AMC 8 · 2023 · #6

Grade 6 algebra

Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In

place-valuemulti-digit-arithmeticoptimization-counting systematic-enumerationcasework ↑ Prerequisites: place-valuemulti-digit-arithmetic

📏 Short solution 💡 2 insights 📊 Diagram

Problem

The four digits 2, 0, 2, 3 are placed one per box into the expression (base)^(exponent) × (base)^(exponent). Find the largest value the expression can take.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

There are only four digits to drop into four slots, so the candidate arrangements are very few — Tool 2 (Systematic List) lets us enumerate the relevant ones. First, Tool 3 (Eliminate) trims the search massively: if 0 sits in a base slot the whole product is 0, so 0 must go in an exponent slot. Then a short list of cases settles it.

1STEP 1

Keep the product big: 0 as a base makes that factor 0, so 0 must sit in an exponent slot.

0² × (anything) = 0, 2⁰ = 1

2STEP 2

A base to the 0 power is 1, so the product reduces to one power made from the leftover 2, 2, 3.

x⁰ = 1, 1 × (base)^(exp) = (base)^(exp)

3STEP 3

List every power from 2, 2, 3: 3²=9, 2³=8, 2²=4.

3²=9, 2³=8, 2²=4

4STEP 4

Biggest is 3², so pair the leftover 2 with 0 as 2⁰=1: 3² × 2⁰ = 9.

3² × 2⁰ = 9 × 1 = 9

Answer

Check the choices: 16 = 2⁴ and 18 = 2 × 9 aren't reachable since we don't have a 4 in our digit set and 3² × 2 isn't the allowed form (the second factor must itself be a power). 9 from 3² × 2⁰ is the largest reachable value, so the answer 9 makes sense.

💡Key takeaway

This AMC 8 problem only needs Grade 6 exponents (any number to the 0 power is 1!) that you already know!

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