AMC 8 · 2014 · #23
Grade 6 number-theorylogicPick an answer.
AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The candidate set is tiny: 2-digit primes whose pairwise sums fit on a calendar (at most 31). Tool #2 (Systematic List) lets us write that short list explicitly, and Tool #3 (Eliminate) knocks out every prime ≥ 19 because pairing it with any other 2-digit prime gives a sum past 31. Tool #7 (Identify Subproblems) splits the question into two cleaner pieces: (a) which three primes are used, and (b) which girl wears which number. Part (b) is decided by chaining the date inequalities, never algebra.
List the 2-digit primes: 11, 13, 17, 19, 23, 29, … — we only need the smallest few, since any pairwise sum must stay ≤ 31.
Checking which numbers from 10 to 31 are prime is exactly the Grade 4 "prime or composite" skill.
4.OA.B.4Make A Systematic ListAny prime ≥ 19 fails: 13 + 19 = 32 ＞ 31. Only 11, 13, 17 survive, and three distinct primes are needed, so the set must be {11, 13, 17}.
Knocking out each answer choice by checking the sum against 31 is the Grade 4 multi-step arithmetic check.
4.OA.A.3Eliminate PossibilitiesCheck {11, 13, 17} works: 11 + 13 = 24, 11 + 17 = 28, 13 + 17 = 30 — all ≤ 31 and distinct, so they match the three different dates.
Splitting "find the primes" from "find who wears which" is the Tool #7 subproblems move; the arithmetic check belongs to the first subproblem.
4.OA.A.3Identify SubproblemsLet A, B, C be their numbers. The date order A + C ＜ B + C ＜ A + B; cancel the shared term in each: A ＜ B and C ＜ A, so C ＜ A ＜ B.
Reading the date order as a chain of inequalities and subtracting the common term is Grade 6 inequality reasoning.
6.EE.B.8Identify SubproblemsBy C ＜ A ＜ B, Caitlin gets the smallest, Ashley the middle, Bethany the largest: C = 11, A = 13, B = 17. So Caitlin wears 11 → (A).
Matching the sorted primes to the sorted variables eliminates the four other answer choices in one stroke.
6.EE.B.8Eliminate PossibilitiesThis AMC 8 problem only needs the Grade 6 idea that you can subtract the same number from both sides of an inequality — the prime-number list and date cap do all the rest of the work!