Tough PS question

[rishianand7]

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

[GMATGuruNY]

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.

[rishianand7]

Hi Mitch,

Thanks a lot for all the solutions that uve been posting so far!

I saw this question in a practice test I gave recently. Do you think it is reflective of the GMAT?
Are such questions asked in the actual thing or is it too vague/hard?

Regards,
Rishi

[Sul]

Hi GMATguru,

I think we are missing on the factors that would come with A+B+C; so if A-1,B-2,C-3 then 6 (along with some other factors) would also be a factor of (ABC+ BCA+ CAB). Hence I think the question is incomplete.

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.

[[email protected]]

Hi Sul and rishianand7,

Sul, notice how the prompt asks for what MUST be factors. There will certainly be some factors that show up depending on what digits A, B and C represent, but the question is only interested in the factors that will ALWAYS be there. Mitch's explanation shows you the factors that show up no matter what A, B and C are; thus, it's the correct answer.

Rishianand7, the individual concepts referenced in this prompt (number properties, factors, digits, etc.) will all show up on the GMAT. However, this is a really high-level packaging of the concepts and you're not likely to see this on test day.

GMAT assassins aren't born, they're made,
Rich