Problem Solving

Q: In how many ways can 2310 be expressed as a product of 3 factors ?

a) 23

b) 32

c) 41

d) 46

e) 56

NB:I have the solution to this question, but its very complex. I am unable to grasp the logic of the solution. So if anybody can give a very simple and lucid solution to this problem

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

EDIT: Here's how I'd attack it, it may be wrong though, but I'll give it a try atleast.

First reduce 2310 down to its prime factors. 2, 3, 5, 7, 11

From here we need the number of ways these numbers can be multiplied together to reach 3 numbers.
Example: 2*11(22) 3*5(15) *7
22*15*7=2310 is one example.

Edit: I think I've found it, but I'm still not sure, you need to figure out the number of combinations when you pick 2 numbers from the 5 prime factors. Then figure out the combinations again for picking 2 prime factors except minus the first 2 you picked. Then the number of combinations for the last number

(combination of 2 out of 5)+(combination of 2 out of 5 minus 2)+combinations for the last number
10+8+5=23? Seems like a long shot this correct, but I'm looking forward to seeing the answer and see what an expert has to say.

UPDATE: I've found separate answers to this question when I googled it. Plus The above technique I did is wrong. I've found out. I only did combinations for a 2-2-1 grouping, neglecting a 3-1-1 grouping

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

Impressive to say the least. Looks like I failed to include 1 in with the original prime numbers, and then didn't calculate pretty much anything right either. It's interesting because I googled the answer and several math "experts" on some websites who are parading "40" around as the answer, without realizing that case 4 you mentioned is just as crucial. Well done eagle!

how did you take a1+a2+a3=1.. why dint you go upto a4 or could have restricted to a1+a2=1.. please help

Anurag@Gurome wrote:

Aman verma wrote:I have the solution to this question, but its very complex. I am unable to grasp the logic of the solution. So if anybody can give a very simple and lucid solution to this problem

2310 = 2*3*5*7*11
Now we can find the number of positive integral solutions of xyz = 2310 as follows.

We can safely assume that

• x = (2^a1)*(3^b1)*(5^c1)*(7^d1)*(11^e1)
  y = (2^a2)*(3^b2)*(5^c2)*(7^d2)*(11^e2)
  z = (2^a3)*(3^b3)*(5^c3)*(7^d3)*(11^e3)

Hence, we can writhe the following equations,

• a1 + a2 + a3 = 1
  b1 + b2 + b3 = 1
  c1 + c2 + c3 = 1
  d1 + d2 + d3 = 1
  e1 + e2 + e3 = 1

Each of the above equations will have 3 positive integral solutions.
Hence, a total 3^5 = 243 solutions

Now, this 243 solutions are ordered solutions, i.e. they contain 2*15*77, 15*2*77 etc.
To find our answer, we need to determine the number of unordered solutions.

Now, each of the sets with three different factors (like 2*15*77) can be arranged among themselves in 3! = 6 ways
And there is only one case when two factors will be equal to each other, i.e. 1*1*2310, which can be arranged in 3 ways.

Hence, number of unordered solutions = [(243 - 3)/6 + 1] = [(240/6) + 1] = [40 + 1] = 41 ways

Note : For a better understanding of the basics, please read this post first.

My goodness!
Such a simple problem is being tortured by you guys!

Hint: how can you express a^K as a product of three factors?

We can think of the factors as being separated by bars
a/aaaaaaaa..../aaaaaa....
: for three factors I have two bars that I can put on k+1 positions: before the factors, k-1 positions between the a-s and one after. ( before corresponds to a^0=1)
In how many ways can I choose to put 2 bars on k+1 positions is combinations of (k+1) taken by 2

One can easily generalize it to a prime decomposition with multiple factors.