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How many prime factors does 1000 has? [#permalink]
17 Nov 2010, 20:45

I am asking this because I am not sure if we need to count the repeats. as 1000 = 2^3*5^3, does it have 2 prime factors, i.e. 2 and 5, or 6 prime factors, i.e. 2,2,2,5,5,5?

Re: How many prime factors does 1000 has? [#permalink]
18 Nov 2010, 05:06

Expert's post

yufenshi wrote:

I am asking this because I am not sure if we need to count the repeats. as 1000 = 2^3*5^3, does it have 2 prime factors, i.e. 2 and 5, or 6 prime factors, i.e. 2,2,2,5,5,5?

When you say prime factors of a number, you only need distinct values. So 1000 has only 2 prime factors 2 and 5, _________________

Re: How many prime factors does 1000 has? [#permalink]
18 Nov 2010, 19:19

2

This post received KUDOS

Expert's post

Werewolf wrote:

kyro wrote:

For a number say 2^A*5^B, The prime factors are 2 and 5, the number of factors are given by (A+1)*(B+1)

I didn't get the concept. Could you please elaborate with an example? Thanks in advance!

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 \(72 = 8*9 = 2^3 x 3^2\) - 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 \(N = 36 * 48 = 2^2 * 3^2 * 2^4 * 3 = 2^6 * 3^3.\) 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 \(N = p^a * q^b * r^c\) ... 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. _________________

Re: How many prime factors does 1000 has? [#permalink]
18 Nov 2010, 19:28

VeritasPrepKarishma wrote:

Werewolf wrote:

kyro wrote:

For a number say 2^A*5^B, The prime factors are 2 and 5, the number of factors are given by (A+1)*(B+1)

I didn't get the concept. Could you please elaborate with an example? Thanks in advance!

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 \(72 = 8*9 = 2^3 x 3^2\) - 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 \(N = 36 * 48 = 2^2 * 3^2 * 2^4 * 3 = 2^6 * 3^3.\) 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 \(N = p^a * q^b * r^c\) ... 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.

That was an awesome explanation! Thank you so much!

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