AMC 10 · 2023 · #4
Grade 8 arithmeticPick an answer.
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.
Write it in primes
Write everything in primes.
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 SubproblemsGather like primes
Gather the like primes.
a^m · aⁿ = a^m+n is the second Grade 8 exponent law — same base, add exponents.
8.EE.A.1Identify SubproblemsPair twos with fives
Pairing gives a power of ten.
Spotting that 2 · 5 = 10 and pairing exponents is the "trailing-zero" pattern — the heart of digit-counting for products with 2s and 5s.
Pairing every two with a five makes a ten, and each ten just adds a zero on the end.
▸ Why?
Every number has exactly one prime recipe, so the supply of twos and fives is fixed in advance.
▸ Why?
Multiplying by ten shifts every digit one place left and pads a zero, so the digit count grows by one.
Compute the leftover
Compute the leftover directly.
Evaluating 3⁵ by chaining multiplications is a Grade 6 "whole-number exponent" calculation.
6.EE.A.1Identify SubproblemsCount the digits
There are 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 12 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.