If a number is divisible by 2^n, do bitwise right shift n times and then bitwise left shift n times to get back the same number.

should do using bit-wise operators

I will run a loop that'll subtract 16 from the number. If result is zero than it is divisible else if it goes less than zero than its is not.

if(16>>4 == 0 ) True; else False

The above condition should be 1 instead of 0. if(16>>4 == 1 ) True; else False

for(int i=0;in) { break; } //number is not divisible by 16 //which will be decided by the flag } }

Sorry for that incompleted answer ! I want to say that simply use a loop upto n and multiply each number with 16 and store in a variable and then go for a check if it is matched by n or not. If matched then flag==true else break;