Answer / sudha

#include<stdio.h>
#include<conio.h>
void main()
{
int isprime=1;
int n,i;
printf("enter a number");
scanf("%d",&n);
if(i=2;i<=n/2;i++)
{
if((n%i)==0)
{
isprime=0;
break;
}
}
if(isprime==1)
{
printf("%d is a prime number",n);
}
else
{
printf("%d is not a prime number",n);
}
getch();
}

Answer / sivavendra

Hey sudha,
plz checkout the above code once again.....

if(i=2;i<=n/2;i++) --- What this code indicates?

Answer / naman patidar

we can reduce no of iterations by changing the condition
(i<=n/2) with (i<=sqrt(n))..
This is a rule that if a number is not dividable by any no
(except 1)less than equal to the sqr root of it then the no
is prime.

int n ; // any no, user input.
int i ;
for(i=2; i<=sqrt(n); i++ )
{
if(n%i==0)
{
printf("not prime");
break;
}
}
if(i > sqrt(n)) // you can use a flag as well here
{
printf("prime no");
}

Answer / md.ershad ezaz

#include<stdio.h>
#include<conio.h>
void main()
{
int i,n,c;
clrscr();
printf("Enter a number\n");
scanf("%d",&n);
c=0;
for(i=1;i<=n;i++)
{
if(n%i==0)
c=c+1;
}
if(c==2)
printf("%d is prime",n);
else
printf("%d is not prime",n);
getch();
}

Answer / nikhilreddy gujjula

#include<stdio.h>
#include<conio.h>
void main()
{
int prime=1;
int n,i;
printf("enter a number");
scanf("%d",&n);
if(i=2;i<=n/2;i++)
{
if((n%i)==0)
{
prime=0;
break;
}
}
if(prime==1)
{
printf("%d is a prime number",n);
}
else
{
printf("%d is not a prime number",n);
}
getch();
}

Answer / padhmavathi loganathan

#include<stdio.h>
int main()
{
int a,i,f;
printf("Enter a number: ");
scanf("%d",&a);
f=0;
i=2;
while(i <= a/2)
{
if(a%i == 0)
{
f=1;
break;
}
i++;
}
if(f==0)
printf("Prime Number")
else
printf("Not Prime Number");
return 0;
}

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