Check if a number is divisible by 3 [closed]

Answer 1

Actually the LSB method would actually make this easier. In C:

MSB method:

/* 
Returns 1 if divisble by 3, otherwise 0
Note: It is assumed 'input' format is valid
*/
int is_divisible_by_3_msb(char *input) {
  unsigned value = 0;
  char *p = input;
  if (*p == '1') {
    value &= 1;
  }
  p++;
  while (*p) {
    if (*p != ',') {
      value <<= 1;
      if (*p == '1') {
        ret &= 1;
      }
    }
    p++;
  }
  return (value % 3 == 0) ? 1 : 0;
}

LSB method:

/* 
Returns 1 if divisble by 3, otherwise 0
Note: It is assumed 'input' format is valid
*/
int is_divisible_by_3_lsb(char *input) {
  unsigned value = 0;
  unsigned mask = 1;
  char *p = input;
  while (*p) {
    if (*p != ',') {
      if (*p == '1') {
        value &= mask;
      }
      mask <<= 1;
    }
    p++;
  }
  return (value % 3 == 0) ? 1 : 0;
}

Personally I have a hard time believing one of these will be significantly different to the other.

Answer 2

Here... something new... how to check if a binary number of any length (even thousands of digits) is divisible by 3.

-->((0))<---1--->()<---0--->(1)        ASCII representation of graph

From the picture.

1. You start in the double circle.
2. When you get a one or a zero, if the digit is inside the circle, then you stay in that circle. However if the digit is on a line, then you travel across the line.
3. Repeat step two until all digits are comsumed.
4. If you finally end up in the double circle then the binary number is divisible by 3.

You can also use this for generating numbers divisible by 3. And I wouldn't image it would be hard to convert this into a circuit.

1 example using the graph...

11000000000001011111111111101 is divisible by 3 (ends up in the double circle again)

Try it for yourself.

You can also do similar tricks for performing MOD 10, for when converting binary numbers into base 10 numbers. (10 circles, each doubled circled and represent the values 0 to 9 resulting from the modulo)

EDIT: This is for digits running left to right, it's not hard to modify the finite state machine to accept the reverse language though.

NOTE: In the ASCII representation of the graph () denotes a single circle and (()) denotes a double circle. In finite state machines these are called states, and the double circle is the accept state (the state that means its eventually divisible by 3)

Answer 3

Here is an simple way to do it by hand. Since 1 = 2² mod 3, we get 1 = 2²ⁿ mod 3 for every positive integer. Furthermore 2 = 2²ⁿ⁺¹ mod 3. Hence one can determine if an integer is divisible by 3 by counting the 1 bits at odd bit positions, multiply this number by 2, add the number of 1-bits at even bit posistions add them to the result and check if the result is divisible by 3.

Example: 57₁₀=111001₂. There are 2 bits at odd positions, and 2 bits at even positions. 2*2 + 2 = 6 is divisible by 3. Hence 57 is divisible by 3.

Here is also a thought towards solving question c). If one inverts the bit order of a binary integer then all the bits remain at even/odd positions or all bits change. Hence inverting the order of the bits of an integer n results is an integer that is divisible by 3 if and only if n is divisible by 3. Hence any solution for question a) works without changes for question b) and vice versa. Hmm, maybe this could help to figure out which approach is faster...

Answer 4

A number is divisible by 3 if the sum of it's digits is divisible by 3.

So you can add the digits and get the sum:

• if the sum is greater or equal to 10 use the same method
• if it's 3, 6, 9 then it is divisible
• if the sum is different than 3, 6, 9 then it's not divisible