AMC 8 · 2023 · #6
Grade 6 algebraProblem
The digits and are placed in the expression below, one digit per box. What is the maximum possible value of the expression?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: Digits available: 2, 0, 2, 3 (each used exactly once); Expression shape: a product of two powers — (base)^(exponent) × (base)^(exponent); Exactly one digit per box
Unknowns: The maximum possible value of the expression
Understand
Restated: 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.
Givens: Digits available: 2, 0, 2, 3 (each used exactly once); Expression shape: a product of two powers — (base)^(exponent) × (base)^(exponent); Exactly one digit per box
Plan
Primary tool: #2 Make a Systematic List
Secondary: #3 Eliminate Possibilities
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.
Execute — Answer: C
6.EE.A.1 Step 1 - To make a product of two powers as large as possible, neither factor should be 0.
- If 0 is placed as a base, then 0^2 = 0 or 0^3 = 0, killing the whole product.
- So 0 must be in an exponent slot.
💡 Knowing how exponents work (any nonzero number to the 0 power is 1) tells us instantly where the 0 must go.
6.EE.A.1 Step 2 - With 0 in an exponent slot, that term equals 1.
- The expression collapses to 1 × (base)^(exponent), so we just need to maximize one power built from two of the remaining digits {2, 2, 3}.
- The leftover digit pairs with the 0 as a base.
💡 The zero-exponent rule simplifies the expression to a single power to maximize.
6.EE.A.1 Step 3 - List every distinct (base, exponent) pair from {2, 2, 3} systematically: (3, 2), (2, 3), and (2, 2).
- Compute each.
💡 Writing every small case down makes the maximum obvious — no case is missed.
3.OA.C.7 Step 4 - The largest of 9, 8, 4 is 9, achieved by 3^2.
- The leftover 2 pairs with the 0 as base, giving 2^0 = 1.
- The maximum expression is 3^2 × 2^0 = 9 × 1 = 9.
💡 Multiplying 9 × 1 is a basic Grade 3 multiplication fact.
6.EE.A.1 To make a product of two powers as large as possible, neither factor should be 0 6.EE.A.1 With 0 in an exponent slot, that term equals 1. The expression collapses to 1 × 6.EE.A.1 List every distinct (base, exponent) pair from {2, 2, 3} systematically: (3, 2), 3.OA.C.7 The largest of 9, 8, 4 is 9, achieved by 3^2. The leftover 2 pairs with the 0 as Review
Reasonableness: Check the choices: 16 = 2^4 and 18 = 2 × 9 aren't reachable since we don't have a 4 in our digit set and 3^2 × 2 isn't the allowed form (the second factor must itself be a power). 9 from 3^2 × 2^0 is the largest reachable value, so the answer 9 makes sense.
Alternative: Tool 6 (Guess and Check) — try a few sensible arrangements (3^2 × 2^0, 2^3 × 2^0, 2^2 × 3^0) and pick the biggest. It lands on 9 just as quickly.
CCSS standards used (min grade 6)
6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Used the meaning of base^exponent and the special cases 0^n = 0, x^0 = 1 to constrain and evaluate arrangements.)3.OA.C.7Fluently multiply and divide within 100 (Used to compute the final product 9 × 1 = 9 as an instant multiplication fact.)
⭐ This AMC 8 problem only needs Grade 6 exponents (any number to the 0 power is 1!) that you already know!
⭐ This AMC 8 problem only needs Grade 6 exponents (any number to the 0 power is 1!) that you already know!