AMC 10 · 2021 · #9
Grade 8 algebraPick an answer.
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.
Insert the missing factor
Three minus two is one, so it is free.
Two factors give 3⁴ - 2⁴ — the exponent doubled twice from 1 to 4, the count of original factors plus one.
6.EE.A.3Solve An Easier Related ProblemMake a difference of squares
Two factors merge into a difference of squares.
Each application of the difference of squares squares the previous exponent — so 1 → 2 → 4 → 8 → …
Each use of the difference of squares squares the previous exponent.
▸ Why?
A difference of two squares collapses into the two numbers added times the two subtracted.
▸ Why?
An exponent counts how many times a factor is used, so squaring a power doubles that count.
Cascade through the factors
Each factor doubles the exponent.
Each collapse doubles the exponent: 1 → 2 → 4 → 8 → 16 → 32 → 64 → 128. Seven doublings produce 2⁷ = 128.
8.EE.A.1Convert To AlgebraTrack the exponent
Seven doublings reach one hundred twenty-eight.
7 original factors → 7 doublings of the exponent starting from 1 → end at 2⁷ = 128.
8.EE.A.1Look For A PatternMatch the choice
The answer is three to the one hundred twenty-eighth minus two to the same.
The telescoping identity always produces aⁿ - bⁿ (a difference), so any sum-form choice is impossible.
8.EE.A.1Eliminate PossibilitiesThis AMC 12 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.