Write a C Program to find the number of integers with exactly 9 divisors
Program:-
#include
int count_no_of_divisors(int num)
{
int count = 0;
for (int i = 1; i <= num; i++)
{
if (num % i == 0)
count = count + 1;
}
return count;
}
void check_9_factors(int n)
{
int c = 0;
for (int i = 1; i <= n; i++)
{
if (count_no_of_divisors(i) == 9)
{
printf(“%d “, i);
c = c + 1;
}
}
printf(“\n\nTotal = %d\n”, c);
}
int main()
{
int n;
printf(“\nEnter the number : “);
scanf(“%d”, &n);
printf(“\nThe number which has exactly 9 divisors : “);
check_9_factors(n);
return 0;
}
int count_no_of_divisors(int num)
{
int count = 0;
for (int i = 1; i <= num; i++)
{
if (num % i == 0)
count = count + 1;
}
return count;
}
void check_9_factors(int n)
{
int c = 0;
for (int i = 1; i <= n; i++)
{
if (count_no_of_divisors(i) == 9)
{
printf(“%d “, i);
c = c + 1;
}
}
printf(“\n\nTotal = %d\n”, c);
}
int main()
{
int n;
printf(“\nEnter the number : “);
scanf(“%d”, &n);
printf(“\nThe number which has exactly 9 divisors : “);
check_9_factors(n);
return 0;
}
Output:-
....
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