List L: ABC, BCA, CAB
In list L above, there are 2 positive integers, where each of A, B, and C is a different nonzero digit. Which of the following is the sum of all the positive integers that MUST be factors of the sum of the integers in list L?
Pick one:
A) 47
B) 114
C) 152
D) 161
E) 488
Let's try to sum the 2 given numbers taking their place values into account:   (100A + 10B + C) + (100B + 10C + A) + (100C + 10A + B)   = 100(A + B + C) + 10(A + B + C) + (A + B + C)   = 111*(A + B + C)   We know 111 = 37 * 3 so the sum will certainly have 1, 3, 37 and 111 as factors. 1 + 3 + 37 + 111 = 152   Note: How do we know that (A + B + C) will not give us a factor that we get every time? Try and take a few cases where A, B and C have different characteristics e.g. case 1: they are all odd, case 2: they are all even with no multiple of 3 etc. We want to see if there are cases where (A+B+C) has no common factors with other cases.   Let's say A, B and C are all odd. 1+3+5 = 9. Factors 3 and 3   A, B and C are all even. 2+4+8 = 14. Factors 2 and 7   Other factors will depend on values of A, B and C. Hence there is no other factor which we MUST get.   161
<result>D</result>
List L: ABC, BCA, CAB
In list L above, there are 2 positive integers, where each of A, B, and C is a different nonzero digit. Which of the following is the sum of all the positive integers that MUST be factors of the sum of the integers in list L?