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

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.

oh yeah, there're only 12 pairs.. i counted wrongly.. wondering how many points would this simple mistake cost me in the real exam =(

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
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save 10% on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews

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.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Save 10% on Veritas Prep GMAT Courses And Admissions Consulting
Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options.

Veritas Prep Reviews