Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
In list L above, there are 3 positive integers, where each o [#permalink]
19 Jul 2011, 14:45
2
This post received KUDOS
5
This post was BOOKMARKED
00:00
A
B
C
D
E
Difficulty:
75% (hard)
Question Stats:
44% (01:54) correct
56% (01:23) wrong based on 69 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
7
This post received KUDOS
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
3
This post received KUDOS
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. _________________
As I’m halfway through my second year now, graduation is now rapidly approaching. I’ve neglected this blog in the last year, mainly because I felt I didn’...
Perhaps known best for its men’s basketball team – winners of five national championships, including last year’s – Duke University is also home to an elite full-time MBA...
Hilary Term has only started and we can feel the heat already. The two weeks have been packed with activities and submissions, giving a peek into what will follow...