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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