Competition · AMC preparation · step 4 of 4

AMC 10 · 2021A · #10

Grade 8 algebra
Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation
difference-of-squaresexponentspattern-recognitiontelescoping-sum pattern-recognitioneasier-related-problem ↑ Prerequisites: difference-of-squares
📏 Medium solution 💡 2 insights
Problem
Simplify the long product (2 + 3)(2² + 3²)(2⁴ + 3⁴)(2⁸ + 3⁸)(2¹⁶ + 3¹⁶)(2³² + 3³²)(2⁶⁴ + 3⁶⁴). The exponents are 1, 2, 4, 8, 16, 32, 64 — successive powers of 2. Express the result in a clean closed form matching one of the choices.

Pick an answer.

(A)
$3^{127} + 2^{127}$
(B)
$3^{127} + 2^{127} + 2 \cdot 3^{63} + 3 \cdot 2^{63}$
(C)
$3^{128}-2^{128}$
(D)
$3^{128} + 2^{128}$
(E)
$5^{127}$

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Look for a Pattern

Tool #9 (Easier Problem) first — try the product with 2, then 3 factors and watch what happens. Tool #5 (Pattern) reveals the doubling exponent in each collapse. Tool #13 (Algebra) supplies the engine, the difference of squares identity (a - b)(a + b) = a² - b². Tool #3 (Eliminate) discards the impostor choices by checking exponents.

1STEP 1

Try a smaller product

Warm up on just two factors: sneak (3 − 2) = 1 in front, then (a−b)(a+b) = a²−b² twice collapses (3+2)(3²+2²) to 3⁴ - 2⁴.

(3 + 2)(3² + 2²) = (3 - 2)(3 + 2)(3² + 2²) = (3² - 2²)(3² + 2²) = 3⁴ - 2⁴
2STEP 2

Find the exponent pattern

The pattern: n factors collapse to 3²ⁿ − 2²ⁿ. Check n = 3 — (3+2)(3²+2²)(3⁴+2⁴) telescopes to 3⁸ - 2⁸.

n factors → 3²ⁿ - 2²ⁿ after multiplying by (3-2) = 1
3STEP 3

Apply the pattern to seven factors

Apply to all 7 factors: prepend (3 − 2) = 1 and chain the identity, doubling the exponent each step, until it collapses to 3¹²⁸ - 2¹²⁸.

(3-2)(3+2) = 3² - 2² → 3⁴ - 2⁴ → 3⁸ - 2⁸ → 3¹⁶ - 2¹⁶ → 3³² - 2³² → 3⁶⁴ - 2⁶⁴ → 3¹²⁸ - 2¹²⁸
4STEP 4

Count the doublings

Count check: 7 factors → 7 doublings of the exponent from 1, so 2⁷ = 128 is the final exponent — confirming 3¹²⁸ − 2¹²⁸, not a 127-power.

1 → 2 → 4 → 8 → 16 → 32 → 64 → 128 (seven doublings)
5STEP 5

Match against the choices

Match the choices: only (C) fits — the identity gives a difference, killing sum-forms (A)(B)(D); (E) fails as 3² + 2² = 13, not 5².

3¹²⁸ - 2¹²⁸ → (C)
Answer
3¹²⁸-2¹²⁸
Two independent sanity checks. First: count of factors → final exponent. 7 factors with exponents 1, 2, 4, 8, 16, 32, 64 — the sum 1+2+4+…+64 = 127 tempts choice (A), but the correct rule is doubling the highest exponent, giving 128 — matches (C). Second: parity / sign — the difference-of-squares chain always produces aⁿ - bⁿ, never a sum, so (D) is wrong. (E) is wrong because (3² + 2²) = 13 ≠ 5² = 25.
💡Key takeaway

This AMC 10 problem only needs Grade 8 exponent rules you already know — sneak in (3 - 2) = 1 at the front, then (a - b)(a + b) = a² - b² doubles the exponent each step. Seven doublings of 1 land on 128, giving 3¹²⁸ - 2¹²⁸.

  • Try a smaller product
  • Find the exponent pattern
  • Apply the pattern to seven factors
  • Count the doublings
  • Match against the choices

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