题目

有()个这样的三位数,满足abc与cba除以7的余数相同,(abc可以相同)

提问时间：2021-01-28

答案

abc＝100a＋10b＋c
cba＝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