Question:

How can you divide a number by 3 without using *, /, +, – % operator?

Input: 48

Output: 16

The method below implements the method using bit-wise operators.

//a function to add 2 numbers without using + int add(int x, int y) { int a, b; do { a = x & y; b = x ^ y; x = a << 1; y = b; } while (a); return b; } int divideby3(int num) { int sum = 0; while (num > 3) { sum = add(num >> 2, sum); num = add(num >> 2, num & 3); } if (num == 3) sum = add(sum, 1); return sum; } //the main function int main(void) { printf("Enter a number:- "); int num; scanf("%d",&num); printf("The number divided by 3 is:- %d",divideby3(num)); return 0; }

For example if your number is 10 then convert to binary

10 =>00001010

if 10 > 3 then shift the binary number 2 bits

Now num will be 00000010 ie, 2

Now sum will be 2

num = (shift the number by 2 bits)+(number BITWISE AND 3)

num = 2+2

Now the number will be 4 then, 4>3 => true so loop will be repeated

4=>00000100 then shift the binary number 2 bits

Now sum=2+1 ie, sum=3 **(this value is returned)**

num=(shift the number(00000100) by 2 bits)+(number BITWISE AND 3)

num=1+0

ie remainder=1