2005 AMC 10A Problem 21

Start with a few small cases to see what is being asked.

Checking forever is not possible, so use the formula for the sum.

1. Formula. $1 + 2 + \cdots + n = \frac{n(n + 1)}{2}$.
2. Divide. We want $6n$ divided by this to be a whole number:
$$6n \div \frac{n(n + 1)}{2} = \frac{12n}{n(n + 1)} = \frac{12}{n + 1}.$$
The $n$ cancels because $n$ is positive.
3. So $n + 1$ must divide $12$. The divisors of $12$ are $1, 2, 3, 4, 6, 12$.
4. Remove the one that is too small. $n$ is positive, so $n + 1$ is at least $2$. That leaves $n + 1 = 2, 3, 4, 6, 12$, so $n = 1, 2, 3, 5, 11$.
5. Check the biggest. $1 + \cdots + 11 = 66$ and $6 \times 11 = 66$. It divides.

That is $5$ values, which matches the table ($n = 4$ fails because $5$ does not divide $12$).

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

Replacing a long sum with its formula turns "does this sum divide that number?" into a simple fraction. Once the question becomes "when is $\frac{12}{n + 1}$ a whole number?", the answer is a list of divisors of $12$, trimmed to the values $n + 1$ can actually take.

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 $\boxed{\textbf{(B)}\ 5}$.