Python Program - Find all factors of a Number
Objective: Write a Python program to find all distinct factors (divisors) of a given natural number. The divisors of few numbers are given below:
Number: 10 Divisors: 1 2 5 10 Number: 15 Divisors: 1 3 5 15 Number: 100 Divisors: 1 2 4 5 10 20 25 50 100
Method 1: Using iteration
One of the basic approach is to iterate from 1 to n and in each iteration check whether the number divides n. If it divides then print it.
#function to print all divisors of a number def printDivisors(n): print("Divisors of", n ,"are: ", end="") for i in range(1, n+1): if(n%i == 0): print(i, end=" ") print() printDivisors(10) printDivisors(50) printDivisors(100)
The above code will give the following output:
Divisors of 10 are: 1 2 5 10 Divisors of 50 are: 1 2 5 10 25 50 Divisors of 100 are: 1 2 4 5 10 20 25 50 100
Method 2: Optimized Code
Instead of checking the divisibility of the given number from 1 to n, it is checked till square root of n. For a factor larger than square root of n, there must the a smaller factor which is already checked in the range of 1 to square root of n.
import math #function to print all divisors of a number def printDivisors(n): print("Divisors of", n ,"are: ", end="") #loop from 1 to math.sqrt(n) for i in range(1, int(math.sqrt(n)) + 1): if(n%i == 0): if(n/i == i): print(i, end=" ") else: print(i, n//i, end=" ") print() printDivisors(10) printDivisors(50) printDivisors(100)
The above code will give the following output:
Divisors of 10 are: 1 10 2 5 Divisors of 50 are: 1 50 2 25 5 10 Divisors of 100 are: 1 100 2 50 4 25 5 20 10
Method 3: Optimized Code with sorted result
In the previous method, results are produced in a irregular fashion (printed in pairs - small number and large number). The result can be sorted by storing the larger number and print them later on. Consider the example below:
import math #function to print all divisors of a number def printDivisors(n): print("Divisors of", n ,"are: ", end="") #creating a list to store larger numbers list = [] #loop from 1 to math.sqrt(n) for i in range(1, int(math.sqrt(n)) + 1): if(n%i == 0): if(n/i == i): print(i, end=" ") else: print(i, end=" ") #storing the large number of a pair list.append(int(n/i)) #printing stored large numbers of pairs for i in list[::-1]: print (i, end =" ") print() printDivisors(10) printDivisors(50) printDivisors(100)
The above code will give the following output:
Divisors of 10 are: 1 2 5 10 Divisors of 50 are: 1 2 5 10 25 50 Divisors of 100 are: 1 2 4 5 10 20 25 50 100
Method 4: Another Optimized Code
To produce the result in sorted order, we can iterate from 1 to square root of n and printing the number which divides n. After that we can iterate back (in reverse order) and printing the quotient of all numbers which divides n.
import math #function to print all divisors of a number def printDivisors(n): print("Divisors of", n ,"are: ", end="") #loop from 1 to math.sqrt(n) i = 0 for i in range(1, int(math.sqrt(n))): if(n%i == 0): print(i, end=" ") for i in range(int(math.sqrt(n)), 0, -1): if(n%i == 0): print(n//i, end=" ") print() printDivisors(10) printDivisors(50) printDivisors(100)
The above code will give the following output:
Divisors of 10 are: 1 2 5 10 Divisors of 50 are: 1 2 5 10 25 50 Divisors of 100 are: 1 2 4 5 10 20 25 50 100
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