設(shè)三連續(xù)自然數(shù)為m-1,m,m+1；4連續(xù)自然數(shù)為n-1,n,n+1,n+2；5連續(xù)自然數(shù)為p-2,p-1,p,p+1,p+2m-1+m+m+1=3m,能被3整除n-1+n+n+1+n+2=4n+1,被4除余1.p-2+p-1+p+p+1+p+2=5p,能被5整除.即求700~1000以內(nèi)能同時(shí)被3和5整除,...n-1+n+n+1+n+2=4n+1，被4除余1。這個(gè)好像是4n+2我有答案：750.810.870.930.990就是不知是怎么出來(lái)的。?????????????????????700~1000??????????3??5??????4????2????4????2????????????47~66?????????10????720??750??780??810??840??870??900??930??960??990??700??800??900?????4??????????????λ?????? ???б?4????2????У?750??810??870??930??990????5????????????? ??