Can a given number be written as a sum of two or more consecutive positive integers?

Question

I need to write a method which takes in an int and returns true if the number can be written as a sum of two or more consecutive positive integers and false otherwise.

boolean IsSumOfConsecutiveInts(int num)

I figured out that all odd numbers (except the number 1) can be written as the sum of 2 consecutive positive integers:

return (num > 1 && num % 2 == 1);

but this doesn't account for numbers that can be written as the sum of more than 2 consecutive positive integers (such as 6 == 1 + 2 + 3).

How can I determine whether a number can be written as a sum of two or more consecutive positive integers?

Answer 1

These numbers are called Polite Numbers.

And, conveniently, the only numbers that aren't polite are the powers of 2.

So, that gives us 2 options. We can either determine that a number is polite, OR we can determine that it is not a power of 2.

I did both; the latter is easier (and more efficient).

1. This determines whether or not a number is polite:

   boolean IsSumOfConsecutiveInts(int num)
   {
       int sumOfFirstIIntegers = 3;
       for (int i = 2; sumOfFirstIIntegers <= num; i++)
       {
           if (i%2 == 0 ? (num%i == i/2) : (num%i == 0))
           {
               return true;
           }
           sumOfFirstIIntegers += i + 1;
       }
       return false;
   }
   

   This one is pretty hard to understand. It took me a while to come up with.

   Basically, i is the number of consecutive integers that we are checking;

   sumOfFirstIIntegers is equal to the sum of the first i integers, so that means that all the numbers that can be expressed as a sum of i consecutive integers are greater than or equal to sumOfFirstIIntegers.

   The last part that deserves discussing is the boolean statement i%2 == 0 ? (num%i == i/2) : (num%i == 0). Let's look at some examples:

   i    all sums of i consecutive positive integers
   2    3, 5, 7, 9...
   3    6, 9, 12, 15...
   4    10, 14, 18, 22...
   5    15, 20, 25, 30...
   

   There are two cases, but in either case, we can express all possible numbers that are a sum of i consecutive integers pretty simply.

   1. When i is even, num must be equal to (i * n) + (i / 2) where n is a non-negative integer. This can of course be written as num % i == i / 2.

   2. When i is odd, num must be equal to i * n, where n is a non-negative integer. Which gives us our second condition num % i == 0.

   In addition to these conditions, num can not be less than the sum of the first i positive integers. Hence, our for loop's conditional: sumOfFirstIIntegers <= num.

2. This determines whether a number is not a power of 2:

   boolean IsSumOfConsecutiveInts(int num)
   {
       return (num & (num - 1)) != 0;
   }
   

   This answer does a good job of explaining why this works.

Note that both of the above solutions have the same result, they are just different ways of thinking about the problem.