Check if a number is divisible by 3

Question

I need to find whether a number is divisible by 3 without using %, / or *. The hint given was to use atoi() function. Any

Answer 1

bool isDiv3(unsigned int n)
{
    unsigned int n_div_3 =
        n * (unsigned int) 0xaaaaaaab;
    return (n_div_3 < 0x55555556);//<=>n_div_3 <= 0x55555555

/*
because 3 * 0xaaaaaaab == 0x200000001 and
 (uint32_t) 0x200000001 == 1
*/
}

bool isDiv5(unsigned int n)
{
    unsigned int n_div_5 =
        i * (unsigned int) 0xcccccccd;
    return (n_div_5 < 0x33333334);//<=>n_div_5 <= 0x33333333

/*
because 5 * 0xcccccccd == 0x4 0000 0001 and
 (uint32_t) 0x400000001 == 1
*/
}

Following the same rule, to obtain the result of divisibility test by 'n', we can : multiply the number by 0x1 0000 0000 - (1/n)*0xFFFFFFFF compare to (1/n) * 0xFFFFFFFF

The counterpart is that for some values, the test won't be able to return a correct result for all the 32bit numbers you want to test, for example, with divisibility by 7 :

we got 0x100000000- (1/n)*0xFFFFFFFF = 0xDB6DB6DC and 7 * 0xDB6DB6DC = 0x6 0000 0004, We will only test one quarter of the values, but we can certainly avoid that with substractions.

Other examples :

11 * 0xE8BA2E8C = A0000 0004, one quarter of the values

17 * 0xF0F0F0F1 = 10 0000 0000 1 comparing to 0xF0F0F0F Every values !

Etc., we can even test every numbers by combining natural numbers between them.

Answer 2

Given a number x. Convert x to a string. Parse the string character by character. Convert each parsed character to a number (using atoi()) and add up all these numbers into a new number y. Repeat the process until your final resultant number is one digit long. If that one digit is either 3,6 or 9, the origional number x is divisible by 3.

Answer 3

Here is your optimized solution that every one should know.................

Source: http://www.geeksforgeeks.org/archives/511

#include<stdio.h>


int isMultiple(int n)
{
    int o_count = 0;
    int e_count = 0;


    if(n < 0)  
           n = -n;
    if(n == 0) 
           return 1;
    if(n == 1)
           return 0;

    while(n)
    {

        if(n & 1)
           o_count++;
        n = n>>1;


        if(n & 1)
            e_count++;
        n = n>>1;
    }

     return isMultiple(abs(o_count - e_count));
}


int main()
{
    int num = 23;
    if (isMultiple(num))
        printf("multiple of 3");
    else
        printf(" not multiple of 3");

    return 0;
}

Answer 4

My solution in Java only works for 32-bit unsigned ints.

static boolean isDivisibleBy3(int n) {
  int x = n;
  x = (x >>> 16) + (x & 0xffff); // max 0x0001fffe
  x = (x >>> 8) + (x & 0x00ff); // max 0x02fd
  x = (x >>> 4) + (x & 0x000f); // max 0x003d (for 0x02ef)
  x = (x >>> 4) + (x & 0x000f); // max 0x0011 (for 0x002f)
  return ((011111111111 >> x) & 1) != 0;
}

It first reduces the number down to a number less than 32. The last step checks for divisibility by shifting the mask the appropriate number of times to the right.

Answer 5

While the technique of converting to a string and then adding the decimal digits together is elegant, it either requires division or is inefficient in the conversion-to-a-string step. Is there a way to apply the idea directly to a binary number, without first converting to a string of decimal digits?

It turns out, there is:

Given a binary number, the sum of its odd bits minus the sum of its even bits is divisible by 3 iff the original number was divisible by 3.

As an example: take the number 3726, which is divisible by 3. In binary, this is 111010001110. So we take the odd digits, starting from the right and moving left, which are [1, 1, 0, 1, 1, 1]; the sum of these is 5. The even bits are [0, 1, 0, 0, 0, 1]; the sum of these is 2. 5 - 2 = 3, from which we can conclude that the original number is divisible by 3.

Answer 6

C# Solution for checking if a number is divisible by 3

namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args)
        {
            int num = 33;
            bool flag = false;

            while (true)
            {
                num = num - 7;
                if (num == 0)
                {
                    flag = true;
                    break;
                }
                else if (num < 0)
                {
                    break;
                }
                else
                {
                    flag = false;
                }
            }

            if (flag)
                Console.WriteLine("Divisible by 3");
            else
                Console.WriteLine("Not Divisible by 3");

            Console.ReadLine();

        }
    }
}