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Intern  Joined: 31 Aug 2009
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What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6,  [#permalink]

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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
Veritas Prep GMAT Instructor V
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Re: What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6,  [#permalink]

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3
6
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
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Intern  Joined: 31 Aug 2009
Posts: 18
Re: What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6,  [#permalink]

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23
10
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
General Discussion
Intern  Joined: 31 Aug 2009
Posts: 46
Re: What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6,  [#permalink]

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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.
Intern  Joined: 13 May 2008
Posts: 7
Re: What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6,  [#permalink]

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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.

So the answer is 13.
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Re: What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6,  [#permalink]

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2
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. _________________
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Posts: 7
Re: What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6,  [#permalink]

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oh yeah, there're only 12 pairs.. i counted wrongly.. wondering how many points would this simple mistake cost me in the real exam =(
Director  Joined: 23 Apr 2010
Posts: 509
Re: What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6,  [#permalink]

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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.
Director  Joined: 23 Apr 2010
Posts: 509
Re: What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6,  [#permalink]

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

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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.
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Director  Joined: 23 Apr 2010
Posts: 509
Re: What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6,  [#permalink]

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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.
Director  Joined: 07 Aug 2011
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GMAT 1: 630 Q49 V27 Re: What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6,  [#permalink]

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1
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
Manager  Joined: 22 Aug 2014
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Re: What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6,  [#permalink]

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1
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)

thus,12 choices!
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Re: What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6,  [#permalink]

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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

we have 3 elements each of 3k, 3k+1, and 3k+2 form

for 7 elements to form 3k, the following is possible:

1) 2*(3k+1) + 2*(3k+2) + 3*3k --> C(3,2)*C(3,2)*C(3,3) = 9
2) 3*(3k+1) + 3*(3k+2) + 1*3k --> C(3,3)*C(3,3)*C(3,1) = 3

total = 12.
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Re: What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6,  [#permalink]

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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.
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What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6,  [#permalink]

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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

so, 15-3=12

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

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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.
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What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6,  [#permalink]

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Hi!
Bunuel , chetan4u , magoosh , kindly share your solution on this as I am sure it will another concept learning for people like me.
Thnks
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Re: What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6,  [#permalink]

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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

$$1, 4, 7$$ : They have a remainder $$1$$ when they are divided by $$3$$.
$$2, 5, 8$$ : They have a remainder $$2$$ when they are divided by $$3$$.
$$3, 6, 9$$ : They are multiples of $$3$$.

$$1 + 2 + 3 + ... + 9 = 45$$. It is a multiple of $$3$$.
We need to choose two numbers whose sum is a multiple of $$3$$ and subtract their sum from $$45$$.
There are two cases. One is choosing two numbers from $$\{ 3, 6, 9 \}$$ and other case is choosing one number from $$\{ 1, 4, 7 \}$$ and one number from $$\{ 2, 5, 8 \}$$.
The number of ways to choose two numbers from $$\{ 3, 6, 9 \}$$ is $${}_3C_2 = 3$$. And the number of ways to choose one number from $$\{ 1, 4, 7 \}$$ and one number from $$\{ 2, 5, 8 \}$$ is $${}_3C_1 * {}_3C_1 = 3*3 = 9$$.
The number of all cases is $$3 + 9 = 12$$.

The answer is C.
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Re: What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6,  [#permalink]

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Set = {1,2,3,4,5,6,7,8,9}
Modulo by 3 = {1,2,0,1,2,0,1,2,0}

so if we remove {0,0} or {1,2} combos from the set, remaining set sum will be divisible by 3

selecting {0,0}, there are three numbers with modulo 0, => 3c2 = 3
selecting {1,2}, there are three numbers with modulo 1 and three numbers with modulo 2, 3C1 * 3C1 = 9

total {0,0} + {1,2} = 12 Re: What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6,   [#permalink] 11 Mar 2018, 04:06

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