Question

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

Answer 1

#include <stdio.h>
#include <string.h> // strcmp

// Function to check if a number is a palindrome in a given base
int isPalindrome(unsigned int num, unsigned int base) {
    if (base < 2) return 0; // Invalid base

    char strnumreversed[65]; // Enough for binary representation of 64-bit number
    int len = 0;
    unsigned int temp = num;

    // Convert number to string in given base
    do {
        unsigned int digit = temp % base;
        strnumreversed[len++] = (digit < 10) ? ('0' + digit) : ('A' + (digit - 10));
        temp /= base;
    } while (temp > 0);
    strnumreversed[len] = '\0';

    // Check palindrome
    char strnum[65];
    for (int i = 0; i < len; ++i) {
        strnum[i] = strnumreversed[len - 1 - i];
    }
    strnum[len] = '\0';

    return strcmp(strnumreversed, strnum) == 0;
}

int main() {
    unsigned int limit = 1000;

    printf("Numbers that are palindromes in both base 10 and base 2:\n");
    for (unsigned int i = 1; i <= limit; i++) {
        if (isPalindrome(i, 10) && isPalindrome(i, 2)) {
            printf("%u (binary: ", i);

            // Print binary representation
            char binary[65];
            int len = 0;
            unsigned int temp = i;
            do {
                binary[len++] = (temp % 2) ? '1' : '0';
                temp /= 2;
            } while (temp > 0);

            // Reverse binary string
            for (int j = len - 1; j >= 0; --j) {
                putchar(binary[j]);
            }
            printf(")\n");
        }
    }

    return 0;
}



/*
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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