C Exercises: Find the Abundant numbers (integers) between 1 to 1000

C Numbers: Exercise-3 with Solution

Write a program in C to find the Abundant numbers (integers) between 1 and 1000.

An abundant number is a number for which the sum of its proper divisors (excluding the number itself) is greater than the number.

Explanation:

• Proper Divisors: These are the divisors of a number excluding the number itself.
• Abundant Number: If the sum of proper divisors of a number is greater than the number, it is called an abundant number.

Visual Presentation:

Here are some uses and applications of Abundant numbers:

• Number Theory:
• Research and Classification: Abundant numbers help in the classification of numbers. They form one of the three main types of numbers, along with deficient numbers and perfect numbers.
• Understanding Divisor Functions: They are used in the study of divisor functions and the properties of integers.
• Cryptography:
• Algorithms and Security: Some cryptographic algorithms leverage number theory, including properties of abundant numbers, to create secure encryption methods.
• Mathematical Curiosities:
• Patterns and Sequences: Abundant numbers are part of various integer sequences and are studied to understand patterns within these sequences.
• Educational Tools:
• Problem-Solving Skills: Problems involving abundant numbers are used to develop problem-solving skills and logical thinking.
• Recreational Mathematics:
• Puzzles and Games: Abundant numbers can be used in creating mathematical puzzles and games, challenging people to find patterns or relationships.

Expected Output :

The Abundant number between 1 to 1000 are:
-----------------------------------------------
12 18 20 24 30 36 40 42 48 54 56 60 66 70 72 78 80...

Sample Solution-1:

C Code:

#include <stdio.h>
#include <string.h>
#include <stdbool.h>
#include <math.h>

// Function to calculate the sum of divisors of a number
int getSum(int n)
{
    int sum = 0;
    for (int i = 1; i <= sqrt(n); i++) // Loop through numbers from 1 to the square root of 'n'
    {
        if (n % i == 0) // Check if 'i' is a divisor of 'n'
        {
            if (n / i == i)
                sum = sum + i; // If 'i' is a divisor and is equal to the square root of 'n', add it to 'sum'
            else
            {
                sum = sum + i; // Add 'i' to 'sum'
                sum = sum + (n / i); // Add 'n / i' to 'sum'
            }
        }
    }
    sum = sum - n; // Subtract the number 'n' from the sum of its divisors
    return sum; // Return the sum of divisors
}

// Function to check if a number is an abundant number
bool checkAbundant(int n)
{
    return (getSum(n) > n); // Return true if the sum of divisors is greater than 'n', otherwise return false
}

// Main function
int main()
{
    int n;
    printf("\n\n The Abundant number between 1 to 1000 are: \n");
    printf(" -----------------------------------------------\n");

    for (int j = 1; j <= 1000; j++) // Loop through numbers from 1 to 1000
    {
        n = j; // Assign the current value of 'j' to 'n'
        if (checkAbundant(n) == true) // Check if 'n' is an abundant number
            printf("%d ", n); // Print the abundant number

    }
    printf("\n");

    return 0;
}

Explanation:

• Includes standard libraries: stdio.h, string.h, stdbool.h, and math.h.
• Defines getSum function to calculate the sum of proper divisors of a number n.
• Initializes sum to 0.
• Loops from 1 to the square root of n.
• Checks if i is a divisor of n.
• Adds i and n / i to sum if they are divisors, avoiding duplication for perfect squares.
• Subtracts n from sum to exclude n itself from its sum of divisors.
• Returns the sum of divisors.
• Defines checkAbundant function to determine if a number n is abundant.
• Returns true if the sum of divisors (calculated by getSum) is greater than n.
• main function prints all abundant numbers between 1 and 1000.
• Iterates through numbers from 1 to 1000.
• Checks if each number is abundant using checkAbundant.
• Prints the number if it is abundant.
• Returns 0 to indicate successful execution.

Sample Solution-2:

C Code:

#include <stdio.h>
// Function to calculate the sum of proper divisors of a number
int sum_of_divisors(int num) {
    int sum = 0;
    for (int i = 1; i <= num / 2; i++) {
        if (num % i == 0) {
            sum += i;
        }
    }
    return sum;
}

int main() {
    printf("Abundant numbers between 1 and 1000 are:\n");

    for (int i = 1; i <= 1000; i++) {
        int sum = sum_of_divisors(i);
        if (sum > i) {
            printf("%d\n", i);
        }
    }
    return 0;
}

Explanation:

• sum_of_divisors Function:
• This function calculates the sum of all proper divisors of a given number.
• It loops through numbers from 1 to num / 2 (since no proper divisor can be greater than half of the number) and adds up the divisors.
• main Function:
• The main function iterates through numbers from 1 to 1000.
• For each number, it calls sum_of_divisors to get the sum of its proper divisors.
• If the sum is greater than the number itself, it prints the number as it is an abundant number.

Output:

 The Abundant number between 1 to 1000 are:                                                                 
 -----------------------------------------------                                                              
12 18 20 24 30 36 40 42 48 54 56 60 66 70 72 78 80...  

Flowchart:

C Programming Code Editor:

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