AMC 10 · 2022 · #10
Grade 7 countingPick an answer.
Tool #9 (Easier Related Problem) re-frames the question: instead of "pair 14 numbers with a 2 × rule", first ask the easier structural question "can two big numbers ever be paired?". The answer (no — see Step 1) collapses the problem to a bijection between L = {1,…,7} and G = {8,…,14}, halving the search space. Tool #2 (Systematic List) then tabulates, for each a ∈ L, the set of legal partners in G. Tool #5 (Pattern) handles the order: assign partners starting from the most constrained a (which has the fewest legal b's) and work down. The multiplication principle multiplies the choice counts. Tool #16 (Change Focus) is a sanity-check angle: the smaller a's (a ≤ 4) are completely unconstrained at the end, so they contribute 4! in one chunk. Tool #3 closes by matching the final integer to the answer list.
Split into low and high
The seven smaller ones are the lower half.
Grade 6 number-ordering: comparing 2x with the largest possible y rules out any in-G pair without trying examples.
Comparing the doubled small value with the largest possible partner rules out whole families at once.
▸ Why?
If even the largest partner falls short, every smaller one falls short too.
▸ Why?
The candidates split into groups that never overlap, so each group can be settled on its own.
Tabulate the legal partners
Write each number's legal partners.
Listing partners row-by-row makes the constraint structure visible at a glance — the tail of L is where the bottleneck lives.
6.NS.C.7Make A Systematic ListStart with the tightest
Pair the three tightest ones first.
Hardest-first ordering means each later count is unambiguous — Grade 7 organized-listing of compound events.
7.SP.C.8Look For A PatternThe other four are free
The remaining four pair in any order.
Tool #16 angle: instead of tracking individual constraints, notice they're ALL automatically met — every remaining b ≥ 8 ≥ 2 · 4, so freedom is total.
7.SP.C.8Change Focus Count The ComplementMultiply
Multiplying gives 144.
Grade 7 fundamental counting: multiply choices across independent stages.
7.SP.C.8Look For A PatternMatch the choice
It matches a choice exactly.
Final compare to the multiple-choice list.
6.EE.B.5Eliminate PossibilitiesThis AMC 12 problem only needs Grade 7 organized-list counting you already know — once you notice that the big numbers 8-14 can never pair with each other, the problem becomes a clean bijection. Pair the hardest first (7 with 14, then 6, then 5), and the four leftover small numbers are free, giving 1 × 2 × 3 × 4! = 144.