Divisiblity of 5 without using % and / operator

Question

how to check whether a number is divisible by 5 or not without using % and / operator. I want a quickest algorithm for this problem.

Answer 1

It finally got unlocked, so I can explain my comment, which incidentally turns out to generate better code than GCC does for x % 5 == 0. See here, fill in

#include <stdint.h>
bool divisible_by_5(uint32_t x)
{
   return x % 5 == 0;
}
bool divisible_by_5_fast(uint32_t x)
{
   return x * 0xCCCCCCCD <= 0x33333333;
}

I'll assume unsigned input, because the OP suggested an algorithm that only works with positive input. This method can be extended to signed input, but it's a little messy.

0xCCCCCCCD is the modular multiplicative inverse of 5, modulo 2³². Multiplying a multiple of k (for example, n * k) by the (modular) multiplicative inverse is equivalent to dividing by k, because

(n * k) * inv(k) =
// use associativity
n * (k * inv(k)) =
// use definition of multiplicative inverse
n * 1 =
// multiplicative identity
n

Modulo powers of two, a number has a modular multiplicative inverse iff it is odd.

Since multiplying by an odd number is invertible and is actually a bijection, it can't map any non-multiples of k to the 0 - (2³²-1)/k range.

So when it's outside that range, it can't have been a multiple of k.

0x33333333 is (2³²-1)/5, so if x * 0xCCCCCCCD higher, x can't have been a multiple of 5.

Answer 2

bool trythis(int number){
  Int start = number;
  Do{
    start = start - 5;
  } while (start > 5)

  If (start == 5 || start == 0) {
    Return true;
  } else return false;
}

Answer 3

A good starting point is to look into how division can be accomplished with multiplication and bit-shifts. This question is one place to look.

In particular, you can follow the attached post to hit upon the following strategy. First, "divide by 5" using multiplication and bit-shifts:

 int32_t div5(int32_t dividend) {
     int64_t invDivisor = 0x33333333;
     return 1 + (int32_t) ((invDivisor * dividend) >> 32);
 }

Then, take the result and multiply by 5:

int result = div5(dividend) * 5;

Then, result == dividend if and only dividend is divisible by 5.

if(result == dividend) {
    // dividend is divisible by 5
}
else {
    // dividend is not divisible by 5
}