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)

问题描述:

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)

81个
x=a y=b z=c
(100x+10y+z)/7=14x+y余2/7x+3/7y+1/7z
(100z+10y+x)/7=14z+y余2/7z+3/7y+1/7x
则2/7x+3/7y+1/7z=2/7z+3/7y+1/7x
得x=z
这样的三位数有9*10=90个

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