abc=100a+10b+ccba=100c+10b+c因为abc与cba除以7的余数相同,则有abc-cba,能被7整除abc-cba=99(a-c)所以,若a=c,a=1,2,3,4,5,6,7,8,9有:101,111,121,131,141,151,161,171,181,191,202,212,222,232,242,252,262,272,282,292,。。。。。。即(i0i)其中i=1,2,3,4,5,6,7,8,9若a≠c,a=1,c=8,108,118,128,138,148,158,168,178,188,198或a=2,c=9,209,219,229,239,249,259,269,279,289,299,或a=8,c=1,801,811,821,831,841,851,861,871,881,891或a=9,c=2,902,912,922,932,942,952,962,972,982,992. 共:90+40=130个 -- by 会员 aronguo94 (2012/3/7 22:55:41)
个人觉得应该还包含0和7的组合 如果题目描述abc为3位数,则a不能为0但能为7,所以还有10组数能被7整除分别是 700,710,720,730,740,750,760,770,780,790 因此加上之前算的130个应该是140个,请讨论
另外如果a,b,c是不同数字则一开始的那90个差为0的数是不能算的,因此题目信息还有待完善 |