2016 AMC 10B Problem 18

Step 1: a formula for the sum.

1. Suppose the run has $k$ numbers and starts at $a$: $a + (a + 1) + \cdots + (a + k - 1)$.
2. The average of evenly spaced numbers is the average of the first and last, $\frac{a + (a + k - 1)}{2}$. The sum is the count times the average:
$$345 = k \cdot \frac{2a + k - 1}{2}, \qquad \text{so} \qquad k(2a + k - 1) = 690.$$

Step 2: turn it into factor pairs.

1. So $690$ is split into two factors: $k$ and $m = 2a + k - 1$.
2. Since $a \ge 1$, the factor $m$ is bigger than $k$. So $k$ is the smaller factor of the pair.
3. Once you pick the pair, the start is $a = \frac{m - k + 1}{2}$. This is a whole number when $m - k$ is odd, that is, when one factor is even and the other odd.
4. $690 = 2 \cdot 3 \cdot 5 \cdot 23$ has only one factor of $2$. So in every factor pair, one factor is even and the other odd. Every pair works.

Step 3: list the pairs.

1. The pair with $k = 1$ is just "$345$" by itself, which is not two or more numbers. Throw it out.
2. The other $7$ pairs each give a run. For example, $k = 2$ gives $172 + 173$, and $k = 23$ gives $4 + 5 + \cdots + 26$.

The answer is $\boxed{\textbf{(E)}\ 7}$.

A sum of consecutive whole numbers is "how many numbers" times "the average," which turns the question into a factoring question. Each factor pair of $2 \times 345$ with one even and one odd factor gives one run. A shortcut to remember: this count equals the number of odd divisors of $345$ (here $8$), minus $1$ for the one-number run.

Suppose $345 = a + (a+1) + \cdots + (a + k - 1)$ with $k \ge 2$ terms and first term $a \ge 1$. The sum of $k$ terms is $k$ times the average of the first and last, so
$$345 = \frac{k\,(2a + k - 1)}{2} \quad\Longrightarrow\quad k\,(2a + k - 1) = 690.$$

The second factor $2a + k - 1$ exceeds $k$ (since $2a - 1 \ge 1$), so $k$ is the smaller factor in a factor pair of $690 = 2 \cdot 3 \cdot 5 \cdot 23$. The factor pairs are
$$(1, 690),\ (2, 345),\ (3, 230),\ (5, 138),\ (6, 115),\ (10, 69),\ (15, 46),\ (23, 30).$$
For each pair, $a = \frac{m - k + 1}{2}$ where $m$ is the larger factor; this is a positive integer exactly when $m - k$ is odd. Since $690$ has a single factor of $2$, one factor in every pair is even and the other odd, so every pair works.

Discard $(1, 690)$, which is the trivial one-term sum. The remaining $7$ pairs each give a valid run, for instance $k = 2$: $172 + 173$, and $k = 23$: $4 + 5 + \cdots + 26$.

The answer is $\boxed{\textbf{(E)}\ 7}$.