Check if a number is multiple of 5 without using / and % operators | GeeksforGeeks

Given a positive number n, write a function isMultipleof5(int n) that returns true if n is multiple of 5, otherwise false. You are not allowed to use % and / operators.

Method 3 (Set last digit as 0 and use floating point trick) 
A number n can be a mulpile of 5 in two cases. When last digit of n is 5 or 10. If last bit in binary equivalent of n is set (which can be the case when last digit is 5) then we multiply by 2 using n<<=1 to make sure that if the number is multpile of 5 then we have the last digit as 0. Once we do that, our work is to just check if the last digit is 0 or not, which we can do using float and integer comparison trick.

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;

}

Method 1 (Repeatedly subtract 5 from n) 
Method 2 (Convert to string and check the last character) 
Convert n to a string and check the last character of the string. If the last character is ’5′ or ’0′ then n is multiple of 5, otherwise not.

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