finding number of factors of 2520

[m_a_t_h]

I'm a grade 11 student and I came across this question on my textbook.
It says to find the number of factors of 2520.

I know that there are 48 factors in total, but is there an easier way for me to find the number of factors than simply counting all the possibilities?
For example, like algebraically?

 

[pka]

I'm a grade 11 student and I came across this question on my textbook.
It says to find the number of factors of 2520.

I know that there are 48 factors in total, but is there an easier way for me to find the number of factors than simply counting all the possibilities?
For example, like algebraically?

Any positive integer has a prime factorization like this

If \(\displaystyle N=p_1^k\cdot p_2^j\cdot p_3^s\cdot p_4^t\) is the prime factorization.

Note that an factor of \(\displaystyle N\) looks \(\displaystyle p_1^a\cdot p_2^b\cdot p_3^c\cdot p_4^d\) where \(\displaystyle 0\le a\le k,~0\le b\le j,~0\le c\le s,~0\le d\le t \)

So (k+1)(j+1)(s+1)(t+1) is the number of factors of \(\displaystyle N\).

Go to the above link to verify that fact.

Use that page to put 2520 in the input window.

 

[Dale10101]

A process approach

I'm a grade 11 student and I came across this question on my textbook.
It says to find the number of factors of 2520.

I know that there are 48 factors in total, but is there an easier way for me to find the number of factors than simply counting all the possibilities?
For example, like algebraically?

Where did you get 48 from? I would be interested. I came up with 7.

 

[pka]

Where did you get 48 from? I would be interested. I came up with 7.

Did you read reply #2?

 

[Dale10101]

OK

Did you read reply #2?

Yes, late the other night, my eyes crossed and I fell asleep, had a bad dream. But looking at it again I see what you are saying, thanks.

 

[m_a_t_h]

Any positive integer has a prime factorization like this

If \(\displaystyle N=p_1^k\cdot p_2^j\cdot p_3^s\cdot p_4^t\) is the prime factorization.

Note that an factor of \(\displaystyle N\) looks \(\displaystyle p_1^a\cdot p_2^b\cdot p_3^c\cdot p_4^d\) where \(\displaystyle 0\le a\le k,~0\le b\le j,~0\le c\le s,~0\le d\le t \)

So (k+1)(j+1)(s+1)(t+1) is the number of factors of \(\displaystyle N\).

Go to the above link to verify that fact.

Use that page to put 2520 in the input window.

sorry for replying so late, but thank you for the solution.

 

[nonnb]

Using the Cartesian product of the set of powers to resolve all factors of a number

Just to clarify pka's excellent answer, and specifically taking the example of 2520

It has the following PRIME Factors : 2 x 2 x 2 x 3 x 3 x 5 x 7

= 2³ x 3² x 5 x 7


You can determine ALL factors of 2520 by taking the Cartesian product of the sets of the powers (0..N) of each prime factor

{2⁰,2¹,2²,2³}

{3⁰,3¹,3²}

{5⁰,5¹}

{7⁰,7¹}

Removing the powers, leaves the following sets:
{1,2,4,8}

{1,3,9}

{1,5}

{1,7}

You now need to do the fairly tedious exercise of multiplying out each combination of one number from each set with one of each value of every other set:
1 x 1 x 1 x 1, 2 x 1 x 1 x 1, 4 x 1 x 1 x 1, 8 x 1 x 1 x 1, .... 8 x 9 x 5 x 7

Which yields the 48 factors:

1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 12 , 14 , 15 , 18 , 20 , 21 , 24 , 28 , 30 , 35 , 36 , 40 , 42 , 45 , 56 , 60 , 63 , 70 , 72 , 84 , 90 , 105 , 120 , 126 , 140 , 168 , 180 , 210 , 252 , 280 , 315 , 360 , 420 , 504 , 630 , 840 , 1260 , 2520

Note that since your question requires just the NUMBER of factors, you can just multiply the number of values in each set together, viz, from this step:

{2⁰,2¹,2²,2³}

{3⁰,3¹,3²}

{5⁰,5¹}

{7⁰,7¹}

(this was pka's point about (3 + 1)(2 + 1)(1 + 1)(1 + 1))
The number of factors is simply:

4 x 3 x 2 x 2 = 48

(you can double check your prime factorization from a site like http://statsfiddle.info/Primes/PrimeFactors/2520)

 

[m_a_t_h]

Just to clarify pka's excellent answer, and specifically taking the example of 2520

It has the following PRIME Factors : 2 x 2 x 2 x 3 x 3 x 5 x 7

= 2³ x 3² x 5 x 7


You can determine ALL factors of 2520 by taking the Cartesian product of the sets of the powers (0..N) of each prime factor

{2⁰,2¹,2²,2³}

{3⁰,3¹,3²}

{5⁰,5¹}

{7⁰,7¹}

Removing the powers, leaves the following sets:
{1,2,4,8}

{1,3,9}

{1,5}

{1,7}

You now need to do the fairly tedious exercise of multiplying out each combination of one number from each set with one of each value of every other set:
1 x 1 x 1 x 1, 2 x 1 x 1 x 1, 4 x 1 x 1 x 1, 8 x 1 x 1 x 1, .... 8 x 9 x 5 x 7

Which yields the 48 factors:

1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 12 , 14 , 15 , 18 , 20 , 21 , 24 , 28 , 30 , 35 , 36 , 40 , 42 , 45 , 56 , 60 , 63 , 70 , 72 , 84 , 90 , 105 , 120 , 126 , 140 , 168 , 180 , 210 , 252 , 280 , 315 , 360 , 420 , 504 , 630 , 840 , 1260 , 2520

Note that since your question requires just the NUMBER of factors, you can just multiply the number of values in each set together, viz, from this step:

{2⁰,2¹,2²,2³}

{3⁰,3¹,3²}

{5⁰,5¹}

{7⁰,7¹}

(this was pka's point about (3 + 1)(2 + 1)(1 + 1)(1 + 1))
The number of factors is simply:

4 x 3 x 2 x 2 = 48

(you can double check your prime factorization from a site like http://statsfiddle.info/Primes/PrimeFactors/2520)

Thank you :-D