Aliquot Parts

[Explain this!]

Is there a fast and simple method that can be used to determine the aliquot parts, whole positive numbers
that can divide into another whole positive number?

For example, the aliquot parts for 10 are 1, 2, 5, 10. One method is to divide
the number by whole numbers from 1 up to and including that whole number.
If the number is large, for example 1000, the division would be long and
tiresome.

Note: I am not a high school or college student.

 

[lev888]

Context? Are you looking for an algorithm?
Start with prime factorization, then go through all unique combinations of prime factors.

 

[Dr.Peterson]

Here's an example of an algorithm, applied to a slightly harder case, 24.

Since 24 = 2^3 * 3^1, any divisor must be 2^m * 3^n, where 0<=m<=3 and 0<=n<=1. So you would loop through all such pairs of exponents:
m=0, n=0 --> 1
m=0, n=1 --> 3
m=1, n=0 --> 2
...

This also immediately gives you the number of divisors.

 

[Explain this!]

Here's an example of an algorithm, applied to a slightly harder case, 24.

Since 24 = 2^3 * 3^1, any divisor must be 2^m * 3^n, where 0<=m<=3 and 0<=n<=1. So you would loop through all such pairs of exponents:
m=0, n=0 --> 1
m=0, n=1 --> 3
m=1, n=0 --> 2
...

This also immediately gives you the number of divisors.

I want to thank you for the reply, but I still need some clarification. Can you use the number 24 example and show the divisors by using your example? This process may help me to understand more fully.

 

[Dr.Peterson]

I showed three of them; you aren't able to continue?

I'll finish the table, with more work visible:

m=0, n=0 --> divisor 2^m * 3^n = 2^0 * 3^0 = 1*1 = 1
m=0, n=1 --> divisor 2^m * 3^n = 2^0 * 3^1 = 1*3 = 3
m=1, n=0 --> divisor 2^m * 3^n = 2^1 * 3^0 = 2*1 = 2
m=1, n=1 --> divisor 2^m * 3^n = 2^1 * 3^1 = 2*3 = 6
m=2, n=0 --> divisor 2^m * 3^n = 2^2 * 3^0 = 4*1 = 4
m=2, n=1 --> divisor 2^m * 3^n = 2^2 * 3^1 = 4*3 = 12
m=3, n=0 --> divisor 2^m * 3^n = 2^3 * 3^0 = 8*1 = 8
m=3, n=1 --> divisor 2^m * 3^n = 2^3 * 3^1 = 8*3 = 24

So those are the 8 divisors of 24: 1, 2, 3, 4, 6, 8, 12, 24. (As I understand it, all but 24 would be called "aliquot parts".)

 

[JeffM]

Harder example: 252

First step: find smallest prime divisor of 252

[MATH]\dfrac{252}{2} = 126 \in \mathbb Z.[/MATH]
Thus, 2 is first in list.

Second step: find smallest prime divisor of 126

[MATH]\dfrac{1252= 63 \in \mathbb Z.[/MATH]
But 2 is already in the list so we do not add it again, but

2 * 2 = 4 is added to the list.

Third step: find smallest prime divisor of 63

[MATH]\dfrac{63}{2} \not \in \mathbb Z.[/MATH]
[MATH]\dfrac{63}{3} = 21 \in \mathbb Z.[/MATH]
Thus 3 is added to the list as are 3 * 2 = 6 and 3 * 4 = 12.

Fourth step: find smallest prime divisor of 21 > 2

[MATH]\dfrac{21}{3} = 7 in \mathbb Z.[/MATH]
3 is already in the list so it is not added again, but we do add

3 * 3 = 9, 3 * 6 = 18, and 3 * 12 = 18.

Fifth step: find smallest prime divisor of 7 > 2

[MATH]\dfrac{7}{3} \not \in \mathbb Z.[/MATH]
[MATH]\dfrac{7}{5} \not \in \mathbb Z.[/MATH]
[MATH]\dfrac{7}{7} = 1.[/MATH]
Add 7 to the list as well as

7 * 2 = 14, 7 * 4 = 28, 7 * 3 = 21, 7 * 6 = 42, 7 * 12 = 84, 7 * 9 = 63, 7 * 18 = 126, and 7 * 36 = 252.

No prime divides 1 evenly so we are done. The list of divisors is

2, 4, 3, 6, 12, 9, 18, 36, 14, 28, 21, 42, 84, 63, 126, and 252.

The coding gets a little trichy in figuring out how to add to the list without duplicates, but it is a simple loop.

 

[Steven G]

For example, the aliquot parts for 10 are 1, 2, 5, 10. One method is to divide
the number by whole numbers from 1 up to and including that whole number.
If the number is large, for example 1000, the division would be long and
tiresome.

Note: I am not a high school or college student.

You said that One method is to divide the number by whole numbers from 1 up to and including that whole number. You can get away using that method and only go up to half of the number as long as you include that number at the end.
For example let's consider 100. No number above 50 (half of 100) , except for 100 itself, goes into 100 evenly. So just try 1,2,3,...50 then list the numbers (from 1 to 50) that go into 100 and do not forget to list 100 as well.

Why does this work? 2*50 =100. Now if you increase the 2, then you must decrease the 50 for the product to still be 100. So no number between 50 and 99 will work. So why does 100 work? That is easy since 1*100 =100

Of course you have already been given better methods than this one.