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Divisors - PS

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by Anurag@Gurome » Tue Dec 06, 2011 12:30 am
karthikpandian19 wrote:If pand q are prime numbers, how many divisors does the product p^3q^6 have?

(A) 9 (B) 12(C) 18(D) 28(E) 36[/list]
When we have to find the number of divisors of a number, then factorize the number into its prime factors, like, n = a^x * b^y, where a, b are distinct prime factors of n, and x, y are powers of prime factors a and b respectively.
Then number of divisors of n = (x + 1)(y + 1)

Here number of divisors of p^3 * q^6 = (3 + 1)(6 + 1) = 4 * 7 = 28

The correct answer is D.
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by karthikpandian19 » Tue Dec 06, 2011 12:57 am
Anurag,

Can this be taken as like this also?
n = a^x * b^y implies number of divisors of n = (x + 1)(y + 1)

Is it, n = a^x implies number of divisors of n = (x + 1)

How does it work for a number 36?

36 = 2^2 * 3^2
Which implies there should be (2+1) (2+1)
ie. 9 divisors

Divisors of 36 - 2,3,4,6,9,12,18,36 (we have only 8, is "1" also to be done included in each instances.?????
Anurag@Gurome wrote:
karthikpandian19 wrote:If pand q are prime numbers, how many divisors does the product p^3q^6 have?

(A) 9 (B) 12(C) 18(D) 28(E) 36[/list]
When we have to find the number of divisors of a number, then factorize the number into its prime factors, like, n = a^x * b^y, where a, b are distinct prime factors of n, and x, y are powers of prime factors a and b respectively.
Then number of divisors of n = (x + 1)(y + 1)

Here number of divisors of p^3 * q^6 = (3 + 1)(6 + 1) = 4 * 7 = 28

The correct answer is D.
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by Anurag@Gurome » Tue Dec 06, 2011 1:02 am
Anurag,

Can this be taken as like this also?
n = a^x * b^y implies number of divisors of n = (x + 1)(y + 1)

Is it, n = a^x implies number of divisors of n = (x + 1)

How does it work for a number 36?

36 = 2^2 * 3^2
Which implies there should be (2+1) (2+1)
ie. 9 divisors

Divisors of 36 - 2,3,4,6,9,12,18,36 (we have only 8, is "1" also to be done included in each instances.?????

Yes, that's correct, when we count we have to include 1 and the number itself.
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by gunjan1208 » Tue Dec 06, 2011 1:22 am
Hi Anurag,

Can we take it as a rule everytime we see this kind of question?

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by Anurag@Gurome » Tue Dec 06, 2011 1:23 am
gunjan1208 wrote:Hi Anurag,

Can we take it as a rule everytime we see this kind of question?

Regards
Yes, definitely.
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by GMATGuruNY » Tue Dec 06, 2011 7:40 am
This question should state that p and q are DISTINCT prime numbers. I posted an explanation here:

https://www.beatthegmat.com/divisors-t85731.html
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by Elena89 » Tue Dec 13, 2011 1:29 am
karthikpandian19 wrote:If pand q are prime numbers, how many divisors does the product p^3q^6 have?

(A) 9 (B) 12(C) 18(D) 28(E) 36[/list]
The question is asking to find the total factors of the product P^3*q^6

since p and q are prime numbers, lets take p=2 and q=3

=> 2^3*3^6

total factors = (total possibilities for 2 to be in a factor of the product)*(total possibilities for 3 to be in a factor of the product)

=> total factors = (none + 3)*(none + 6)
= (1 + 3)*(1 + 6)
= 4*7
= 28

Hence the correct option: D) 28
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