AMC 10 · 2023 · #13
Grade 7 arithmeticPick an answer.
Multiple-choice with only five candidate values for G is the textbook signal for Tool #3 (Eliminate Possibilities). We split the work with Tool #7: extract two clean tests that any valid G must pass — (i) the 1.4-ratio of wins forces G to be a multiple of 12, and (ii) round-robin forces G to be a triangular number C(N, 2). Then test each answer choice against both filters. The unique survivor is the answer.
Write the win ratio
The ratio forces a divisibility.
A clean way to read "40% more" is the ratio 7:5; whenever a total splits in ratio 7:5, the total must be a multiple of 12.
6.RP.A.3Identify SubproblemsThe shape of the total
The total must be a binomial coefficient.
Round-robin = pick a pair. "Pick 2 from N" is the triangular-number formula (N(N-1))/2.
A round robin plays one game per pair, so the game count is a count of pairs.
▸ Why?
Each pair is counted once from either team, so naming both and halving counts every game once.
▸ Why?
Naming the two teams in order is two independent picks, which is why the raw count is a product.
Filter the choices
The two conditions leave two choices.
Checking divisibility by 12 is a fast filter that knocks out three of the five options in one pass.
4.OA.B.4Eliminate PossibilitiesCheck the last two
Only one is a binomial coefficient.
Two consecutive integers whose product is 72 jump out as 8 and 9; 96 has no such pair.
6.EE.B.6Eliminate PossibilitiesVerify the actual line-up
Verifying with real counts gives 36.
If the survivor passes the original ratio condition exactly, the elimination is airtight.
6.RP.A.3Eliminate PossibilitiesTwo clean conditions on the total — multiple of 12 from the win ratio, and a "C(N, 2) pairs" count from the round-robin — knock out four of the five choices, leaving (B) 36. When a multiple-choice problem hands you the answers, build small filters that each candidate must pass and the answer falls out.