<p>Given an integer <code>n</code>, return <em>the number of ways you can write </em><code>n</code><em> as the sum of consecutive positive integers.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 5
<strong>Output:</strong> 2
<strong>Explanation:</strong> 5 = 2 + 3
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 9
<strong>Output:</strong> 3
<strong>Explanation:</strong> 9 = 4 + 5 = 2 + 3 + 4
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 15
<strong>Output:</strong> 4
<strong>Explanation:</strong> 15 = 8 + 7 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def consecutiveNumbersSum(self, n: int) -> int:
        ret = 0
        k = 1 # numer of terms
        
        # we want n = sum of consecutive integers
        # where there are k numbers such that
        # n = a + (a + 1) + (a + 2) + ... + (a + k - 1)

        # n = (k * a) + ((k - 1) * k) // 2
        # thus, we can try each k, and we know it is valid if
        # (n - ((k - 1) * k) // 2) % k == 0
        # this tells us that there exists a value a for such a k

        while (k * (k - 1)) // 2 < n:
            rem = n - (k * (k - 1)) // 2
            if rem % k == 0:
                ret += 1
            k += 1
        
        return ret

        
16500