Pythondef is_prime(num):    if num <= 1:        return False    for i in range(2, int(num ** 0.5) + 1):        if num % i == 0:            return False    return Truenum = int(input("Enter a number: "))if is_prime(num):    print(f"{num} is a prime number.")else:    print(f"{num} is not a prime number.")Javaimport java.util.Scanner;public class Main {    public static boolean isPrime(int num) {        if (num <= 1) {            return false;        }        for (int i = 2; i <= Math.sqrt(num); i++) {            if (num % i == 0) {                return false;            }        }        return true;    }    public static void main(String[] args) {        Scanner scanner = new Scanner(System.in);        System.out.print("Enter a number: ");        int num = scanner.nextInt();        if (isPrime(num)) {            System.out.println(num + " is a prime number.");        } else {            System.out.println(num + " is not a prime number.");        }    }}C++#include #include using namespace std;bool isPrime(int num) {    if (num <= 1) {        return false;    }    for (int i = 2; i <= sqrt(num); i++) {        if (num % i == 0) {            return false;        }    }    return true;}int main() {    int num;    cout << "Enter a number: ";    cin >> num;    if (isPrime(num)) {        cout << num << " is a prime number." << endl;    } else {        cout << num << " is not a prime number." << endl;    }    return 0;}C:#include #include int isPrime(int num) {    if (num <= 1) {        return 0;    }    for (int i = 2; i <= sqrt(num); i++) {        if (num % i == 0) {            return 0;        }    }    return 1;}int main() {    int num;    printf("Enter a number: ");    scanf("%d", &num);    if (isPrime(num)) {        printf("%d is a prime number.\n", num);    } else {        printf("%d is not a prime number.\n", num);    }    return 0;}Explanation:As the language was not mentioned in the question, I have written the code in different languages.PythonJavaC++CLogic of the code:Taking input from the userpassing the number through the function is_prime to check if the number is prime or notIf the number is less than or equal to 1, then return false, which means it is not a prime numberlooping from 2 to square root of n and we are checking if the each number is divisible with the number given.if the condition comes out to positive for atleast once, that means, it is divisible by some number i. Hence, return true which means it is a prime number. the loop ends if the no divisor is found, hence return false which means it is not a prime number.Sample test cases are verified and attached below for reference.