In how many ways can 105 be expressed as the sum of at least two consecutive positive integers?
In any sequence of consecutive integers, the average, $a$, is either an integer (if there is an odd number of integers in the sequence) or halfway between two integers (if there is an even number of integers in the sequence).

Furthermore, the sum of this sequence is the product of the average $a$ with the number of integers in the sequence, $n$.

We note that the minimum value for a sequence of 14 consecutive positive integers is $1+2+3+\cdots+13+14=\frac{1}{2}(14)(15)=105$ and any sequence of more than 14 consecutive positive integers has a larger sum.

In other words, $n \leq 14$.

We proceed by considering the equation $a n=105$ and solving for $a$ to obtain $a=\frac{105}{n}$.

For each $n$ from 2 to 14 , we calculate $\frac{105}{n}$ to see if it is an integer or is halfway between two integers:

| $n$ | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $\frac{105}{n}$ | $52 \frac{1}{2}$ | 35 | $26 \frac{1}{4}$ | 21 | $17 \frac{1}{2}$ | 15 | $13 \frac{1}{8}$ | $11 \frac{2}{3}$ | $10 \frac{1}{2}$ | $9 \frac{6}{11}$ | $8 \frac{3}{4}$ | $8 \frac{1}{13}$ | $7 \frac{1}{2}$ |

Each of the seven possibilities where $a=\frac{105}{n}$ gives a result that is an integer or halfway between two integers gives a sequence of positive integers whose sum is 105 :

$52+53 \quad 34+35+36 \quad 19+20+21+22+23 \quad 15+16+17+18+19+20 \quad 12+13+14+15+16+17+18$

$6+7+8+9+10+11+12+13+14+15 \quad 1+2+3+4+5+6+7+8+9+10+11+12+13+14$ Therefore, there are 7 sequences of consecutive positive integers with a sum of 105 .