List L: ABC, BCA, CAB
In list L above, there are 3 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? Choose the correct choice:
A) 47	B) 114	C) 152	D) 161	E) 488
Sum = (100A + 10B + C) + (100B + 10C + A) + (100C + 10A + B) = 111*(A + B + C).   So, we have that the sum will be a multiple of 111=3*37.   As for A + B + C: if A=1, B=2 and C=4, then A + B + C = 7, so the sum will also be divisible by 7 BUT if A=1, B=2 and C=8, then A + B + C = 11, so the sum will also be divisible by 11. This implies that A + B + C will not produce the same factors for all possible values of A, B and C.   Therefore, we can say that 111*(A + B + C) MUST be divisible only by 111 --&gt; so, by 1, 3, 37, and 111 --&gt; 1 + 3 + 37 + 111 = 152.   Answer: 152.
<result>C</result>
List L: ABC, BCA, CAB
In list L above, there are 3 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?