jdciaravino wrote:Man this is tough, I assume it solvable by prime factorization and then usage of combinations/permutations. Still trying to crack it. if I found this on a test I'd be screwed, taking me well over 2 mins to solve it
I don't think you would ever encounter such a complex question in the GMAT. This one tests so many things, its next to impossible to solve it within 2 minutes and think of all the possibilities. Anyway, here's how I did it.
First we know that 2310 = 1*2*3*5*7*11
Now we need to find in how many ways can we express the number as a product of 3 numbers.
I divide this into 4 cases. I will be solving it using the fundamental principle of counting.
Case 1: Divide the numbers into 3 pairs. (and 2310 is then the product of these 3 pairs). If we were to arrange the numbers, we would have 6 choices for the first number, 5 for the second etc.
Here we have for the first pair, 6*5 ways of arranging, for the second pair 4*3 ways of arranging, for the third 2*1 ways of arranging.
Then we have something like this : 6*5| 4*3 | 2*1
Now remember we are only selecting, and not arranging. So within each group, the order shouldn't matter. Hence the no of ways is (6*5)/(2!) * ((4*3)/2!) * ((2*1)/2!). But wait, we are not finished yet. The pairs (of 2, 2, and 2) should not have any order among themselves either. So the three "slots" _ _ | _ _ | _ _ will need to be compensated for as well. Hence we divide what we have with number of ways of arranging these 3 slots = 3!. Hence we get:
No. of ways of creating 3 pairs out of 6 numbers = (1/3!) * ( (6*5)/(2!) * ((4*3)/2!) * ((2*1)/2!))
= 6!/(8*6) = 5!/8 = 15.
Just solving this by itself will be a 700+ level GMAC question. Anyway, let's continue.
Case 2: Divide the numbers into 3 groups of 3, 2, 1.
In the same way as above, we have the number of ways of doing this = (1/3!)* ( (6*5*4)/3! * (2*1)/2!) = 6!/(2*3!*3!) = 6*5*4*3/(6*6) = 10.
Case 3: Divide the numbers into 3 groups of 4, 1, 1.
Now this one is relatively benign to calculate. We only have to select 4 numbers and the other 2 select themselves. No. of ways of doing so = 6C4 = (6*5)/2 = 15.
Case 4: And here's the kicker. This case can be easily missed by many a people. You might ask yourself, "Hey, there should be no other case. We have already considered all possible ways of grouping". Alas, there is one more. And it has only choice. Can you guess it?
It is
1*1*2310.
Hence number of ways of case 4 = 1
So, the overall number of factors = 15+10+15+1 = 41.
C is the correct answer.
As I said before, even the parts of this question require GMAT level sub-questions to be solved. And there is an additional trick of recognizing not only that 1 is a factor, but also that 1*1*2310 is a possible solution. I would be highly surprised if this showed up in an exam. But then again, the folks at GMAC would know better. And maybe, there is an easier way of doing this which renders this an easier question.
Let me know if this helps
