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In list L above, there are 3 positive integers, where each o [#permalink]
19 Jul 2011, 14:45

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57% (01:21) wrong based on 50 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: Problem solving - digits [#permalink]
19 Jul 2011, 20:29

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Expert's post

bschool83 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?

47 114 152 161 488

Let's try to sum the 3 given numbers taking their place values into account: (100A + 10B + C) + (100B + 10C + A) + (100C + 10A + B) = 100(A + B + C) + 10(A + B + C) + (A + B + C) = 111*(A + B + C)

We know 111 = 37 * 3 so the sum will certainly have 1, 3, 37 and 111 as factors. 1 + 3 + 37 + 111 = 152

Note: How do we know that (A + B + C) will not give us a factor that we get every time? Try and take a few cases where A, B and C have different characteristics e.g. case 1: they are all odd, case 2: they are all even with no multiple of 3 etc. We want to see if there are cases where (A+B+C) has no common factors with other cases. Let's say A, B and C are all odd. 1+3+5 = 9. Factors 3 and 3 A, B and C are all even. 2+4+8 = 14. Factors 2 and 7 Other factors will depend on values of A, B and C. Hence there is no other factor which we MUST get. _________________

Re: Problem solving - digits [#permalink]
20 Jul 2011, 19:52

VeritasPrepKarishma wrote:

bschool83 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?

47 114 152 161 488

Let's try to sum the 3 given numbers taking their place values into account: (100A + 10B + C) + (100B + 10C + A) + (100C + 10A + B) = 100(A + B + C) + 10(A + B + C) + (A + B + C) = 111*(A + B + C)

We know 111 = 37 * 3 so the sum will certainly have 1, 3, 37 and 111 as factors. 1 + 3 + 37 + 111 = 152

Note: How do we know that (A + B + C) will not give us a factor that we get every time? Try and take a few cases where A, B and C have different characteristics e.g. case 1: they are all odd, case 2: they are all even with no multiple of 3 etc. We want to see if there are cases where (A+B+C) has no common factors with other cases. Let's say A, B and C are all odd. 1+3+5 = 9. Factors 3 and 3 A, B and C are all even. 2+4+8 = 14. Factors 2 and 7 Other factors will depend on values of A, B and C. Hence there is no other factor which we MUST get.

Karishma - 2 questions 1) How will we find out all the possible values of (A+B+C) in 2mns ? 2) A, B and C are all even. 2+4+8 = 14. Factors 2 and 7 In this case : 111*(A + B + C) should be having the following factors 1, 3, 37 and 111 and 2,7 So sum of factors = 152+2+7=161 So are we not getting 2 ambiguous answers as per the question ? Please correct me if I am wrong. _________________

Re: Problem solving - digits [#permalink]
22 Jul 2011, 02:13

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Expert's post

krishp84 wrote:

Karishma - 2 questions 1) How will we find out all the possible values of (A+B+C) in 2mns ?

Getting 111*(A + B + C) is less than a minute job. After this, we are looking for 2 cases such that (A+B+C) share no common factor. Why? Because the question specifically says 'MUST be factors'. If I can quickly find two cases where (A+B+C) do not have any common factors, I know there is no other factor which MUST be a factor of the sum of the list. Then I will know that only the factors of 111 MUST be factors of the sum of the list. To find the two cases, I will try and take the values of A, B and C as varied as possible in the two cases. Case 1: All odd to get odd A+B+C = 1+3+5 = 9 Now in the next case I don't want 3 to be a factor. I try to find some even values 2+4+6 = 12 NO. 2+4+8 = 14 Yes. 3 is not a factor here. So there is no other number that definitely needs to be a factor of the sum of the list.

krishp84 wrote:

2) A, B and C are all even. 2+4+8 = 14. Factors 2 and 7 In this case : 111*(A + B + C) should be having the following factors 1, 3, 37 and 111 and 2,7 So sum of factors = 152+2+7=161 So are we not getting 2 ambiguous answers as per the question ? Please correct me if I am wrong.

No, in this case, there will be many other factors. 6, 21 etc... Anyway, the question says "MUST be factors". It is not NECESSARY that 2 and 7 will be factors of A+B+C (as shown above in the case where A+B+C = 9). Hence, they will not be included. _________________

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...