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In list L above, there are 3 positive integers, where each [#permalink]
17 Dec 2012, 15:41

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55% (03:12) correct
45% (02:04) wrong based on 100 sessions

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?

Re: In list L above, there are 3 positive integers, where each [#permalink]
18 Dec 2012, 02:02

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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.

Re: In list L above, there are 3 positive integers, where each [#permalink]
18 Dec 2012, 02:21

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Expert's post

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. _________________

Re: In list L above, there are 3 positive integers, where each [#permalink]
20 Dec 2012, 10:24

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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]
12 Nov 2014, 09:22

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