Aprime number is a number that is only evenly divisble by itself and 1. for example, the number 5 is prime because it can only be evenlly divided by 1 and 5. the number 6, however, is not prime because it can be divided evenly by 2 and 3. write a boolean function named is_prime which takes an integer as an argument and returns true if the argument is a prime number, or false otherwise. use the function in a program that prompts the user to enter a number and then prints whether the number is prime.
Let's be honest, there are plenty of solutions on the internet. Here's one where I added a program to use it. Note that this one is optimized to not consider even numbers, and to only iterate to the squareroot of the number you're testing.
def is_prime(n):
if n == 2 or n == 3: return True
if n % 2 == 0 or n < 2: return False
for i in range(3, int(n**0.5)+1, 2):
if n % i == 0:
return False
return True
x = int(input("Enter a number: "))
print("{} is {}a prime.".format(x, "" if is_prime(x) else "not "))
Java program given below
Explanation:
Java Code: Save this file as Main.java and keep This file and GenericQueue.java in same folder
class Main {
public static boolean isPrime(int n){
for (int i=2; i<= n/2; i++){
if (n % i== 0)
return false;
}
return true;
}
public static void main(String[] args) {
GenericQueue<Integer> queue = new GenericQueue<Integer>();
int n = 0, p = 2;
while(n <= 50)
{
if(isPrime(p))
{
queue.enqueue(p);
n++;
}
p++;
}
System.out.println(queue);
}
}
What program is this in?
// here is code in C++.
#include <bits/stdc++.h>
using namespace std;
// function to check prime
bool isPrime(int n)
{
// if input is less than 1
if(n<1)
return false;
// if input is 1,2 or 3
else if (n == 1||n ==2 ||n==3)
{
return true;
}
// if input is greater than 3
else
{
for(int i=2; i<n; i++)
{
if(n%i==0)
return false;
}
return true;
}
}
// driver function
int main()
{
// variable
int num;
cout << "Enter a number: ";
// read the input number
cin >> num;
// call function to check prime
if (isPrime(num)==true)
// if prime
cout << num << " is prime.";
else
// if not prime
cout << num << " is not prime.";
return 0;
}
Explanation:
Read a number from user and assign it to "num".Then call the function isPrime() with "num" as parameter.In this function, if the input is less than 1 then it will return "false".If input is 1 or 2 or 3 then it will return "true".For an input greater than 3, if the number is divisible by only 1 and itself then function will return "true" otherwise it will return "false".
Output:
Enter a number: 17
17 is prime.
Delete your historyhope this
Or mine means that itself and 1 can only go into it. Conposit means that multiple numbers can go into it like 24. 12,6,8,2 can all go into it
public class Main { public static void main (String [] args) { for(int i = 2; i < 10000; i++){ if(isPrime1(i)){ System.out.print(i + " "); } } System.out.println(); for(int i = 2; i < 10000; i++){ if(isPrime2(i)){ System.out.print(i + " "); } } } public static boolean isPrime1(int n){ for(int i=2; i <= n/2; i++){ if(n % i == 0){ return false; } } return true; } public static boolean isPrime2(int n){ for(int i=2; i <= Math.sqrt(n); i++){ if(n % i == 0){ return false; } } return true; } }
Explanation:
Firstly, create the first version of method to identify a prime number, isPrime1. This version set the limit of the for loop as n/2. The for loop will iterate through the number from 2 till input n / 2 and check if n is divisible by current value of i. If so, return false to show this is not a prime number (Line 22 - 26). Otherwise it return true to indicate this is a prime number.
In the main program, we call the isPrime1 method by passing the i-index value as an argument within a for-loop that will iterate through the number 2 - 10000 (exclusive). If the method return true, print the current i value). (Line 5 - 9)
The most direct way to ensure all the prime numbers below 10000 are found, is to check the prime status from number 2 - 9999 which is amount to 9998 of numbers.
Next we create a second version of method to check prime, isPrime2 (Line 31 - 40). This version differs from the first version by only changing the for loop condition to i <= square root of n (Line 33). In the main program, we create another for loop and repeatedly call the second version of method (Line 13 - 17). We also get the same output as in the previous version.