A number is a prime number if that number has precisely two distinct divisors, one and itself. First ten prime numbers are

2, 3, 5, 7, 11, 13, 17, 19, 23, 29

So, if we can find that N has two divisors than it’s a prime number else not, this is actually brute force approach and the complexity is O (N). How we do that, starting from 1 to N we have check if N is divisible by 1, 2, 3, ….., N each time it divide we increment our divisor count by one and at the end we will check if the divisor count is 2 or not.

Can we do better, yes we can. Look carefully the only even prime is 2. If we add an if condition that if the number is 2 return true else false if the number is even, because other even numbers can’t not be a prime number. For even numbers the complexity becomes O (1). So what about odd numbers? How can we improve that? We can reduce the complexity for odd number O (N / 2). See we don’t need to divide by even numbers because the Number N is an odd number, so it will never be divide by an even number. So we have to check if N is divisible by 1, 3, 5, 7, 9, 11, 13, 15 …. N.

We never satisfied! We need more yes the ultimate complexity of an odd number to check whether it’s prime or not is O (√N).
For finding if the number has any divisors other then 1 and itself it will appear under the square root of N, we don’t need to check up to N.

Java implementation of this method is,

public static boolean IsPrime(long num) {
      if(num < 2)
          return false;

     if(num == 2)
       	return true;

      if( (num & 1) == 0)  // the number is even    
          return false;
      long sqrt = (int) Math.sqrt(num); 
      for(int i = 3; i <= sqrt; i += 2) {
          if(num % i == 0)
             return false;
    return true;  

S. Mahbub – Uz – Zaman
Tuesday, September 13, 2011



We all know if a number is divisible by two that is an even number else odd. But in computer divide is not an easy task, it comes up with a cost.

We know, computer works with bits so if we can convert this odd, even check using bits that will be faster than our common method. Look carefully the following numbers.

Look how the last bit is changed, for every odd number the last bit is always one and for even its zero. Using this pattern we can check if a number is odd or even. Here we will use bitwise operator & (and), we know when both bit is one the answer is one else zero. So if we and with one with an odd number the answer will be always one else zero if the number is even.
Java implementation of this method is,

// this method returns true if odd
public static boolean isOdd(int i) {

    return (i & 1) == 1;


Binary representation of 6 (even) is 110 and 1 is 001
6 & 1 = 110 & 001 = 000
Binary representation of 7 (odd) and 1 is 001
7 & 1 = 111 & 001 = 001

S. Mahbub – Uz – Zaman
Monday, September 12, 2011