C Exercises: Find the Deficient numbers (integers) between 1 to 100
C Numbers: Exercise-5 with Solution
Write a program in C to find the Deficient numbers (integers) between 1 to 100.
Sample Solution:
C Code:
#include <stdio.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 a deficient number
bool checkDeficient(int n)
{
return (getSum(n) < n); // Return true if the sum of divisors is less than 'n', otherwise return false
}
// Main function
int main()
{
int n, ctr = 0; // Declare variables 'n' for the number, 'ctr' to count deficient numbers
printf("\n\n The Deficient numbers between 1 to 100 are: \n");
printf(" ------------------------------------------------\n");
for (int j = 1; j <= 100; j++) // Loop through numbers from 1 to 100
{
n = j; // Assign the current value of 'j' to 'n'
if (checkDeficient(n) == true) // Check if 'n' is a deficient number
{
printf("%d ", n); // Print the deficient number
ctr++; // Increment the counter for deficient numbers
}
}
printf("\n The Total number of Deficient numbers are: %d \n", ctr); // Print the total count of deficient numbers
return 0;
}
Sample Output:
The Deficient numbers between 1 to 100 are: ------------------------------------------------ 1 2 3 4 5 7 8 9 10 11 13 14 15 16 17 19 21 22 23 25 26 27...
Visual Presentation:
Flowchart:
C Programming Code Editor:
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