AMC 10 · 2015 · #13

Grade 6 logiccounting

Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In

paritycombinations-basicinvariant-monovariant logical-deductionextremal-construction ↑ Prerequisites: paritycombinations-basic

📏 Long solution 💡 3 insights

Problem

Every pair of teams meets once and each game pays out a fixed total. Find the claim that is not forced.

Pick an answer.

(A)

There must be an even number of odd scores.

(B)

There must be an even number of even scores.

(C)

There cannot be two scores of }0\text{.

(D)

The sum of the scores must be at least }100\text{.

(E)

The highest score must be at least }12\text{.

How to solve

Strategy Eliminate Possibilities

Four of these statements are theorems and one is not, so the honest job is to prove four of them and break the fifth with a real tournament. Almost everything hinges on one fact I get by looking at a game instead of at a team: a win pays 2 + 0 and a draw pays 1 + 1, so every game puts exactly 2 points into the league no matter how it ends. That freezes the grand total, which settles the sum statement outright and pins down the parity of the whole list. For the last statement I push to an extreme: a claim that the top score must be large is at its weakest when the scores are spread as evenly as the rules allow, so I check whether a perfectly flat list of scores is legal.

1STEP 1

Every game pays out two points

Every game pays the same total.

2+0=2, 1+1=2

2STEP 2

Count the games, freeze the total

That freezes the grand total at 132.

games=(12 · 11)/2=66, sum of scores=66 · 2=132

3STEP 3

Statement D: the sum is exactly 132

One claim is just that total.

sum = 132 ≥ 100

4STEP 4

Statements A and B: parity of the list

Two more follow from its parity.

132 = (sum of even scores) + (sum of k odd scores) → k is even, 12-k is even

5STEP 5

Statement C: two zeros collide

Another follows from one head-to-head game.

x+y ≥ 2 (from their head-to-head game alone), 0+0=0 ＜ 2

6STEP 6

Statement E turns on the average

The last turns on the average score.

132/12=11; every score ≤ 11 → sum ≤ 12 · 11=132=sum → every score=11

7STEP 7

Build the all-draw tournament

An all-draw tournament breaks it, choice (E).

11 · 1=11 ＜ 12, 12 · 11=132

Answer

The highest score must be at least 12.

The all-draw list passes every demand the other four statements make, which is strong evidence it is a real tournament and not a slip: its total is 132, which is at least 100, so (D) holds; it has 0 odd scores and 12 even scores, and 0 and 12 are both even, so (A) and (B) hold; and it contains no score of 0 at all, so (C) holds. Only the top-score claim breaks, which is what a single false statement should look like. The bound is also tight in exactly the right way. The smallest the highest score can ever be is 11, since 11 is the average, and the largest it can be is 22, when one team wins all 11 of its games and collects 2 points eleven times. So the highest score always lands somewhere from 11 to 22, and the statement asked for at least 12 — missing the true floor by exactly one point. That one-point gap is the whole trap, and it is a narrow one: in ordinary mixed tournaments the top score does reach 12 or more, and only the perfectly flat all-11 list slips underneath.

💡Key takeaway

When the total is locked in, the biggest number in the list can never drop below the average, so to break a claim that the top value must be big, spread everything out as evenly as the rules allow.

Every game pays out two points
Count the games, freeze the total
Statement D: the sum is exactly 132
Statements A and B: parity of the list
Statement C: two zeros collide
Statement E turns on the average
Build the all-draw tournament