public class TenDigitPrimes {
/*
This program finds:
1. The first 10-digit prime number.
2. The last 10-digit prime number.
Approach:
- Use long to safely hold 10-digit values.
- Implement a primality test using trial division up to sqrt(n).
This is efficient for checking individual numbers in this range.
- Search upward from the smallest 10-digit number for the first prime.
- Search downward from the largest 10-digit number for the last prime.
- Skip even numbers to reduce unnecessary work.
*/
// Check whether a number is prime
public static boolean isPrime(long n) {
if (n < 2) return false;
if (n % 2 == 0) return n == 2;
long limit = (long) Math.sqrt(n);
for (long d = 3; d <= limit; d += 2) {
if (n % d == 0) {
return false;
}
}
return true;
}
// Find the first 10-digit prime
public static long first10DigitPrime() {
long n = 1_000_000_000L; // smallest 10-digit number
if (n % 2 == 0) {
n++; // move to next odd number
}
while (!isPrime(n)) {
n += 2; // check only odd numbers
}
return n;
}
// Find the last 10-digit prime
public static long last10DigitPrime() {
long n = 9_999_999_999L; // largest 10-digit number
if (n % 2 == 0) {
n--; // move to previous odd number
}
while (!isPrime(n)) {
n -= 2; // check only odd numbers
}
return n;
}
public static void main(String[] args) {
long first = first10DigitPrime();
long last = last10DigitPrime();
System.out.println("First 10-digit prime: " + first);
System.out.println("Last 10-digit prime: " + last);
}
}
/*
run:
First 10-digit prime: 1000000007
Last 10-digit prime: 9999999967
*/
