Info

Hi everyone I was searching an efficient way to check if a number is multiple of 5. So I searched on google and found this solution on geeksforgeeks.org.

There were 3 solutions of my problem.

1. First solution was to subtract 5 until reaching zero,
2. Second solution was to convert the number to string and check last character to be 5 or 0,
3. Third solution was by doing some interesting operations on bitwise level.

I'm interested in third solution as I can fully understand the first and the second.

Here's the code from geeksforgeeks.

bool isMultipleof5(int n) 
{ 
    // If n is a multiple of 5 then we
    // make sure that last digit of n is 0
    if ( (n & 1) == 1 )
        n <<= 1;

    float x = n;
    x = ( (int)(x * 0.1) ) * 10;

    // If last digit of n is 0 then n
    // will be equal to (int)x
    if ( (int)x == n )
        return true;
    return false;
}

I understand only some parts of the logic. I haven't even tested this code. So I need to understand it to use freely.

As said in mentioned article this function is multiplying number by 2 if last bit is set and then checking last bit to be 0 and returns true in that case. But after checking binary representations of numbers I got confused as last bit is 1 in case of any odd number and last bit is 0 in case of any even number. So...

Actual question is

What's the logic of this function?

Any answer is appreciated!

Thanks for all!

CodePudding user response：

1. The most straightforward way to check if a number is a multiple of 5 is to simply

if (n % 5 == 0) {
    // logic...
}

What the bit manipulation code does is:

1. If the number is odd, multiply it by two. Notice that for multiples of 5, the ones digit will end in either 0 or 5, and doubling the number will make it end in 0.
2. We create a number x that is set to n, but with a ones digit set to 0. (We do this by multiplying n by 0.1, which removes the ones digit, and then multiply by 10 in order to add a 0, which has a total effect of just changing the ones digit to 0).
3. We know that originally, if n was a multiple of 5, it would have a ones digit of 0 after step 1. So we check if x is equal to it, and if so, then we can say n was a multiple of 5.