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# Divisors of Prime numbers

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Divisors of Prime numbers [#permalink]  11 Jan 2010, 11:26
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Question Stats:

30% (01:41) correct 69% (00:23) wrong based on 13 sessions
If p and q are prime numbers, how many divisors does the product p^3 X q^6 have?

A) 9
B) 12
C) 18
D) 28
E) 36
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Re: Divisors of Prime numbers [#permalink]  11 Jan 2010, 12:00
For p^mq^n, the number of divisors is (m+1)(n+1).

Thus the number of divisors is (3+1)(6+1) = 28

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Re: Divisors of Prime numbers [#permalink]  12 Jan 2010, 04:10
mrblack

Thanks! does this formula hold for only prime numbers or is it true for all numbers in general?
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Re: Divisors of Prime numbers [#permalink]  12 Jan 2010, 12:56
prime numbers only. if you have 6^2, make it 2^2 and 3^2 => therefore you have (2+1)(2+1)=9 factors
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Re: Divisors of Prime numbers [#permalink]  13 Jan 2010, 02:39
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zaarathelab wrote:
mrblack

Thanks! does this formula hold for only prime numbers or is it true for all numbers in general?

We can use the counting approach to prove the correctness of the formula

(Should keep in mind that P and Q are primes so that they cannot be further factorized)

Let us take P^3;

We can select three Ps
Or We can select two Ps
Or We can select one P
Or We can select no P

So 4 possibilities
Similarly Q^6 would have 7 possibilities

Total possibilities are 4 * 7 =24

This principle can be generalized
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Re: Divisors of Prime numbers [#permalink]  14 Jan 2010, 05:34
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mrblack and jade, correct me if i am wrong

The calculate the number of divisors of 36 -----(2^2 * 3^2) = (2+1)(2+1) = 9

Similarly to calculate for 36 * 48 ------(2^2 * 3^2) and (2^4 * 3) = (2+1)(2+1)*(4+1)(3+1) =180
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Re: Divisors of Prime numbers [#permalink]  14 Jan 2010, 06:32
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zaarathelab wrote:
mrblack and jade, correct me if i am wrong

The calculate the number of divisors of 36 -----(2^2 * 3^2) = (2+1)(2+1) = 9

Similarly to calculate for 36 * 48 ------(2^2 * 3^2) and (2^4 * 3) = (2+1)(2+1)*(4+1)(3+1) =180

Zaarathelab,

A slight issue:

48 = ( 2^4*3^1) => (4+1) * (1+1) = 10
So 36*48 = 9 *10 = 90.

Cheers
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Re: Divisors of Prime numbers [#permalink]  15 Jan 2010, 00:45
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zaarathelab wrote:
mrblack and jade, correct me if i am wrong

The calculate the number of divisors of 36 -----(2^2 * 3^2) = (2+1)(2+1) = 9

Similarly to calculate for 36*48 ------(2^2 * 3^2) and (2^4 * 3) = (2+1)(2+1)*(4+1)(3+1) =180

Yes 36 has 9 divisors

But 36*48 does not have 180 divisors(The colored step is wrong. reduce every thing to prime factors before proceeding)

36*48 is (2^2*3^2)*(2^4*3) = 2^6*3^3

So the number of divisors is (6+1)*(3+1)=28

The below is the list of all 28 divisors of 36*48
1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, 48, 54, 64, 72, 96, 108, 144, 192, 216, 288, 432, 576, 864, 1728
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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.

Thank you.
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Re: Divisors of Prime numbers [#permalink]  18 Oct 2010, 06:49
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?
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Re: Divisors of Prime numbers [#permalink]  18 Oct 2010, 07:01
Hi Guys,

Lets take an example: find the number of factors of 100

try to represent 100 in least possible prime numbers

100= 2^2 * 5^2=p^a*q^b

no of factors is given by the formula (a+1)* (b+1)=(2+1)(2+1)=9

similarly find the number of factors of p^3 * q^6--->a=3,b=6

(3+1)(6+1)=28 Ans D.
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Re: Divisors of Prime numbers [#permalink]  18 Oct 2010, 23:53
swaraj and VeritasPrepKarishma, thank you. I got it.
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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.

HTH
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Re: Divisors of Prime numbers [#permalink]  19 Oct 2010, 04:37
mrblack wrote:
For p^mq^n, the number of divisors is (m+1)(n+1).

Thus the number of divisors is (3+1)(6+1) = 28

thank you... i understood how to solve it....
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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.

As for your doubt: 2^2=4 thus 4 must have (2+1)=3 factors: 1, 2, and 4.

FOR MORE ON NUMBER THEORY CHECK: math-number-theory-88376.html

Hope it helps.
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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.
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Prime Numbers with Exponents [#permalink]  30 Dec 2010, 18:41
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
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Re: Prime Numbers with Exponents [#permalink]  30 Dec 2010, 20:09
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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

The detailed theory for this has been given here:
http://www.veritasprep.com/blog/2010/12/quarter-wit-quarter-wisdom-writing-factors-of-an-ugly-number/

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.
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Re: Prime Numbers with Exponents [#permalink]  30 Dec 2010, 20:32
Thank you VeritasPrepKarishma! Your explanations are awesome!
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Re: Prime Numbers with Exponents [#permalink]  31 Dec 2010, 00:52
Expert's post
Merging similar topics.
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Re: Prime Numbers with Exponents   [#permalink] 31 Dec 2010, 00:52
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