In how many ways can 75 be expressed as the sum of two or more consecutive positive integers?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
When we add up a number of consecutive integers, the sum of these integers will be equal to the number of integers being added times the average of the integers being added.

Let $N$ be the number of consecutive integers being added, and let $A$ be the average of the consecutive integers being added. Thus, $N A=75$.

Before we determine the possibilities for $N$, we make the observation that $N$ must be less than 12 , since the sum of the first 12 positive integers is 78 , and thus the sum of any 12 or more consecutive positive integers is at least 78.

Case 1: $N$ is odd.

In this case, $A$ (the average) is an integer (there is a "middle number" among the integers being added). Therefore, since $N A=75, N$ must be an odd positive factor of 75 which is bigger than 1 and less than 12, ie. $N$ is one of 3 or 5 . So there are two possibilities when $N$ is odd, namely $24+25+26=75$ and $13+14+15+16+17=75$.

Case 2: $N$ is even.

In this case $A$ will be half-way between two integers.

Set $N=2 k$ and $A=\frac{2 l+1}{2}$, where $k$ and $l$ are integers.

Then

$$
\begin{aligned}
2 k\left(\frac{2 l+1}{2}\right) & =75 \\
k(2 l+1) & =75
\end{aligned}
$$

Thus $k$ is a factor of 75 , and so $N=2 k$ is 2 times a factor of 75 , ie. $N$ could be 2,6 or 10 . So the possibilities here are $37+38=75,10+11+12+13+14+15=75$, and $3+4+5+6+7+8+9+10+11+12=75$.

So there are 5 ways in total.

ANSWER: (E)