My test cases are not satisfied with this code

#1

#include
using namespace std;
int main() {
int n;
cin>>n;
int i;
for (i=2;i<=n-1;i++)
{
if(n%i==0)
{
cout<<" not prime";
break;

	}
}
if (i==n)
{
	cout<<"prime"<<endl;
}
return 0;

}

pls provide with the solution

#2

Take as input a number N, print “Prime” if it is prime if not Print “Not Prime”.

you have to print Prime not prime
P should be capital

also run loop till squareroot of n to optimised it
use this condition
for (i=2;i*i<=n;i++)

also instead of int use long long int as no are large

Code with Modifications

i hope this helps
if yes hit a like and don’t forgot to mark doubt as resolved :grinning:
if you have more doubts regarding this feel free to ask

#3

i have not able to solve pls provide with the sol

Screenshot 2020-07-09 at 7.43.27 PM
Screenshot 2020-07-09 at 7.43.27 PM1440×900 119 KB

#4

i have already provided the solution you can see it in previous reply

in this code the mistake is that
you have to check this condition outside the loop
if(i==N){
cout<<“Prime”<<endl;
}

#5

why are we doing i*i instead of just i …pls explain the reason

#6

@suryansh771
Both way is correct but i*i way is more efficient, it is O(sqrt(n))
In number theory,
Composite numbers have the property that it is divisible atleast one divisor <=sqrt(n) and greater than 1
Thus prime number should not have such property, if it does then its not a prime

Proof is also easy if you are interested(try yourself by contradiction),

#7

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