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???
Divisors
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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.
- Deepak
(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.
- Deepak
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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: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
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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Hi Ian,Ian Stewart wrote: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: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
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?
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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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.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?
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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