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.

_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:

PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.

What are GMAT Club Tests?

25 extra-hard Quant Tests

GMAT Club Premium Membership - big benefits and savings