2005 AMC 10A Problem 21

The sum is $\frac{n(n+1)}{2}$. The requirement is that
$$\frac{6n}{\,n(n+1)/2\,} = \frac{12n}{n(n+1)} = \frac{12}{n+1}$$
be an integer, i.e. $n + 1$ divides $12$.

The divisors of $12$ are $1, 2, 3, 4, 6, 12$. Since $n \geq 1$, we need $n + 1 \geq 2$, leaving $n + 1 \in \{2, 3, 4, 6, 12\}$ and $n \in \{1, 2, 3, 5, 11\}$.

Quick check of the largest: $1 + \dots + 11 = 66$ and $6 \cdot 11 = 66$, so it divides. There are $5$ values.

The answer is $\boxed{\textbf{(B)}\ 5}$.

Replacing the sum by its closed form turns a divisibility statement about a sequence into a divisibility statement about $n + 1$. Whenever a condition reduces to "$\frac{c}{f(n)}$ is an integer," the answer is a divisor count of $c$, filtered by whatever range $f(n)$ can take.

• Counting all six divisors of 12 and answering 6 (not a choice) or forgetting to exclude n + 1 = 1.
• Simplifying $\frac{6n}{n(n+1)/2}$ to $\frac{6}{n+1}$ (dropping the factor of $2$), which gives only $n \in \{1, 2, 5\}$ and answer (A).