四位数ABCD,倒序排列DCBA,和
= 1000A + 100B + 10C + D + 1000D + 100C + 10B + A
= 1001A + 110B + 110C + 1001D
= 11×(91A + 10B + 10C + 91D)
根据题意有:
1001A + 110B + 110C + 1001D = (A + B + C + D)×100 + 9190
901A + 10B + 10C + 901D = 9190
901(A + D) + 10(B + C) = 9190
显然A + D = 10,B + C = 18
要使四位数最小,则A = 1,D = 9,B = C = 9
这个四位数最小就是1999.
1999 + 9991 = 11990 = (1 + 9 + 9 + 9)×100 + 9190