Problem 2 · AMC 8 Stretch
Core
Counting & Probability
reduce-and-expandpattern-recognitionorganizing-data
At the end of the 7th inning of last night's baseball game, the score was tied 8–8. How many different scores were possible at the end of the 6th inning (the inning before)?
Show answer
Answer: 81 possible scores
Show hints
Hint 1 of 4
The tie 8–8 is a big number. Shrink it! How many scores could come before a 0–0 tie? A 1–1 tie? A 2–2 tie? Work out the tiny cases first.
Still stuck? Show hint 2 →
Hint 2 of 4
Before an \(n\)–\(n\) tie, each team's score could be anything from 0 up to \(n\). Make a little grid of (Team A score, Team B score) and count the boxes. For \(n = 0, 1, 2, 3\) you should get 1, 4, 9, 16.
Still stuck? Show hint 3 →
Hint 3 of 4
Those counts 1, 4, 9, 16 are the perfect squares! Each team has \(n+1\) possible scores (0, 1, ..., \(n\)), and the teams are independent, so multiply: \((n+1)\times(n+1)\).
Show solution
Approach: Reduce and expand — small ties reveal the perfect-square count
- Reduce the tie to tiny cases. Tie 0–0: only 0–0 came before → 1 possibility.
- Tie 1–1: each team had 0 or 1, a 2-by-2 grid → 4 possibilities. Tie 2–2: a 3-by-3 grid → 9. Tie 3–3: a 4-by-4 grid → 16.
- The counts 1, 4, 9, 16 are the perfect squares. For a tie \(n\)–\(n\), Team A has \(n+1\) choices (0 to \(n\)) and so does Team B, and they are independent, so the count is \((n+1)^2\).
- For the 8–8 tie, \(n = 8\), so the count is \((8+1)^2 = 9^2 = 81\) possible scores.
Mark:
· log in to save