2008 AMC 10A Problem 3

1. The brackets are nested, so start with the innermost one: $\langle 6\rangle$.
2. The divisors of $6$ are $1, 2, 3, 6$. Leave out $6$ itself and add the rest:
$$\langle 6\rangle = 1 + 2 + 3 = 6.$$
3. The answer came back as $6$. So the next bracket is again $\langle 6\rangle$, which is again $6$.
4. The same happens for the outer bracket:
$$\langle\langle\langle 6\rangle\rangle\rangle = \langle\langle 6\rangle\rangle = \langle 6\rangle = 6.$$

The answer is $\boxed{\textbf{(A)}\ 6}$.

With brackets inside brackets, always work from the inside out and actually compute the first step. Here the first step gives back the number you started with, so nothing ever changes. Numbers like $6$, equal to the sum of their divisors other than themselves, are called perfect numbers ($28$ is the next one).

Work from the inside out. The divisors of $6$ other than $6$ itself are $1, 2, 3$, so
$$\langle 6\rangle = 1 + 2 + 3 = 6.$$

The operation sends $6$ back to $6$, so the next two applications change nothing:
$$\langle\langle\langle 6\rangle\rangle\rangle = \langle\langle 6\rangle\rangle = \langle 6\rangle = 6.$$

The answer is $\boxed{\textbf{(A)}\ 6}$.