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 350,000 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:

Re: What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6, [#permalink]
14 Sep 2009, 23:22

i took the highest possible sum (which is 42) divided by 3 which is 14. That's the max set. then worked backward to see if i can make multiples of 3 from 1-14. Basically, can you make 42 out of those numbers? yes. can you make 39,(ie 3 less so take out 4 replace by 1.) ? i did it under 2-min. but not elegant at all. anybody have better solution?

a slightly quicker way was finding the max number you can making using all 7,6,5,4,3,2 etc.. and see how the delta between those sets behave.

Re: What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6, [#permalink]
14 Sep 2009, 23:45

The sum of all elements is 45 which is divisible by 3.

So the sum of any 7-element subset will be divisible by 3 only if the sum of the remaining 2 elements is divisible by 3 too.

There are 13 ways of choosing 2 elements from that set so that their sum is divisible by 3 (i guess there are not so many of them so i just list them all out). There might be another way to count them though.

Re: What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6, [#permalink]
15 Sep 2009, 00:05

1

This post received KUDOS

thailandvc wrote:

i took the highest possible sum (which is 42) divided by 3 which is 14. That's the max set. then worked backward to see if i can make multiples of 3 from 1-14. Basically, can you make 42 out of those numbers? yes. can you make 39,(ie 3 less so take out 4 replace by 1.) ? i did it under 2-min. but not elegant at all. anybody have better solution?

a slightly quicker way was finding the max number you can making using all 7,6,5,4,3,2 etc.. and see how the delta between those sets behave.

I am not sure of this but I figured it out the answer as 12. Sum of elements = 45(9+8+...+1) Removing 2 elements at a time : 45 - 1,2 = 42 45 - 1,5 = 39 45 - 1,8 = 36 45 - 2,4 = 39 (2,1 already appeared) 45 - 2,7 = 36

and so... I ended up getting on 12. What is the OA? And unfortunately no. This was not a 2 min solution for the first time. But this seemed a more organised way to approach. Hope I am right here. _________________

GMAT offended me. Now, its my turn! Will do anything for Kudos! Please feel free to give one.

Re: What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6, [#permalink]
21 Sep 2009, 14:34

9

This post received KUDOS

1

This post was BOOKMARKED

there is actually a very quick way to answer this problem.

Remember, whenever you divide any number by 3, there are 3 possible remainders: 0, 1, or 2

Now let's figure out how many 7 digit subsets are possible from the original set. Using the combination formula, \(9C7\) we get 36. The only way that the sum of a subset will be divisible by 3 is if the remainder is 0, which is ONE THIRD OF THE TIME! So 1/3 of 36 = 12

Re: What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6, [#permalink]
02 Jan 2011, 06:42

Can I ask someone to take a look at this one? The solution provided by manifestdestiny is a very good one. But I would like someone to explain me the method suggested by gmatcouple.

Re: What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6, [#permalink]
02 Jan 2011, 09:56

Expert's post

3

This post was BOOKMARKED

nonameee wrote:

Can I ask someone to take a look at this one? The solution provided by manifestdestiny is a very good one. But I would like someone to explain me the method suggested by gmatcouple.

Also, I don't know how this could be proved:

Quote:

manifestdestiny: which is ONE THIRD OF THE TIME

Thank you.

gmatcouple solution: {1, 2, 3, 4, 5, 6, 7, 8, 9} - Sum = 45 is Divisible by 3 Now, if the sum of 7 of these numbers has to be divisible by 3, the sum of the remaining 2 numbers should also be divisible by 3. [e.g. if I take out 1 and 2 from the 9 numbers, the sum of 1 and 2 is 3 (a multiple of 3). So the leftover sum of the 7 numbers will be 42 (another multiple of 3) (Which simply implies from the fact that when two multiples of 3 are added, we get another multiple of 3)]

Now, all positive integers are of one of 3 forms: (3n) or (3n + 1) or (3n + 2)... where n is a whole number... e.g. 9 is of the form 3n, 10 is of the form 3n+1, 11 is of the form 3n + 2, 12 is again of the form 3n and so on....

Of the 9 consecutive numbers above, 3 are of the form 3n, 3 are of the form (3n + 1) and 3 are of the form (3n + 2) 3n: 3, 6, 9 3n + 1: 1, 4, 7 3n + 2: 2, 5, 8

To choose 3 numbers from these 9 such that their sum is a multiple of 3, we can either take 2 numbers which are of the form 3n (e.g. 3 + 6) or we can take 1 number of the form (3n + 1) and one number of the form (3n + 2) (e.g. 1 and 2)

2 numbers of the form 3n: 3C2 = 3 ways 1 number of the form (3n + 1) and one number of the form (3n + 2): 3C1 * 3C1 = 9 ways So there are a total of 12 ways of picking 2 numbers whose sum is a multiple of 3. Or you could enumerate all of them which is tricky since you could make a mistake in counting.

manifestdestiny: which is ONE THIRD OF THE TIME

{1, 2, 3, 4, 5, 6, 7, 8, 9} Take 2 numbers at a time: 1, 2 - Sum 3 (form 3n) 1, 3 - Sum 4 (form 3n + 1) 1, 4 - Sum 5 (form 3n + 2) 1, 5 - Sum 6 (form 3n) 1, 6 - Sum 7 (form 3n + 1) and so on Since you can select 2 numbers in 9C2 = 36 ways, a third of them will have sum of the form 3n, a third will have the sum of the form (3n + 1) and a third will have the sum of the form (3n + 2). Hence there are 12 ways of selecting 2 numbers such that their sum is of the form 3n

Note: This happens because the numbers are consecutive. It may not be true if the numbers are not consecutive. e.g. If we have 3 numbers as given below: 1, 4, 5 and we pick 2 at a time: 1+4 = 5; 1+5 = 6; 4+5 = 9 Here, 2 of the 3 sums are divisible by 3 _________________

Re: What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6, [#permalink]
05 Jan 2011, 03:58

Karishma, thank you for your reply. I understand your explanation of gmatcouple's solution. However, I don't quite understand the explanation of manifestdestiny's solution. I can see that you use the same logic as in gmatcouple's solution.

Actually, I think both your explanations are pretty much the same: instead of counting 7-member subsets divisible by 3, we count 2-member subsets divisible by 3 (this can be done because there's a bijection between these subsets).

Re: What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6, [#permalink]
05 Jan 2011, 05:04

Expert's post

nonameee wrote:

Karishma, thank you for your reply. I understand your explanation of gmatcouple's solution. However, I don't quite understand the explanation of manifestdestiny's solution. I can see that you use the same logic as in gmatcouple's solution.

Actually, I think both your explanations are pretty much the same: instead of counting 7-member subsets divisible by 3, we count 2-member subsets divisible by 3 (this can be done because there's a bijection between these subsets).

manifestdestiny says in his solution that remainder will be 0 in 1/3 of the 36 ways in which you can select 7 out of 9 digits. Selecting 7 out of 9 is same as selecting 2 out of 9 and putting them away. You split the 9 into 2 groups - 7 digits and 2 digits. It doesnt matter which one you are analyzing since, as you said, there is a bijection between the two sets (each element of the two sets is a set). It is always easier to wrap your head around 2 digits than it is to do the same for 7 digits. Remainder 0 is same as form 3n. I don't need to find remainders in every step to explain so I stick with the forms. I have given you the pattern above to explain why a THIRD of them will be divisible by 3 (which is the concept manifestdestiny uses.) Conceptually, both have given the same solution. gmatcouple executes it, manifestdestiny arrives at the answer conceptually. _________________

Re: What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6, [#permalink]
05 Jan 2011, 05:10

Karishma, thanks a lot for your explanation. I just thought that manifestdestiny arrived at his solution through analyzing 7-number subsets rather than 2-number subsets.

Re: What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6, [#permalink]
15 Nov 2014, 06:47

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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6, [#permalink]
02 Jan 2015, 08:59

1

This post was BOOKMARKED

VeritasPrepKarishma wrote:

nonameee wrote:

Can I ask someone to take a look at this one? The solution provided by manifestdestiny is a very good one. But I would like someone to explain me the method suggested by gmatcouple.

Also, I don't know how this could be proved:

Quote:

manifestdestiny: which is ONE THIRD OF THE TIME

Thank you.

gmatcouple solution: {1, 2, 3, 4, 5, 6, 7, 8, 9} - Sum = 45 is Divisible by 3 Now, if the sum of 7 of these numbers has to be divisible by 3, the sum of the remaining 2 numbers should also be divisible by 3. [e.g. if I take out 1 and 2 from the 9 numbers, the sum of 1 and 2 is 3 (a multiple of 3). So the leftover sum of the 7 numbers will be 42 (another multiple of 3) (Which simply implies from the fact that when two multiples of 3 are added, we get another multiple of 3)]

Now, all positive integers are of one of 3 forms: (3n) or (3n + 1) or (3n + 2)... where n is a whole number... e.g. 9 is of the form 3n, 10 is of the form 3n+1, 11 is of the form 3n + 2, 12 is again of the form 3n and so on....

Of the 9 consecutive numbers above, 3 are of the form 3n, 3 are of the form (3n + 1) and 3 are of the form (3n + 2) 3n: 3, 6, 9 3n + 1: 1, 4, 7 3n + 2: 2, 5, 8

To choose 3 2 numbers from these 9 such that their sum is a multiple of 3, we can either take 2 numbers which are of the form 3n (e.g. 3 + 6) or we can take 1 number of the form (3n + 1) and one number of the form (3n + 2) (e.g. 1 and 2)

2 numbers of the form 3n: 3C2 = 3 ways 1 number of the form (3n + 1) and one number of the form (3n + 2): 3C1 * 3C1 = 9 ways So there are a total of 12 ways of picking 2 numbers whose sum is a multiple of 3. Or you could enumerate all of them which is tricky since you could make a mistake in counting.

manifestdestiny: which is ONE THIRD OF THE TIME

{1, 2, 3, 4, 5, 6, 7, 8, 9} Take 2 numbers at a time: 1, 2 - Sum 3 (form 3n) 1, 3 - Sum 4 (form 3n + 1) 1, 4 - Sum 5 (form 3n + 2) 1, 5 - Sum 6 (form 3n) 1, 6 - Sum 7 (form 3n + 1) and so on Since you can select 2 numbers in 9C2 = 36 ways, a third of them will have sum of the form 3n, a third will have the sum of the form (3n + 1) and a third will have the sum of the form (3n + 2). Hence there are 12 ways of selecting 2 numbers such that their sum is of the form 3n

Note: This happens because the numbers are consecutive. It may not be true if the numbers are not consecutive. e.g. If we have 3 numbers as given below: 1, 4, 5 and we pick 2 at a time: 1+4 = 5; 1+5 = 6; 4+5 = 9 Here, 2 of the 3 sums are divisible by 3

_________________

Thanks, Lucky

_______________________________________________________ Kindly press the to appreciate my post !!

Re: What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6, [#permalink]
31 Mar 2015, 08:00

1

This post received KUDOS

what i understood from the question was that how many ways are possible that we choose 7 numbers from the set and the sum of those 7 numbers is multiple of 3? This is what I did-

Max sum is 45 when we take 9 numbers,so we have to eliminate only 2 numbers such that sum of 7 numbers is multiple of 3.Now,max sum=45=15*3 so possible sums-42(3*14),39(13*3),36,33,30,27....

now to make 42 we have to eliminate 3(2,1) To make 39,we have to eliminate 6(5,1)(4,2) To make 36,we have to eliminate 9-(8,1) (7,2) (6,3) (4,5) To make 33,we have to eliminate 12-(8,4) (7,5) (9,3) To make 30 ,we have to eliminate 15-(8,7) (9,6)

Re: What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6, [#permalink]
08 Apr 2015, 05:54

Min sum of set elements: 1+2+3+4+5+6+7 = 28, max sum of set elements: 3+4+5+6+7+8+9 = 37. We have only 3 possible sums meeting requirements: 30 33 36. For each sum there are only for possible arrangements. So the total number of sets for which the sum of elements has to be divisible by 3 is 12.

What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6, [#permalink]
09 Apr 2015, 01:47

1,2,3,4,5,6,7,8,9

we have 3 multiples (3,6,9) and 6 non-multiples (1,2,4,5,7,8). All sums of 7 numbers to be multiple we should have three multiples and sum of 4 non-multiples to be multiple of 3

3C3*6C4=1*15=15

but 3 combinations of non-multiples do not give multiple of 3 in sum: 1,2,4,7 ; 1,2,5,8 ; 2,4,5,8

Re: What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6, [#permalink]
09 Apr 2015, 20:27

Expert's post

thailandvc wrote:

What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6, 7, 8, 9} for which the sum of those 7 elements is a multiple of 3 ?

(A) 10 (B) 11 (C) 12 (D) 13 (E) 14

Quote:

Qu 1: had the question been sum to be divisible by 5 . in that case we could not use this trick. is there a variant of this technique to deal with such odd situation ?

The sum of all elements is 45. If you want to remove two elements such that the sum stays a multiple of 5, the sum of the elements removed must be a multiple of 5. Say if you remove 10, you will be left with 35. In how many ways can you make 5? 1+4, 2+3 In how many ways can you make 10? 1+9, 2+8, 3+7, 4+6 In how many ways can you make 15? 6+9, 7+8 You cannot make 20 and higher multiplies.

Total 8 ways

Quote:

Qu :2 : if we have to think on lines of unit digit , the unit digit of the sum of 7 numbers should be 0 or 5 . how should we go ahead with this method.

Explain how you intend to do it and I will tell you whether it is correct. _________________

Hey, everyone. After a hectic orientation and a weeklong course, Managing Groups and Teams, I have finally settled into the core curriculum for Fall 1, and have thus found...

MBA Acceptance Rate by Country Most top American business schools brag about how internationally diverse they are. Although American business schools try to make sure they have students from...

After I was accepted to Oxford I had an amazing opportunity to visit and meet a few fellow admitted students. We sat through a mock lecture, toured the business...