2002 AMC 10A Problem 3

There are only a few ways to group four $2$s, so the safest plan is to list all of them and work each one out.

1. How many groupings are there? Think about which power is done last. It splits the four $2$s into a left group and a right group: $1$ and $3$, or $2$ and $2$, or $3$ and $1$. A group of three $2$s can itself be grouped $2$ ways, and a group of one or two $2$s only $1$ way. So there are $2 + 1 + 2 = 5$ groupings.

1. Work out each one, always starting with the innermost parentheses:

1. The first row is the standard order. Why do so many rows agree? Because $2^4$ and $4^2$ are both $16$, so two different-looking middle steps land on the same number.

1. The only values that appear are $65536$ and $256$. Apart from the standard value $65536$, exactly one other value, $256$, is possible.

The answer is $\boxed{\textbf{(B)}\ 1}$.

Changing the order of powers can change the answer, but when there are only a handful of orders, listing them all is the quickest and most reliable proof. The lucky fact $2^4 = 4^2 = 16$ is what squeezes five groupings down to two values. Read the question carefully: it asks for other values, not other groupings.

A chain of four numbers with three operations can be fully parenthesized in exactly five ways. Write $x^y$ as $x \wedge y$ to keep the groupings readable:

1. $2 \wedge (2 \wedge (2 \wedge 2))$: the standard order, $2^{16} = 65536$.
2. $2 \wedge ((2 \wedge 2) \wedge 2)$: inner part is $4^2 = 16$, so again $2^{16}$.
3. $((2 \wedge 2) \wedge 2) \wedge 2$: $(4^2)^2 = 16^2 = 256$.
4. $(2 \wedge (2 \wedge 2)) \wedge 2$: $(2^4)^2 = 16^2 = 256$.
5. $(2 \wedge 2) \wedge (2 \wedge 2)$: $4^4 = 256$.

The key coincidence is $2^{(2^2)} = (2^2)^2 = 16$, which is why groupings 1 and 2 agree and groupings 3 and 4 agree. Every non-standard grouping gives $256$, so exactly one value other than $65536$ occurs.

The answer is $\boxed{\textbf{(B)}\ 1}$.