AMC 10 · 2019 · #16
Grade 7 probabilityPick an answer.
Tool #7 (Subproblems): only parity matters — strip the values to O (odd) and E (even). Sub-question 1: which 3 × 3 O/E patterns make every row and column have odd sum? Sub-question 2: how many actual number-placements correspond to each valid pattern? Tool #2 (List): enumerate the valid parity patterns. Tool #16 (Complement-style insight): since 5 odds + 4 evens are fixed totals, the parity pattern is heavily constrained; the four evens must form a 2 × 2 rectangle.
Keep only the parity
Only parity matters.
An odd row sum needs an odd number of odd entries — and 5 odds split as 3+1+1.
A row sum is odd exactly when the row holds an odd number of odd entries.
▸ Why?
Odd numbers pair up into even totals, so only an unpaired odd one can tip the sum.
▸ Why?
The row sum is its entries added together, so the parity is decided entirely by what is inside it.
Where the evens must sit
The four evens form a two by two block.
Two non-special rows × two non-special columns = exactly 4 cells for the 4 evens.
2.OA.C.3Make A Systematic ListCount the layouts
Choosing rows and columns gives nine.
Independent choices: pick the all-odd row and the all-odd column.
7.SP.C.8Make A Systematic ListFillings per layout
Odds and evens then arrange freely.
Odd numbers fill odd cells in any order; same for evens — independent choices.
7.SP.C.8Identify SubproblemsCompute the probability
Dividing by all arrangements gives one fourteenth.
Favorable ÷ total — and the factorials cancel cleanly.
7.SP.C.8Identify SubproblemsThis AMC 12 problem only needs Grade 7 parity-and-counting you already know — once you see the 4 evens must form a 2 × 2 rectangle, the count is C(3, 2)C(3, 2) = 9 patterns and P = 9/(C(9, 4)) = 1/14. The answer is (B).