A + B + C + D = 9P
A - B + C - D = 1 1Q
Because A+B+C+D, A-B+C-D must be homoparity, and a+b+c+d >; A - B + C - D
Therefore, it is only possible to:
A + B + C + D = 18
A - B + C - D = 0
Do you understand:
A + C = 9
B + D = 9
A and c, b and d, parity must be different. It is also deduced that the parity of a and d, b and c must be different.
According to the "three division" of numbers divisible by 7, 13, there are:
100B + 10C + D - A
= 99B + B + D + 1 1C - A - C
= 99B + 1 1C + 9 - 9
= 1 1(9B+C) is divisible by 13, that is, 9B+C is divisible by 13 because
9B + C = 13、26、39、52、65、78
(b, c) = (1, 4) or (2,8) or (4,3) or (5,7) or (7,2) or (8,6), then correspondingly:
(d, a) = (8,5) or (7, 1) or (5,6) or (4,2) or (2,7) or (1, 3).
Have again
100A + 10B + C - D
= 99A + A + C + 1 1B - B - D
= 99A + 1 1B + 9 - 9
= 1 1(9A+B) is divisible by 7, that is, 9A+B is divisible by 7, that is, 2A+B is divisible by 7.
Replace only the above solution:
①
(B，C) = (7，2)
(D，A) = (2，7)
or
②
(B，C) = (8，6)
(D，A) = ( 1，3)
Meet. Because the numbers can't be repeated, only the solution ② is consistent.
Therefore, A = 3, B = 8, C = 6, D = 1.
ABCD = 386 1