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

A stack of four 2's, 2^{2^{2²}}, is worth 65536 when the exponents are grouped in the standard right-to-left order. Insert the parentheses in every other possible order and count how many different values (not counting the given one) can come out.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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 brackets in exactly five ways; write all five out before evaluating any of them.

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

2STEP 2

Evaluate each from the inside out

Evaluating each from the innermost bracket out gives 65536, 65536, 256, 256, 256.

65536, 65536, 256, 256, 256

3STEP 3

Discard the repeats and the original

Only two distinct values exist; drop the given 65536 and 1 new value remains, 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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