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 to 1000.
Sample Solution:
C Code:
# include <stdio.h>
# include <string.h>
# include <stdbool.h>
# include <math.h>
int getSum(int n)
{
int sum = 0;
for (int i=1; i<=sqrt(n); i++)
{
if (n%i==0)
{
if (n/i == i)
sum = sum + i;
else
{
sum = sum + i;
sum = sum + (n / i);
}
}
}
sum = sum - n;
return sum;
}
bool checkAbundant(int n)
{
return (getSum(n) > n);
}
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++)
{
n=j;
if(checkAbundant(n)==true)
printf("%d ",n);
}
printf("\n");
}
Sample 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...
Pictorial Presentation:
Flowchart:

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Previous: Write a program in C to check whether a given number is Abundant or not.
Next: Write a program in C to check whether a given number is Deficient or not.
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C Programming: Tips of the Day
Why do C and C++ compilers allow array lengths in function signatures when they're never enforced?
It is a quirk of the syntax for passing arrays to functions.
Actually it is not possible to pass an array in C. If you write syntax that looks like it should pass the array, what actually happens is that a pointer to the first element of the array is passed instead.
Since the pointer does not include any length information, the contents of your [] in the function formal parameter list are actually ignored.
Ref : https://bit.ly/3fhlvdH
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