AMC 10 · 2002 · #3

Grade 6 arithmetic

Archetype Branching State Enumeration └ Sub-archetype Numeric Operation Tree

exponentsorder-of-operations systematic-enumeration ↑ Prerequisites: order-of-operationsexponents

📏 Medium solution 💡 2 insights

Problem

Exponents are grouped right to left by convention, so a tower of four 2's means

2^{(2^{(2^2)})} = 2^{16}

, which equals 65536. If only the order in which the exponentiations are performed changes, how many other values are possible?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

"How many values are possible" is a counting question over a small, finite set of arrangements, so Tool #2 (Make a Systematic List) is the backbone: there are only a handful of ways to parenthesize a four-high tower, and writing every one of them down guarantees none is missed or double-counted. Tool #7 (Identify Subproblems) does the arithmetic safely — each grouping is evaluated one small power at a time from the innermost bracket outward, so no single step is bigger than squaring a two-digit number. Tool #3 (Eliminate Possibilities) finishes the job: after listing the results we strike out the duplicates and the original value, and what remains is the count the problem wants.

1STEP 1

List every parenthesization

A four-high tower of 2's can be bracketed in exactly five ways, and the given $2^{(2^{(2^2)})}$ is only the first.

2^{(2^{(2²)})}, 2^{((2²)²)}, (2²)^{(2²)}, ((2²)²)², (2^{(2²)})²

2STEP 2

Evaluate each from the inside out

Working each from the innermost bracket, two groupings give $2^{16}=65536$ and the other three all collapse to 256.

65536, 65536, 256, 256, 256

3STEP 3

Discard the repeats and the original

Only 65536 and 256 ever appear, and 65536 is the value already given, so exactly one other value survives — choice (B).

{65536, 256}∖{65536}={256} → 1 other value → (B)

Answer

There are only five ways to parenthesize the tower and one of them is the given expression, so at most 4 other values could ever exist — the count must be between 0 and 4, matching the answer choices. It cannot be 0, because 256 ≠ 65536 shows a genuinely different value does appear. Finding exactly one new value is believable: an exponent tower of 2's collapses to just two sizes, the huge 2¹⁶=65536 when a 16 lands up in the exponent, and the modest 256 whenever the final step is a squaring. Two sizes, minus the original, leaves one.

💡Key takeaway

Parentheses decide the order of the exponent steps, and a four-high tower of 2's only ever collapses to two sizes — 65536 or 256 — so beyond the value given there is just one other.

List every parenthesization
Evaluate each from the inside out
Discard the repeats and the original

List every parenthesization

Evaluate each from the inside out

Discard the repeats and the original

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