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How to determine whether an n‑bit binary number is divisible by 5 in Java

1 Answer

0 votes
public class Main {

    // function to compute whether a binary number is divisible by 5
    // it processes the bits left to right and keeps track of the remainder modulo 5
    // for each bit b:
    //     remainder = (remainder * 2 + b) % 5
    public static boolean isDivisibleByFive(String bin) {

        // remainder modulo 5 while scanning bits
        int remainder = 0;

        // scan each bit of the binary number
        for (int i = 0; i < bin.length(); i++) {

            // convert '0' or '1' to integer 0 or 1
            int b = bin.charAt(i) - '0';

            // update remainder using modulo arithmetic
            remainder = (remainder * 2 + b) % 5;
        }

        // divisible if final remainder is zero
        return remainder == 0;
    }

    public static void main(String[] args) {

        // read an n-bit binary number as a string
        String bin = "01000110";  // 70

        boolean divisible = isDivisibleByFive(bin);

        System.out.println("Binary number: " + bin);
        System.out.println("Divisible by 5: " + (divisible ? "yes" : "no"));

        /*
          Example walk-through for bin = 01000110:

          Start: remainder = 0

          bit = 0 → remainder = (0*2 + 0) % 5 = 0
          bit = 1 → remainder = (0*2 + 1) % 5 = 1
          bit = 0 → remainder = (1*2 + 0) % 5 = 2
          bit = 0 → remainder = (2*2 + 0) % 5 = 4
          bit = 0 → remainder = (4*2 + 0) % 5 = 3
          bit = 1 → remainder = (3*2 + 1) % 5 = 2
          bit = 1 → remainder = (2*2 + 1) % 5 = 0
          bit = 0 → remainder = (0*2 + 0) % 5 = 0

          Final remainder = 0 → divisible by 5
        */
    }
}


/*
run:

Binary number: 01000110
Divisible by 5: yes

*/

 



answered Jun 27 by avibootz
edited Jun 28 by avibootz

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