AMC 10 · 2013 · #4

Grade 4 logiccounting
Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation
paritydivisibility-ruleslogical-deductionsequences-arithmetic caseworkconvert-to-algebra ↑ Prerequisites: parity
📏 Medium solution 💡 2 insights
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Problem
A softball team played ten games. Their own scores in those games were 1,2,3,4,5,6,7,8,9,10 — each of those totals used in exactly one game. In exactly five of the games they lost by exactly one run. In each of the other five games their score was exactly double their opponent's score. Find the total number of runs the opponents scored across all ten games.

Pick an answer.

(A)
35
(B)
40
(C)
45
(D)
50
(E)
55

AMC 10 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Eliminate Possibilities

The opponent's score in a game is easy to compute once you know which type of game it was, so Tool #7 (Identify Subproblems) splits the opponent total into two group totals: the five one-run losses and the five doubling games. The real work is that the problem never says which scores land in which group, and that is where Tool #3 (Eliminate Possibilities) is primary. Halving an odd score would give a fraction of a run, so odd scores are eliminated from the doubling group; then counting shows five doubling games have only five even scores available, which forces the split completely. Tool #4 (Introduce a Variable) is held in reserve for the review: naming the loss-game score total lets the whole answer be written as one expression, a check that runs on different machinery.

1STEP 1

Two game types, two opponent formulas

Split by game type: in a one-run loss the opponent scored s + 1; in a doubling win, half of s. Five games of each type.

one-run loss: o = s + 1 doubling game: s = 2o → o = s/2
2STEP 2

A doubling game needs an even score

The opponent scores half of s, so s must be even — only 2, 4, 6, 8, 10 work; odd scores 1, 3, 5, 7, 9 are the one-run losses.

o = s/2 is a whole number ⇔ s is even ⇔ s ∈ {2,4,6,8,10}
3STEP 3

Five evens fill five slots exactly

Five distinct doubling scores drawn from a five-value set leaves no freedom: doubling is 2, 4, 6, 8, 10, losses 1, 3, 5, 7, 9.

{doubling scores} ⊆ {2,4,6,8,10}, |{doubling scores}| = 5 = |{2,4,6,8,10}| → {doubling scores} = {2,4,6,8,10}
4STEP 4

Add the two group totals

Losses give opponents 2+4+6+8+10 = 30; doubling games give 1+2+3+4+5 = 15, so the total is 45 — choice (C).

losses: 2+4+6+8+10 = 30 doubling: 1+2+3+4+5 = 15 30 + 15 = 45 → (C)
Answer
45
Write out all ten games and check them one at a time. The losses are 1–2, 3–4, 5–6, 7–8, 9–10: each is a loss by exactly one run. The wins are 2–1, 4–2, 6–3, 8–4, 10–5: in each the team's score is exactly double. That is five of each type, the team's scores are exactly 1 through 10 with no repeats, and the opponent runs add to 2+4+6+8+10+1+2+3+4+5 = 45. The size is sensible too: the team scored 55 runs in all, and a team that wins five games by doubling and loses five by a single run should give up a bit less than it scores. A rough bound confirms the scale: whatever the split, the opponent total is (T+65)/2 where T is the team's score total in the loss games, and T ranges between 1+2+3+4+5 = 15 and 6+7+8+9+10 = 40, so the opponent total must lie between 40 and 52.5. That alone rules out (A) 35 and (E) 55 before any parity reasoning, and 45 sits comfortably inside.
💡Key takeaway

Half of an odd number of runs is not a whole number, so the five odd scores had to be the one-run losses — and with only five even scores left over, there is exactly one way the other five games can go.

  • Two game types, two opponent formulas
  • A doubling game needs an even score
  • Five evens fill five slots exactly
  • Add the two group totals

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