If p and q are prime numbers, how many divisors does the product P^3 *q^6 have?
(A) 9
(B) 12
(C) 18
(D) 28
(E) 36
(A) 9
(B) 12
(C) 18
(D) 28
(E) 36
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Thanks man. Does this formula apply to finding the factors in general? Like when you do prime factorization you can use it then to find the factors?Frankenstein wrote:Hi,
if a number N = a^p. b^q.c^r.... where a,b,c are prime factors, then the number of factors is given by:
(p+1)*(q+1)*(r+1)*...
Here the number is p^3*q^6
So, number of factors is (3+1)8(6+1) = 28
Hence, D
Yes. That is my purpose of writing the formula. I have mentioned about the prime factors thing as well.MBA.Aspirant wrote: Thanks man. Does this formula apply to finding the factors in general? Like when you do prime factorization you can use it then to find the factors?
It might be helpful to understand the reasoning behind the formula offered by Frankenstein.MBA.Aspirant wrote:Thanks man. Does this formula apply to finding the factors in general? Like when you do prime factorization you can use it then to find the factors?Frankenstein wrote:Hi,
if a number N = a^p. b^q.c^r.... where a,b,c are prime factors, then the number of factors is given by:
(p+1)*(q+1)*(r+1)*...
Here the number is p^3*q^6
So, number of factors is (3+1)8(6+1) = 28
Hence, D
The question needs to say that p and q are distinct prime numbers and that we are to count the number of positive divisors. Assuming p and q are distinct, we add 1 each exponent and multiply:If p and q are prime numbers, how many divisors does the product P^3 *q^6 have?
(A) 9
(B) 12
(C) 18
(D) 28
(E) 36
Thanks GMATGuru for your help. The part about the non-distinct factors was very helpful. I just faced it with 64 where the factors are/is 2^6. the number of factors then is (6+1) not (3+1) * (3+1)GMATGuruNY wrote:It might be helpful to understand the reasoning behind the formula offered by Frankenstein.MBA.Aspirant wrote:Thanks man. Does this formula apply to finding the factors in general? Like when you do prime factorization you can use it then to find the factors?Frankenstein wrote:Hi,
if a number N = a^p. b^q.c^r.... where a,b,c are prime factors, then the number of factors is given by:
(p+1)*(q+1)*(r+1)*...
Here the number is p^3*q^6
So, number of factors is (3+1)8(6+1) = 28
Hence, D
To determine the number of positive factors of an integer:
1) Prime-factorize the integer
2) Add 1 to each exponent
3) Multiply
For example:
72 = 2³ * 3².
Adding 1 to each exponent and multiplying, we get (3+1)*(2+1) = 12 factors.
Here's why:
To determine how many factors can be created from 72 = 2³ * 3², we need to determine the number of choices we have of each prime factor and to count the number of ways these choices can be combined:
For 2, we can use 2�, 2¹, 2², or 2³, giving us 4 choices.
For 3, we can use 3�, 3¹, or 3², giving us 3 choices.
Multiplying the number of choices we have of each factor, we get 4*3 = 12 possible factors.
Regarding the following question:
The question needs to say that p and q are distinct prime numbers and that we are to count the number of positive divisors. Assuming p and q are distinct, we add 1 each exponent and multiply:If p and q are prime numbers, how many divisors does the product P^3 *q^6 have?
(A) 9
(B) 12
(C) 18
(D) 28
(E) 36
(3+1)(6+1) = 4*7 = 28.
If p and q are not distinct, then p³q� = p³p� = p�, yielding only 9+1 = 10 factors.
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