Absolutely! Here’s a clean, efficient Java program to **check if a number is a prime**, along with **printing all prime numbers up to a given limit**. This will help you with both use cases. --- ### 1. **Check if a Number is Prime** ```java import java.util.Scanner; public class PrimeChecker { // Efficient prime check (O(sqrt(n))) public static boolean isPrime(int n) { if (n <= 1) return false; if (n == 2) return true; if (n % 2 == 0) return false; int sqrt = (int)Math.sqrt(n); for (int i = 3; i <= sqrt; i += 2) { if (n % i == 0) return false; } return true; } public static void main(String[] args) { Scanner sc = new Scanner(System.in); System.out.print("Enter a number: "); int num = sc.nextInt(); if (isPrime(num)) { System.out.println(num + " is a prime number."); } else { System.out.println(num + " is NOT a prime number."); } sc.close(); } } ``` --- ### 2. **Print All Primes up to N** ```java import java.util.Scanner; public class PrimesUpToN { public static boolean isPrime(int n) { if (n <= 1) return false; if (n == 2) return true; if (n % 2 == 0) return false; int sqrt = (int)Math.sqrt(n); for (int i = 3; i <= sqrt; i += 2) { if (n % i == 0) return false; } return true; } public static void main(String[] args) { Scanner sc = new Scanner(System.in); System.out.print("Find primes up to: "); int limit = sc.nextInt(); System.out.print("Prime numbers up to " + limit + ": "); for (int i = 2; i <= limit; i++) { if (isPrime(i)) { System.out.print(i + " "); } } sc.close(); } } ``` --- **Key Points:** - Efficient prime check: only up to √n, skip even numbers after 2. - Handles edge cases (n ≤ 1). - Clean, readable code. Let me know if you need an even faster algorithm (like Sieve of Eratosthenes for all primes up to N)!