int number;
int i=2;
while (i<number)
{
if(number%i==0)
{
printf("Not a prime no.");
break;
}
else
printf("number entered is prime");
getch();
}
#include
#include
bool is_prime(unsigned num) {
unsigned max, factor;
if (num<2) return false;
if (!(num%2)) return num==2;
max = (unsigned) sqrt((double)num) + 1.0;
for (factor=3; factor
return true;
}
unsigned next_prime(unsigned num) {
do { ++num; } while (!isprime(num));
return num;
}
unsigned nth_prime (unsigned n) {
// Returns the nth prime. Returns 0 when n is zero.
// Note: does not guard against underflow!
unsigned prime;
if (!n) return 0;
prime=2;
while (--n) {prime = next_prime(prime);}
return prime;
}
int main() {
printf ("The 100th prime is %u\n", nth_prime(100));
return 0;
}
234
Begin Read num for(i=2; i<num; i++) if(num%2==0) then print "the number is not a prime no."; else if print "the number is prime"; end if Stop
Loop through some numbers - for example, 2 through 100 - and check each one whether it is a prime number (write a second loop to test whether it is divisible by any number between 2 and the number minus 1). If, in this second loop, you find a factor that is greater than 1 and less than the number, it is not a prime, and you can print it out.
#include<stdio.h> #include<conio.h> int i=1, count=0,n; clrscr(); printf("Enter Any Number"); scanf("%d", &n); while(i<n) if(n%i==0) { count++; i++; } if(count==2) printf("Given Number is Prime"); else printf("Given Number is not Prime"); getch(); }
flow t prime numberchar
If you just want a hint: One way to check whether a number is prime is by dividing it by any number between 2 and the square root of your number. If the number divides by any of these, it is not prime. If you want the code: import math for num in range(1,101): if all(num%i!=0 for i in range(2,int(math.sqrt(num))+1)): print num
You can check each individual number, whether it is a prime number. For numbers below 100, it is enough to check whether they are divisible by 2, by 3, by 5, and by 7. If a number is divisible by none of these, it is a prime number.
First write a program to generate the prime number. After one prime number was generated, divide the big int number by the prime number. If the remainder is zero then quotient is the second prime number ( also it is important to check whether the quotient is prime number or not because sometimes you will get wrong answer). Repeat the process until you get the result.
Find a prime number, add 2 to the number. Check if the new number is prime. IE : 3 is prime. 3+2 =5. 5 is prime. (3,5) are twin primes.
29 is a prime number, meaning it has no smaller factors. For any number up to 120, to check whether it is prime or not, it is sufficient to check whether it is divisible by the first four prime numbers (2, 3, 5 and 7).
You just have to work out it,take each number below it and check whether it is prime or not.
A number is prime if it only has two distinct factors.
It is 29 because 29*23 = 667
You can write out this algorithm. This will then be programmed into the device to make determining prime numbers easier.
Take each number in turn, call it "n", and check whether it has any factors f, such that 1 < f < n. If it doesn't, it is a prime number.Take each number in turn, call it "n", and check whether it has any factors f, such that 1 < f < n. If it doesn't, it is a prime number.Take each number in turn, call it "n", and check whether it has any factors f, such that 1 < f < n. If it doesn't, it is a prime number.Take each number in turn, call it "n", and check whether it has any factors f, such that 1 < f < n. If it doesn't, it is a prime number.
You take two consecutive odd numbers and check both of them to see whether they are prime or not.
[object Object]
#include<iostream.h> #include<conio.h> void prime(int n) { clrscr(); int num; cout<<"enter the numbers"<<endl; cin>>num; prime(num); getch(); } void prime(int n) { int prime=1,i; for(i=2;i<=n/2;i++) if(n%i==1) prime=0; if(prime==1) cout<<"the number"<<n>>"is prime"; else cout<<"the number"<<n<<"is not prime"; }