Last Updated on June 5, 2024 by Abhishek Sharma

Python is a powerful and easy-to-learn programming language, ideal for tackling a wide range of problems, from simple calculations to complex algorithms. One interesting problem in the realm of number theory is checking whether a given number is an abundant number. An abundant number is a number for which the sum of its proper divisors (excluding itself) is greater than the number itself. Understanding how to identify such numbers can be a fun and educational exercise in Python programming, involving loops, conditionals, and basic arithmetic operations. In this article, we will explore the concept of abundant numbers and provide a step-by-step guide to creating a Python program that can check if a number is abundant.

## What is an abundant number?

An abundant number is a number for which the sum of its proper divisors (excluding the number itself) is greater than the number itself. For example, 12 is an abundant number because the sum of its proper divisors (1, 2, 3, 4, 6) is 16, which is greater than 12.

### Understanding the Abundant Number in Depth

An abundant number, n, can be expressed as the sum of its proper divisors (excluding n itself). If the sum of the divisors is greater than n, the number is classified as abundant. For example, let’s consider the number 12. Its proper divisors are 1, 2, 3, 4, and 6, with a sum of 16. Since 16 is greater than 12, 12 is an abundant number.

We can solve this problem using two methods :

- Using range until number
- Using range until sqrt()

## Algorithm to Check Abundant Number using range until Number

- Initialize a variable,
`num`

, with the number to be checked for abundance. - Initialize a variable,
`divisor_sum`

, to 0. - Iterate from 1 to
`num // 2`

(inclusive) using a for loop. - Check if the current number is a divisor of
`num`

by using the modulo operator (`num % i == 0`

). - If the current number is a divisor, add it to
`divisor_sum`

. - After the loop, compare
`divisor_sum`

with`num`

. If`divisor_sum`

is greater than`num`

, the number is abundant; otherwise, it is not. - Return the result (either true or false), indicating whether the number is abundant or not.

## Python code to check if a number is abundant using range until Number

n = 12 sum=1 for i in range (2, n): if(n%i==0): sum=sum+i if(sum>n): print(n,'is Abundant Number') else: print(n,'is not Abundant Number')

**Output**

`12 is Abundant Number`

## Algorithm to Check Abundant Number using range until sqrt()

- Import the
`math`

module to access mathematical functions. - Set the value of
`n`

as the number to be checked for abundance. - Initialize a variable
`sum`

as 1, since 1 can divide any number. - Initialize a variable
`i`

as 2. - Enter a while loop with the condition
`i <= math.sqrt(n)`

. - Inside the while loop:
- Check if
`n`

is divisible by`i`

using the modulo operator (`n % i == 0`

). - If
`n`

is divisible by`i`

, proceed to the next step. - Check if the quotient (
`n // i`

) is equal to the divisor (`i`

). - If the quotient is equal to the divisor, add only one of them to the
`sum`

. - If the quotient is not equal to the divisor, add both the divisor and quotient to the
`sum`

. - Increment
`i`

by 1.

- Check if
- After the while loop:
- Check if the
`sum`

is greater than`n`

. - If
`sum`

is greater than`n`

, print that`n`

is an abundant number. - If
`sum`

is not greater than`n`

, print that`n`

is not an abundant number.

- Check if the

## Python code to check if a number is abundant using range until sqrt()

import math n = 12 sum=1 # 1 can divide any number i=2 while(i<=math.sqrt(n)): if(n%i==0): #if number is divisible by i add the number if(n//i==i): # if quotient is equal to divisor add only one of them sum=sum+i else: sum=sum + i + n/i i=i+1 if(sum>n): print(n,"is Abundant Number") else: print(n,"is not Abundant Number")

**Output**

`12 is Abundant Number`

**Conclusion**

Checking for abundant numbers in Python is an excellent exercise to practice basic programming skills, including loops, conditionals, and working with lists or sets. By understanding and implementing this concept, programmers can deepen their understanding of number theory and improve their problem-solving abilities. Python’s simplicity and powerful features make it an ideal language for such tasks, allowing beginners and experienced programmers alike to write clear and efficient code. As you continue to explore more complex algorithms and mathematical concepts, exercises like this will build a strong foundation for further learning and development.

## Frequently Asked Questions (FAQs) Related To Check Abundant Number in Python:

Here are some of the FAQs related To Check Abundant Number in Python:

**1. What are proper divisors?**

Proper divisors of a number are all positive divisors of the number, excluding the number itself. For example, the proper divisors of 12 are 1, 2, 3, 4, and 6.

**2. How do I check if a number is abundant in Python?**

To check if a number is abundant, you need to:

- Find all the proper divisors of the number.
- Calculate the sum of these divisors.
- Compare the sum to the original number to determine if it is greater.

**3. Can you provide an example of an abundant number?**

Sure! Let's take the number 12:

- Proper divisors of 12: 1, 2, 3, 4, and 6.
- Sum of proper divisors: 1 + 2 + 3 + 4 + 6 = 16.
- Since 16 is greater than 12, 12 is an abundant number.

**4. Are all even numbers abundant?**

No, not all even numbers are abundant. For example, 4 and 6 are not abundant numbers. Whether a number is abundant depends on the sum of its proper divisors, not just on whether it is even or odd.

**5. Can a number be both prime and abundant?**

No, a prime number cannot be abundant. By definition, a prime number has no proper divisors other than 1. The sum of its proper divisors is always 1, which is less than the number itself, so it cannot be abundant.