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 pairwise co-prime.

A. 1027

B. 3125

C. 243

D. 729

E. None of these

Question is quite ambiguous but here is what I think:

So we are looking for all cases of N =

x *

y *

z like:

N =

(1) *

(1) *

(2^5 * 3^2 * 5^4 * 7 * 11^3);

N =

(1) *

(2^5) *

(3^2 * 5^4 * 7 * 11^3);

N =

(1) *

(2^5 * 3^2) *

(5^4 * 7 * 11^3);

N =

(2^5) *

(3^2) *

(5^4 * 7 * 11^3);

N =

(2^5 * 3^2) *

(5^4 * 7) *

(11^3);

...

In this case factors

x *

y *

z would be co-prime (wont share any common factor but 1) as each prime will be only in one factor.

As there are total of 5 primes in N then there are following cases possible:

1. {1}*{1}*{factor with all

five primes} - 1 (1*1*N);

2. {1}*{factor with

one prime}*{factor with

four primes} -

C^1_5*C^4_4=5 (

C^1_5 - choosing which prime will be in one-prime factor, the rest primes go to the third factor);

3. {1}*{factor with

two prime}*{factor with

three primes} -

C^2_5*C^3_3=10 (

C^2_5 - choosing which 2 primes will be in two-prime factor, the rest primes go to the third factor);

4. {factor with

one prime}*{factor with

one prime}*{factor with

three primes} -

C^3_5=10 (

C^3_5 - choosing which 3 primes will be in three-prime factor, from the 2 primes left one will go to the first factor and another to the second);

5. {factor with

one prime}*{factor with

two primes}*{factor with

two primes} -

C^1_5*\frac{C^2_4}{2}=15 (

C^1_5 - choosing which 1 primes will be in one-prime factor, 4 primes left can be split among two factors (into two groups) in

\frac{C^2_4}{2} ways, dividing by 2 as the order of the factors (groups) does not matter);

Total:

1+5+10+10+15=41.

No such answer among answer choices.

So I guess it's meant that the order of the factors is important, for example N=

(2^5) *

(3^2) *

(5^4 * 7 * 11^3) is different from N=

(3^2) *

(2^5) *

(5^4 * 7 * 11^3), then case 1 can be arranged in 3 ways and cases 2, 3, 4, and 5 in 3! ways:

3*1+3!(5+10+10+15)=243.

So answer: C.

But if the order of the factors is important then the problem can be solved easier: each of the five primes can be part of the first, the second or the third factor so each prime has 3 options. Total ways to distribute 5 primes would be

3*3*3*3*3=3^5=243.

Answer: C.

Anyway not a GMAT question, so I wouldn't worry about it at all.

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