Question

How to find a list of numbers up to a limit that is a palindrome in base 10 and base 2 in Java

Answer 1

// Function to check if a number is a palindrome in a given base
public class PalindromeChecker {

    public static boolean isPalindrome(int num, int base) {
        if (base < 2) return false; // Invalid base

        StringBuilder strnumreversed = new StringBuilder();
        int temp = num;

        // Convert number to string in given base
        do {
            int digit = temp % base;
            strnumreversed.append(digit < 10 ? (char) ('0' + digit) : (char) ('A' + (digit - 10)));
            temp /= base;
        } while (temp > 0);

        // Check palindrome
        String strnum = strnumreversed.toString();
        String reversed = strnumreversed.reverse().toString();

        return strnum.equals(reversed);
    }

    public static void main(String[] args) {
        int limit = 1000;

        System.out.println("Numbers that are palindromes in both base 10 and base 2:");
        for (int i = 1; i <= limit; i++) {
            if (isPalindrome(i, 10) && isPalindrome(i, 2)) {
                System.out.print(i + " (binary: ");

                // Print binary representation
                StringBuilder binary = new StringBuilder();
                int temp = i;
                do {
                    binary.append((temp % 2) == 1 ? '1' : '0');
                    temp /= 2;
                } while (temp > 0);
                binary.reverse();

                System.out.println(binary + ")");
            }
        }
    }
}



/*
run:

Numbers that are palindromes in both base 10 and base 2:
1 (binary: 1)
3 (binary: 11)
5 (binary: 101)
7 (binary: 111)
9 (binary: 1001)
33 (binary: 100001)
99 (binary: 1100011)
313 (binary: 100111001)
585 (binary: 1001001001)
717 (binary: 1011001101)

*/

 

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