由 lcflcflcf 於 星期六 十月 15, 2005 6:58 pm
1)aaa
2)aaa-bcd=efg
3)efg-hij=mno
4)(bcd)1000000+(hij)1000+mno
=(aaa-efg)1000000+(efg-mno)1000+mno
=aaa000000-efg(1000000)+efg(1000)-mon(1000)+mno
=(111a)(1000000)-efg(999000)-mno(999)
=111(a*1000000)-111(efg000)-111(9mno)
=111[a000000-efg000-9mno]
=37*3[a000000-efg000-9mno]
=37k
所以可被37整除