IMO: C
OA ???

Are you sure you copied the problem correctly? I don't recognize the term "pair wise co-prime".

Tani Wolff - Kaplan wrote:Are you sure you copied the problem correctly? I don't recognize the term "pair wise co-prime".

Pair wise co primes, the way I understand is,

2^5 --> (5,2) are co primes because GCD is 1. Similarly, (3,2) and (5,4) are co primes because GCD is 1.

OR

32*9*625 --> (32,9), (9,625) & (625,32) must have the GCD as 1.

So we can have many different ways of representing the numbers as co primes. I guess answer is E.??

but the greatest common denominator of any two primes is 1, by definition.

But co primes need not have the two numbers as primes and can have the GCD as 1, correct?

ex GCD(5,6) is 1.

Ian Stewart wrote:

pzazz12 wrote:A number N when expressed as product of prime factors gives N= 2^5 * 3^2* 5^4 * 7 * 11^3. Find the number of ways the number N can be expressed as a product of three factors such that the factors are pair wise co-prime.

A.1027
B.3125
C.243
D.729
E.None of these

This is not a GMAT question; you don't need to know what 'coprime' means, and you won't see anything similar on the real test. It's also badly worded; the question does not make clear whether the order of the factors we are multiplying is important. We can, for example, write 15 as 3*5 or as 5*3; would these be counted as two different ways to express 15 as a product in this question, or as just one way?

So it's not a good question, and this solution is for interest only - if you're preparing for the GMAT, you can ignore it: two numbers are 'coprime', or 'relatively prime', if their GCD is 1. So 15 and 16, for example, are relatively prime. Here, if we assume the order of our factors matters, and if we want a*b*c = N, and a, b and c have no prime factors in common, then only one of them can be divisible by 2, so only one of them can be divisible by the 2^5 we need, and we have three choices for which of our letters will be divisible by 2^5. Similarly, only one of them can be divisible by 3, so we have three choices for which of our letters will be divisible by 3^2, and so on. So for each prime, we have three choices, and the answer will be 3^5 = 243.

nice work...
i solved it by guess work
since total no of factors for n wud be = 6 * 3 * 5 * 2 * 4 = 729
and the desired number of combinations wud be less that this number
243 was the only choice..

I don't know the OA?

what is the correct option?

I think the correct expression of "pairs" should be "sets" or "triads" for the question to be meaningfully interesting. A pair of the base and its power to be coprime is simply a ridiculous misunderstanding.

However, come to the solution.

There are 5c2 = 10 triads with co-primacy.

Simply, you have to take the (highest degree of any 2 bases) namely (2^5, 3^2, 5^4, 7^1 and 11^3) to get the firts 2 factors of N and the 3rd factor will be the product of the rest. that is,

(factor#1)(factor#2)(3rd factor) =N

triad one : 2^5 * 3^2 * (5^4*7*11^3) = N
triad two : 2^5 * 5^4 * (3^2*7*11^3) = N
triad thri : 2^5 * 7 * (3^2*5^4*11^3) = N
triad four : 2^5 * 11^3 * (3^2*7*11^3) = N
triad five : 3^2 * 5^4 * (2^5*7*11^3) = N
triad six : 3^2 * 7 * (2^5*5^4*11^3) = N
triad svn : 3^2 * 11^3 * (2^5*5^4*7) = N
triad eigt : 5^4 * 7 * (2^5*3^2*11^3) = N
triad nine: 5^4 * 11^3 * (2^5*3^2*7) = N
triad ten : 7 * 11^3 * (2^5*3^2*5^4) = N

As we can not use any prime in more than one factor and make the product equal N, we have no other options. So. these ten sets make the exhaustive list of all such coprime sets of factors. There is no other way! The answer should be 10.

It's from Towhid, Institute of Business Administration, the University of Dhaka.