ABCD-ABC=DCDC
考虑个位数相等,有D=2×C或D=2×C-10
因ABCD-ABC=9×ABC+D
有9×ABC+D=DCDC
有9×ABC=DCDC-D=9×(112×D+11×C)+(D+2×C)……(2)
因此D+2×C能被9整除
1.当D=2×C
D+2×C=4×C,能被9整除,C=9,D=18,显然不行
2.当D=2×C-10
D+2×C=4×C-10,能被9整除,则2×C-5也能被9整除,尝试得C=7,D=4
代回去,容易得到
A=5,B=2
ABCD-ABC=DCDC
考虑个位数相等,有D=2×C或D=2×C-10
因ABCD-ABC=9×ABC+D
有9×ABC+D=DCDC
有9×ABC=DCDC-D=9×(112×D+11×C)+(D+2×C)……(2)
因此D+2×C能被9整除
1.当D=2×C
D+2×C=4×C,能被9整除,C=9,D=18,显然不行
2.当D=2×C-10
D+2×C=4×C-10,能被9整除,则2×C-5也能被9整除,尝试得C=7,D=4
代回去,容易得到
A=5,B=2