Pre-requisite: Input number n

Algorithm:

1)  Set count=0
2)  Do bit wise AND with n and 1. 
    n& 1

    let n be a7a6a5a4a3a2a1a0
    1->00000001
    So doing bitwise AND (refer to published article on bitwise operators) 
    will result in all bits 0 except the LSB which will be a0. 
    
    If a0 is 0 then the Integer has trailing zero else don’t have
    Thus,
    IF
        n& 1 = =0
        count++
    ELSE
        Break; //no more trailing zero since current LSB is 1
    END IF-ELSE
3)  Right shift n by 1
    n=n>>1
4)  IF n==0
        Break
    ELSE
        REPEAT step 2, 3
5)  Print

Example with Explanation:

Trailing zeroes in 12: 12 → 00001100

    So first iteration:
    n=12 //00001100
    n & 1 =0 
    count=1
    n=n>>1 (n=6) //00000110

    So second iteration:
    n=6 //00000110
    n & 1 =0 
    count=2
    n=n>>1 (n=3) //00000011

    So third iteration:
    n=3 //00000011
    n & 1 =1 
    break
    print count=2
    No of trailing zeroes: 2

C implementation

#include <stdio.h>

int main()
{
	unsigned int n;
	printf("enter the integer\n");
	scanf("%d",&n);
	int count=0;
	while(n!=0){
		if(n & 1 == 1) //if current bit is 1
			break; //no more trailing zero
		n=n>>1; //right shift
		count++; //if trailing zero, increase count
	}

	printf("no of trailing zero ");
	//print no of trailing zero
	printf("in its binary representation: %d \n",count); 

	return 0;
}

Output (first run)

enter the integer
12
no of trailing zero in its binary representation: 2

Output (second run)

enter the integer
13
no of trailing zero in its binary representation: 0