Competition · AMC preparation · step 4 of 4
AMC 10 · 2022A · #14
Grade 7 countingPick an answer.
AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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 small and large
No pair fits inside the big half G = {8,…,14}, since x ≥ 8 forces 2x ＞ 14; so it reduces to a matching f : L → G with f(a) ≥ 2a.
Grade 6 number-ordering: comparing 2x with the largest possible y rules out any in-G pair without trying examples.
6.NS.C.7Solve An Easier Related ProblemList the legal partners
List each a's legal partners in G — smaller a has more, and a = 7 is tightest with only 14 allowed.
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 ListAssign the most constrained first
Assign hardest-first: 7→14 (1 way), 6→{12,13} (2 ways), then 5 always has 3 legal partners left, whichever 6 took.
Hardest-first ordering means each later count is unambiguous — Grade 7 organized-listing of compound events.
Assigning the most constrained item first makes every later count unambiguous.
▸ Why?
Once the tight choice is fixed, the options it blocks are permanently ruled out.
▸ Why?
The remaining stages are then free of each other, so their counts multiply.
Pair the rest freely
Every leftover big number is ≥ 8 ≥ 2×4, so {1,2,3,4} match the last four freely: 4! = 24 ways.
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 the stages
Multiply the independent stages: 1 × 2 × 3 × 4! = 144.
Grade 7 fundamental counting: multiply choices across independent stages.
7.SP.C.8Look For A PatternMatch against the choices
144 matches choice (E).
Final compare to the multiple-choice list.
6.EE.B.5Eliminate PossibilitiesThis AMC 10 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.
- Split into small and large
- List the legal partners
- Assign the most constrained first
- Pair the rest freely
- Multiply the stages
- Match against the choices
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