Python Program to Find Factors of a Number

To find factors of a number in Python, use a loop with for i in range(1, n+1): and check if n % i == 0 to collect all factors.

Examples

Input6

OutputFactors of 6 are: 1 2 3 6

Input13

OutputFactors of 13 are: 1 13

Input1

OutputFactors of 1 are: 1

How to Think About It

To find factors of a number, think about all numbers from 1 up to that number. For each, check if dividing the number by it leaves no remainder. If yes, that number is a factor.

Algorithm

1

Get the input number n

2

Create an empty list to store factors

3

Loop i from 1 to n inclusive

4

Check if n modulo i equals 0

5

If yes, add i to the factors list

6

Return or print the list of factors

Code

python

n = int(input('Enter a number: '))
factors = []
for i in range(1, n + 1):
    if n % i == 0:
        factors.append(i)
print('Factors of', n, 'are:', *factors)

Output

Enter a number: 6
Factors of 6 are: 1 2 3 6

Dry Run

Let's trace the number 6 through the code to find its factors.

1

Input

n = 6

2

Initialize factors list

factors = []

3

Loop i from 1 to 6

Check each i if divides 6 evenly

4

Check i=1

6 % 1 == 0, add 1 to factors

5

Check i=2

6 % 2 == 0, add 2 to factors

6

Check i=3

6 % 3 == 0, add 3 to factors

7

Check i=4

6 % 4 != 0, skip

8

Check i=5

6 % 5 != 0, skip

9

Check i=6

6 % 6 == 0, add 6 to factors

10

Print factors

Factors of 6 are: 1 2 3 6

i	n % i == 0?	Action	Factors List
1	True	Add 1	[1]
2	True	Add 2	[1, 2]
3	True	Add 3	[1, 2, 3]
4	False	Skip	[1, 2, 3]
5	False	Skip	[1, 2, 3]
6	True	Add 6	[1, 2, 3, 6]

Why This Works

Step 1: Loop through numbers

We check every number from 1 to n because factors must be within this range.

Step 2: Check divisibility

Using n % i == 0 tells us if i divides n without remainder, meaning i is a factor.

Step 3: Collect factors

We add all such i values to a list to show all factors at the end.

Alternative Approaches

Check up to square root

python

import math
n = int(input('Enter a number: '))
factors = []
for i in range(1, int(math.sqrt(n)) + 1):
    if n % i == 0:
        factors.append(i)
        if i != n // i:
            factors.append(n // i)
factors.sort()
print('Factors of', n, 'are:', *factors)

This method is faster for large numbers because it checks fewer values but needs sorting at the end.

Using list comprehension

python

n = int(input('Enter a number: '))
factors = [i for i in range(1, n + 1) if n % i == 0]
print('Factors of', n, 'are:', *factors)

This is a concise and readable way but still checks all numbers up to n.

Complexity: O(n) time, O(n) space

Time Complexity

The program checks every number from 1 to n, so it runs in linear time O(n).

Space Complexity

It stores all factors found in a list, which in the worst case can be up to O(n) if the number is 1.

Which Approach is Fastest?

Checking up to the square root reduces time to O(√n), making it faster for large numbers compared to checking all up to n.

Approach	Time	Space	Best For
Check all up to n	O(n)	O(n)	Small numbers, simplicity
Check up to sqrt(n)	O(√n)	O(n)	Large numbers, better performance
List comprehension	O(n)	O(n)	Concise code, readability