個‘1’) =(10^(p-1)-1)/9
由於p不小於 7，即p不會是2或5，又即p和10互質，由費馬小定理可知10^(p-1)除p餘1，所以(10^(p-1)-1)被p整除，而且p不會是3，所以把(10^(p-1)-1)除以9不影響它被p的整除性，所以命題成立。

註：費馬小定理即當p是質數且a和p互質時，(a^(p-1))-1可被p整除

Let be the equivalent class of k mod p ,
then Fp = { <0> , <1> , <2> , ... , , } is a finite field of order p
and Rp = { <1> , <2> , ... , , } is a multiplicative ring of order p-1.
Since p>11 , <10> is well-defined and <1> is not equal to <10>.
Now, < 111...111 > = <10>^(p-2) +... + <10>^2 + <1> = <0> because
1) Every elements in Rp\{ <0> , <1> } are of order p-1 and hence
2) <10>^(p-2) ,... , <10>^2 and <1> are just p-1 distinct " non-zero " elements in Rp and hence in Fp.
3) Sum of all " non-zero " elements in Fp is equal to <0>.

Since <1 11...111 > = <0> , p must divide 111...111

2006-11-26 23:07:42 補充：
Hi-ya, not fast enough........

2006-11-30 15:31:45 補充：
SORRY" Let < k > be the equivalent class of k mod p , "