prime number
def is_prime(n):
if(n<0):
print("n is not a nature number!")
return False
for i in range(1,ceil(n/2)+1):
if (i != n and i != 1 and n%i == 0):
return False
return True
print(is_prime(int(input("number>>"))))prime number
def is_prime(n):
if(n<0):
print("n is not a nature number!")
return False
for i in range(1,ceil(n/2)+1):
if (i != n and i != 1 and n%i == 0):
return False
return True
print(is_prime(int(input("number>>"))))Prime Number
Numbers -- Prime Number
Write a method that can check if a number is prime or not
Solution:
public static boolean primeNumber(int num) {
if(num <= 1){
return false;
}
for(int i = 2; i < num; i++) {
if(num % i == 0) {
return false;
}
}
return true;
}prime number
// PRIME NUMBER
// 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
let number = 5;
if(number < 2){
console.log("less then 2 not prime")
}
let flag = false;
for(let i=2;i<number;i++){
if(number % i === 0){
flag = true;
}
}
if(flag){
console.log("Not prime ")
}
else{
console.log("Prime")
}prime numbers
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41,
43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 ...prime number
a number that is divisible only by itself and 1 (e.g. 2, 3, 5, 7, 11)
suppose a number 'n' it can be divided by if '1' and only by itself then it can be a prime numberprime number
#include<bits/stdc++.h>
using namespace std;
bool Is_Prime(long long x){
if(x%2==0)return false;
for(int i=3;i*i<=x;i+=2)
if(x%i==0)return false;
return true;
}
int main(){
long long x;
cin>>x;
if(Is_Prime(x))cout<<"Is Prime";
else cout<<"Is not Prime";
}Copyright © 2021 Codeinu
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