Here is my Python solution!

def andProduct(a, b):
    a = bin(a)[2:]
    b = bin(b)[2:]
    if len(a) != len(b):
        return 0
    for i in range(len(a)):
        if a[i] != b[i]:
            return int(a[:i] + (len(a) - i) * "0", 2)
    return int(a, 2)

import java.io.; import java.util.;

public class Solution {

public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    while(n--!=0){
        long a=sc.nextLong();
        long b=sc.nextLong();
        long and=a;
        if(a%4==0){
            for(long i=a+4;i<=b;i+=4){
                and&=i;
            }
        }
        else if(a%2==0){
            for(long i=a+2;i<=b;i+=2){
                and&=i;
            }
        }
        else{
            for(long i=a+1;i<=b;i++){
                and&=i;
            }
        }
        System.out.println(and);
    }
}

}

# greedy, needs to be compiled to pass the time limit with PyPy3
def andProduct(a, b):
    return reduce(lambda x, count: count & x, range(a, b+1, (a%2==0)+1))

def andProduct(a, b):
    differing_position = (a ^ b).bit_length()
    
    mask = ~((1 << differing_position) - 1)
    
    return a & mask

The same idea as the above but handling the binary representation as a string throws a runtime error on test-case 3, however, I can't understand what is wrong

    position = b.bit_length() - (a^b).bit_length()

    return int(bin(a)[2:][:position] + '0'*(a.bit_length() - position), 2)

The question is missing edge cases where a=b and a

The question is missing edge cases where a=b and a