rishianand7 wrote: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?
47
114
152
161
188
Integer ABC = 100A + 10B + C.
Integer BCA = 100B + 10C + A.
Integer CAB = 100C + 10A + B.
Sum of all 3 integers =
= (100A + 100B + 100C) + (100B + 10C + A) + (100C + 10A + B)
= (100A + 10A + A) + (100B + 10B + B) + (100C + 10C + C)
= 111A + 111B + 111C
= 111(A+B+C).
Thus, every factor of 111 must be a factor of ABC + BCA + CAB.
Factors of 111:
1 * 111
3 * 37.
Sum of all these factors = 1+3+37+111 = 152.
The correct answer is
C.
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