Puzzle Zero: Consecutive Sums

Solution: The number is 1,048,576, which is 2^20.  There are several ways to arrive at this value.  One way, which is mathematical,
is to prove the following two statements:

(1) "A number of the form 2^n (with n > 0) cannot be written as the sum of consecutive integers."

and

(2) "A number that cannot be written as the sum of consecutive integers has the form 2^n (with n > 0)"

These, taken together, prove that the only numbers that can't be formed as the sum of consecutive integers are numbers
that are of the form 2^n for some n.  There's only one of these in the range 1,000,000 - 2,000,000, and so it must be
the answer.  This proof is actually a bit tricky, but an outline is given below:

Proof of (1): Suppose that a number has the form 2^n.  Consider any sum of an even number of consecutive integers, say 2k of them.  Then
if the sum starts at some number t, then it has the form

  t + (t + 1) + (t + 2) + ... + (t + 2k - 1)
= 2kt + 1 + 2 + ... + (2k - 1)
= 2kt + (2k - 1)(2k) / 2
= 2kt + 2k^k - k
= k (2t + 2k - 1)

This second term, however, must be an odd number, which can't possibly divide 2^n.  This means that if 2^n is the sum of an even number of
consecutive integers, there must be an odd number of them, say, 2k + 1.  But then if the sum starts at t, we get that it is equal to

  t + (t + 1) + (t + 2) + ... + (t + 2k)
= (2k + 1)t + 1 + 2 + ... + 2k
= (2k + 1)t + 2k (2k + 1) / 2
= (2k + 1)t + k (2k + 1)
= (2k + 1)(t + k)

But this has 2k + 1, an odd number, as a divisor, which is impossible.  Thus in both cases the sum is divisible by an odd number, and
2^n can't be written out as the sum of consecutive numbers.

Proof of (2).  Suppose that some number can't be written out as the sum of consecutive integers.  We need to show that it has the form
2^n for some n > 0.  The number can't be odd, since then it could be written out as 2k + 1, which is the sum of k and k + 1.  This means
that the number has 2 as a factor.  Now suppose that it has any other number factor, which must necessarily be an odd factor, which we'll
denote as 2k + 1.  Then using the math from above, the sum of the 2k + 1 numbers starting at t has the value (2k + 1)(t + k).  By choosing
an appropriate value of t, we can cause this value to sum up to any multiple of 2k + 1, meaning that our number would have to have an odd
factor, which is impossible.  Thus the number has two as a factor and no other odd factors, so it has the form 2^n for some n > 0.