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Re: Divisors of Prime numbers [#permalink]
18 Oct 2010, 04:33
Can I ask Bunuel or some other math guru to look at this problem?
I initially thought that 28 is the correct answer. But after some deliberation I don't think it's correct. E.g., 2^2 doesn't have 4 divisors, but only 3: 1, 2, 4.
Re: Divisors of Prime numbers [#permalink]
18 Oct 2010, 06:49
1
This post received KUDOS
Expert's post
The method of finding the number of factors (also called divisors) is based on the concept of finding the basic factors (which make up all other factors) and then combining them in different ways to make as many different factors as possible.
Let us say the question asks us to find the number of factors of 72. We know that 72 = 8x9 = 2^3 x 3^2 - This is called prime factorization. We have essentially brought down 72 to its basic factors. We find that 72 has three 2s and two 3s. We can combine them in various ways e.g. I could take one 2 and two 3s and make 2x3x3 = 18. Similarly, I could take three 2s and no 3 to make 2x2x2 = 8 Since we have three 2s, we can choose a 2 in four ways (take no 2, take one 2, take two 2s or take three 2s) Since we have two 3s, we can choose a 3 in three ways (take no 3, take one 3 or take two 3s) Every time we make a different choice, we get a different factor of 72. Since we can choose 2s in 4 ways and 3s in 3 ways, together we can choose them in 4x3 = 12 ways. Therefore, 72 will have 12 factors.
This is true for any positive integer N. If N = 36 x 48 = 2^2 x 3^2 x 2^4 x 3 = 2^6 x 3^3. To make factors of N, we have seven ways to choose a 2 and four ways to choose a 3. Therefore, we can make 7 x 4 = 28 combinations of these prime factors to give 28 factors of 36x48. Note: I could write 36 x 48 as 72 x 24 or 64 x 27 or 8 x 8 x 27 or many other ways. The answer doesn't change because it is still the same number N.
Generalizing, if N = p^a x q^b x r^c ... where p, q, r are all distinct prime numbers, the total number of factors of N (including 1 and N ) is (a + 1)(b + 1) (c + 1)... Remember, the '+1' is because of an option of dropping that particular prime number from our factor.
On that note, if I tell you that a positive integer N has total 7 factors, what can you say about N?
Re: Divisors of Prime numbers [#permalink]
19 Oct 2010, 00:18
nonameee wrote:
Can I ask Bunuel or some other math guru to look at this problem?
I initially thought that 28 is the correct answer. But after some deliberation I don't think it's correct. E.g., 2^2 doesn't have 4 divisors, but only 3: 1, 2, 4.
Thank you.
It is asking about divisors/factors (all) not distinct (unique) divisors/factors. HTH _________________
"Don't be afraid of the space between your dreams and reality. If you can dream it, you can make it so." Target=780 http://challengemba.blogspot.com Kudos??
Re: Divisors of Prime numbers [#permalink]
19 Oct 2010, 13:29
Expert's post
nonameee wrote:
Can I ask Bunuel or some other math guru to look at this problem?
I initially thought that 28 is the correct answer. But after some deliberation I don't think it's correct. E.g., 2^2 doesn't have 4 divisors, but only 3: 1, 2, 4.
Thank you.
Finding the Number of Factors of an Integer
First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.
The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.
Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)
Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.
Back to the original question: If p and q are prime numbers, how many divisors does the product \(p^3*q^6\) have?
According to above the number of distinct factors of \(p^3*q^6\) would be \((3+1)(6+1)=28\).
Answer: D.
As for your doubt: 2^2=4 thus 4 must have (2+1)=3 factors: 1, 2, and 4.
Re: Divisors of Prime numbers [#permalink]
19 Oct 2010, 13:38
To determine the number of positive factors of any integer:
1) Prime-factorize the integer 2) Add 1 to each exponent 3) Multiply
A question in OG11 (I think) asked for the number of positive factors of 441.
Since 441 = 3^2 * 7^2, we get (2+1)(2+1) = 9 factors.
Here's the reasoning. To determine how many factors can be created from 3^2 * 7^2, we need to determine the number of choices we have of each prime factor:
For 3, we can use 3^0, 3^1, or 3^2, giving us 3 choices. For 7, we can use 7^0, 7^1, or 7^2, giving us 3 choices.
Multiplying, we get 3*3 = 9 possible factors. _________________
Re: Prime Numbers with Exponents [#permalink]
30 Dec 2010, 20:09
1
This post received KUDOS
Expert's post
m990540 wrote:
Got another one that I'm stumped on. Thanks for the help in advance!
If P and Q are prime numbers, how many divisors does the product of (P^3)(Q^6) have?
A 9 B 12 C 18 D 28 E 36
When you need to find the number of divisors of a number, you use this approach: Break down the number into its prime factors. e.g. \(N = a^p*b^q*c^r\)... where a, b and c are all distinct prime factors of N. p, q and r are the powers of the prime factors in N Total number of divisors of N = (p+1)(q+1)(r+1)...
e.g. Total number of factors of 36 (\(= 2^2*3^2\)) is (2+1)(2+1) = 9
In this question, you need to find the total number of divisors of \((P^3)(Q^6)\) where P and Q are prime. Total number of factors = (3+1)(6+1) = 28
Note: They should have mentioned that P and Q are distinct prime numbers. If P and Q are not distinct e.g. \(3^3*3^6 = 3^9\) and its total number of divisors is (9+1) = 10. But from the options, it is obvious they intend you to take them as distinct. Still, erroneous question. _________________
Re: Divisors of Prime numbers [#permalink]
13 Apr 2014, 12:52
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Re: If p and q are prime numbers, how many divisors p^3*q^6 does [#permalink]
14 Apr 2014, 00:11
1
This post received KUDOS
Expert's post
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
Finding the Number of Factors of an Integer:
First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.
The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.
Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)
Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.
Back to the original question:
According to the above, \(p^3*q^6\) will have \((3+1)(6+1)=28\) different positive factors.
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