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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Karishma

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