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# In list L above, there are 3 positive integers, where each

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In list L above, there are 3 positive integers, where each [#permalink]

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17 Dec 2012, 16:41
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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?

A. 47
B. 114
C. 152
D. 161
E. 488
[Reveal] Spoiler: OA

Last edited by Bunuel on 18 Dec 2012, 02:50, edited 1 time in total.
Renamed the topic and edited the question.

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Re: In list L above, there are 3 positive integers, where each [#permalink]

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18 Dec 2012, 03:02
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MOKSH 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?

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 --> so, by 1, 3, 37, and 111 --> 1 + 3 + 37 + 111 = 152.

Hope it's clear.

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Re: In list L above, there are 3 positive integers, where each [#permalink]

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18 Dec 2012, 03:21
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The list L consists of :
ABC, BCA, CAB.

Sum of these numbers will be $$111(A+B+C)$$.
Since A, B, C are distinct non-zero digits, hence minimum value of A+B+C=6. Hence the sum will be 111*6 or 111*7 or 111*8.........

So atleast $$111$$ will be a factor of the sum of the integers.

On prime factorization, we will get 37 and 3 as the prime factors.

Sum of the factors:
{a^ (p+1) - 1}{b^ (q+1) - 1}{c^ (r+1) - 1} / (a-1)*(b-1)*(c-1)

Here a=37, b=3, p=1, q=1

On applying the formula:
we get $$(1368*8)/(36*2)$$
or $$152$$.
Hope that helps.
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Re: In list L above, there are 3 positive integers, where each [#permalink]

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20 Dec 2012, 11:16
very well explained bunnel
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Re: In list L above, there are 3 positive integers, where each [#permalink]

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20 Dec 2012, 11:24
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To explain still elaborately
take any number of the format given:123+231+312=666==>111(1+2+3)
So now the question asks for which of the following will be the factors of all the three postive integers.
Only 111 can be the factor of all 3 +ve integers
So the factors of 111 are 1,3,37,111..
Making a sum ofall the factors gives the answer E
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Re: In list L above, there are 3 positive integers, where each [#permalink]

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21 Dec 2012, 04:12
Ans:

the sum of three numbers would be 100(A+B+C) +10(A+B+C) (A+B+C)= 111(A+B+C) ,

so we need to find the sum of factors of 111 which would be 1+3+37+111=152,

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Re: In list L above, there are 3 positive integers, where each [#permalink]

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13 Jul 2013, 06:47
Bumping for review and further discussion.
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Re: In list L above, there are 3 positive integers, where each [#permalink]

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12 Nov 2014, 10:22
Hello from the GMAT Club BumpBot!

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Re: In list L above, there are 3 positive integers, where each [#permalink]

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12 Dec 2015, 01:27
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: In list L above, there are 3 positive integers, where each [#permalink]

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01 Jan 2017, 22:30
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: In list L above, there are 3 positive integers, where each   [#permalink] 01 Jan 2017, 22:30
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