AMC 10 · 2023 · #5
Grade 8 arithmeticPick an answer.
AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Tool #7 (Identify Subproblems) splits the work cleanly: (A) rewrite every base in prime form using exponent laws, then (B) pair up 2s and 5s into copies of 10, leaving a small leftover, then (C) count the digits of (leftover) × 10¹⁵. Tool #5 (Look for a Pattern) supports step (C) — multiplying any integer by 10¹⁵ adds exactly 15 zeros to the end, so the digit count is (digits of the leftover) + 15. Algebra (#13) is the wrong frame here; this is a pure exponent / place-value problem.
Each base as primes: 8 = 2³, 15 = 3·5. Apply the exponent laws to flatten to 2¹⁵ · 5¹⁰ · 3⁵ · 5⁵.
Pushing exponents through products and powers is the core Grade 8 "properties of integer exponents" move — the expression turns into a clean prime factorization.
8.EE.A.1Identify SubproblemsAdd the exponents of the two 5-powers (5¹⁰ · 5⁵ = 5¹⁵), collapsing to 2¹⁵ · 5¹⁵ · 3⁵.
a^m · aⁿ = a^m+n is the second Grade 8 exponent law — same base, add exponents.
8.EE.A.1Identify SubproblemsPair each 2 with a 5 using aⁿ bⁿ = (ab)ⁿ to build fifteen tens, leaving 10¹⁵ · 3⁵.
Spotting that 2 · 5 = 10 and pairing exponents is the "trailing-zero" pattern — the heart of digit-counting for products with 2s and 5s.
8.EE.A.1Look For A PatternEvaluate the leftover by repeated multiplication: 3⁵ = 243.
Evaluating 3⁵ by chaining multiplications is a Grade 6 "whole-number exponent" calculation.
6.EE.A.1Identify SubproblemsMultiplying 243 by 10¹⁵ appends fifteen zeros, so the numeral is 243 followed by 15 zeros — 3 + 15 = 18 digits.
Multiplying by a power of 10 just shifts the digits to the left and pads with zeros — the Grade 5 "powers of 10 and decimal point" pattern.
5.NBT.A.2Look For A PatternThis AMC 10 problem only needs Grade 8 "integer exponent rules" — pair every 2 with a 5 to make tens, then the leftover times 10¹⁵ tells you the digit count.