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Divisors

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Divisors

by kanha81 » Fri Apr 17, 2009 4:07 pm
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) 363

What should be my approach? How can I understand this problem better?
This is what I started with:

(p^3) * (q^6) = (p)^3 * (q^2)^3 = (p*q^2)^3

---> 3 * (2^2)
[C] ---> 2 * (3^2)
[D] ---> 7 * (2^2)

Now what???
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by DeepakR » Fri Apr 17, 2009 4:21 pm
Assume p = 2 and q = 3 hence
(p^3) * (q^6) = (2^3)*(3^6) = 8*27*27

The following numbers will exactly divide the above number and are the divisors:
2,3,4,6,8,9,27,(27^2),(8*27^2)

Hence i think it might be A.) 9.

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by ssmiles08 » Fri Apr 17, 2009 5:27 pm
I think 1 is a divisor as well. so 9+9 = 18.

What is the OA?
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Re: Divisors

by Ian Stewart » Fri Apr 17, 2009 5:50 pm
kanha81 wrote: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) 363
If p and q are *different* primes, then the answer will be 28. We don't need to list of all of the divisors here - that's a bit time consuming - but if you're interested, they are:

1, q, q^2, q^3, q^4, q^5, q^6
p, pq, pq^2, pq^3, pq^4, pq^5, pq^6
p^2, (p^2)q, (p^2)q^2, (p^2)q^3, (p^2)q^4, (p^2)q^5, (p^2)q^6
p^3, (p^3)q, (p^3)q^2, (p^3)q^3, (p^3)q^4, (p^3)q^5, (p^3)q^6

That's just all of the combinations of products (p^a)(q^b), where a = 0, 1, 2 or 3, and b = 0, 1, 2, 3, 4, 5, or 6. Because we have 4 choices for a, and 7 choices for b, we have 4x7 = 28 divisors in total. Similarly, you can count all the divisors of a number very quickly once you have its prime factorization:

-look only at the exponents
-add one to each exponent, and multiply the resulting numbers together

Here we have (p^6)(q^3). Add one to each power and multiply: 7x4 = 28.

___________

It is genuinely important that the primes p and q are different, something the question above does not make clear. If p and q are not different -- that is, if p = q -- then (p^6)(q^3) = p^9, which has only ten divisors. The question is poorly worded, at least if that's the wording from the original source. Where is it from?
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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Re: Divisors

by kanha81 » Fri Apr 17, 2009 5:59 pm
Ian Stewart wrote:
kanha81 wrote: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) 363
If p and q are *different* primes, then the answer will be 28. We don't need to list of all of the divisors here - that's a bit time consuming - but if you're interested, they are:

1, q, q^2, q^3, q^4, q^5, q^6
p, pq, pq^2, pq^3, pq^4, pq^5, pq^6
p^2, (p^2)q, (p^2)q^2, (p^2)q^3, (p^2)q^4, (p^2)q^5, (p^2)q^6
p^3, (p^3)q, (p^3)q^2, (p^3)q^3, (p^3)q^4, (p^3)q^5, (p^3)q^6

That's just all of the combinations of products (p^a)(q^b), where a = 0, 1, 2 or 3, and b = 0, 1, 2, 3, 4, 5, or 6. Because we have 4 choices for a, and 7 choices for b, we have 4x7 = 28 divisors in total. Similarly, you can count all the divisors of a number very quickly once you have its prime factorization:

-look only at the exponents
-add one to each exponent, and multiply the resulting numbers together

Here we have (p^6)(q^3). Add one to each power and multiply: 7x4 = 28.

___________

It is genuinely important that the primes p and q are different, something the question above does not make clear. If p and q are not different -- that is, if p = q -- then (p^6)(q^3) = p^9, which has only ten divisors. The question is poorly worded, at least if that's the wording from the original source. Where is it from?
Hi Ian,

Thanks for the response and detailed clarification. This is from www.ready4gmat.com website.

Is this some sort of rule to count the different powers and then multiply?

Is it same as Integral divisors?
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Re: Divisors

by Ian Stewart » Fri Apr 17, 2009 6:40 pm
kanha81 wrote: Hi Ian,

Thanks for the response and detailed clarification. This is from www.ready4gmat.com website.

Is this some sort of rule to count the different powers and then multiply?

Is it same as Integral divisors?
Yes, it is a general rule. Try it for any number you're familiar with. For example, 12 = (2^2)(3^1). Add one to each power and multiply: 2x3=6. So 12 should have 6 divisors, and indeed it does: 1, 2, 3, 4, 6, 12.

You won't see the phrase 'integral divisors' on the GMAT; they'll talk about 'positive divisors' if they want to refer to all of the divisors of a number.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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