2016 AMC 10B Problem 18

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 $\boxed{\textbf{(E)}\ 7}$.

Sums of consecutive integers translate into a factorization $k \cdot (\text{odd-or-even partner}) = 2n$ with the two factors of opposite parity. Counting such factorizations is the same as counting odd divisors of $n$: here $345 = 3 \cdot 5 \cdot 23$ has $8$ odd divisors, and dropping the trivial single-term case leaves $7$. Remember this shortcut, but be able to derive it through the factor-pair table.

• Stopping the search at small $k$ and missing long runs such as $23$ terms ($4$ through $26$), or forgetting to throw out the one-term case $k = 1$.
• Requiring $k$ to divide $345$ itself, which only finds the odd-length runs and misses $k = 2, 6, 10$.