unsolved mathematical problem

a,b,c,d are integers => 6-a, 6-b, 6-c, 6-d are also integer

Case I:
Without loss of generality, assume
(6-a)(6-b) = (6-c)(6-d) = 2 or -2
then
6-a =1
6-b=2
6-c=-1
6-d=-2

=>a = 5, b = 4, c = 7, d = 8
=> a+b+c+d=24

Case II:
Without loss of generality, assume
(6-a)(6-b) = -1 or 1(rej)
(6-c)(6-d) = -4 or 4(rej)
then
6-a = 1
6-b = -1
and
6-c = 1 (rej) or -1 (rej) or -2
6-d = 4 (rej) or -4( rej) or 2

=> a = 5, b = 7, c = 8, d = 4
=> a+b+c+d = 24

so the ans is C

Since a,b,c and d are integers, so (6 - a), (6 - b), (6 - c) and (6 - d) are also integers. For the product of 4 integers to be equal to 4, the only possibility is (2)(-2)(1)(-1), that means a,b,c and d are 4, 5, 7 and 8 , so their sum is 24 (C).

2011-05-27 20:49:11 補充：
Correction : To be exact, a,b,c and are DIFFERENT and POSITIVE integers.