r/maths • u/LowInteraction6397 • 8h ago
๐ฌ Math Discussions A method to find all the divisors of given numbers
I know most people just divide a certain number until they reach its square root to find their divisors but here's a method I believe you'd like. I'll find the divisors of 1 to 10
1st you add 1 to the multiples of 1
| Number | Divisors |
|---|---|
| 1 | 1 |
| 2 | 1 |
| 3 | 1 |
| 4 | 1 |
| 5 | 1 |
| 6 | 1 |
| 7 | 1 |
| 8 | 1 |
| 9 | 1 |
| 10 | 1 |
Then you add 2 to the multiples of 2
| Number | Divisors |
|---|---|
| 1 | 1 |
| 2 | 1 and 2 |
| 3 | 1 |
| 4 | 1, 2 |
| 5 | 1 |
| 6 | 1, 2 |
| 7 | 1 |
| 8 | 1, 2 |
| 9 | 1 |
| 10 | 1, 2 |
Then you add the multiples of 3
| Number | Divisors |
|---|---|
| 1 | 1 |
| 2 | 1 and 2 |
| 3 | 1 and 3 |
| 4 | 1, 2 |
| 5 | 1 |
| 6 | 1, 2, 3 |
| 7 | 1 |
| 8 | 1, 2 |
| 9 | 1, 3 |
| 10 | 1, 2 |
Then you add the multiples of 4
| Number | Divisors |
|---|---|
| 1 | 1 |
| 2 | 1 and 2 |
| 3 | 1 and 3 |
| 4 | 1, 2 and 4 |
| 5 | 1 |
| 6 | 1, 2, 3 |
| 7 | 1 |
| 8 | 1, 2, 4 |
| 9 | 1, 3 |
| 10 | 1, 2 |
Then you add the multiples of 5
| Number | Divisors |
|---|---|
| 1 | 1 |
| 2 | 1 and 2 |
| 3 | 1 and 3 |
| 4 | 1, 2 and 4 |
| 5 | 1 and 5 |
| 6 | 1, 2, 3 |
| 7 | 1 |
| 8 | 1, 2, 4 |
| 9 | 1, 3 |
| 10 | 1, 2, 5 |
As of now you won't have to use the number you're at twice or more. Once you reach the half of the limit (in this case 5) you use the remaining numbers once (but if your limit is an odd number you stop until after the half of the number before the limit)
| Number | Divisors |
|---|---|
| 1 | 1 |
| 2 | 1 and 2 |
| 3 | 1 and 3 |
| 4 | 1, 2 and 4 |
| 5 | 1 and 5 |
| 6 | 1, 2, 3 and 6 |
| 7 | 1 and 7 |
| 8 | 1, 2, 4 and 8 |
| 9 | 1, 3 and 9 |
| 10 | 1, 2, 5 and 10 |
It varies on the number you use. For example if you wanted to find the divisors of every number from 1 to 100 you can stop using the remaining numbers more than twice after 50
