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Number Prop

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Number Prop [#permalink] New post 14 Sep 2009, 23:18
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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


Anyone has a 2-min solution?
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Re: Number Prop [#permalink] New post 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.
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Re: Number Prop [#permalink] New post 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.

So the answer is 13.
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Re: Number Prop [#permalink] New post 15 Sep 2009, 00:05
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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Re: Number Prop [#permalink] New post 15 Sep 2009, 00:37
oh yeah, there're only 12 pairs.. i counted wrongly.. wondering how many points would this simple mistake cost me in the real exam =(
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Re: Number Prop [#permalink] New post 21 Sep 2009, 14:34
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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
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Re: Number Prop [#permalink] New post 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.

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

Quote:
manifestdestiny: which is ONE THIRD OF THE TIME


Thank you.
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Re: Number Prop [#permalink] New post 02 Jan 2011, 09:56
Expert's post
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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Re: Number Prop [#permalink] New post 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).
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Re: Number Prop [#permalink] New post 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.
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Re: Number Prop [#permalink] New post 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: Number Prop   [#permalink] 05 Jan 2011, 05:10
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