AMC 10 · 2003 · #15

Grade 4 number-theory

Archetype Pair-Sum Invariant └ Sub-archetype Universe Total Minus Known

complementary-countingdivisibility-rules complementary-countingeasier-related-problem ↑ Prerequisites: divisibility-rules

📏 Medium solution 💡 2 insights

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Problem

There are $100$ players in a single tennis tournament. The tournament is single elimination, meaning that a player who loses a match is eliminated. In the first round, the strongest $28$ players are given a bye, and the remaining $72$ players are paired off to play. After each round, the remaining players play in the next round. The match continues until only one player remains unbeaten. The total number of matches played is

$\qquad\textbf{(A) } \text{a prime number}$

$\qquad\textbf{(B) } \text{divisible by 2}$

$\qquad\textbf{(C) } \text{divisible by 5}$

$\qquad\textbf{(D) } \text{divisible by 7}$

$\qquad\textbf{(E) } \text{divisible by 11}$

Pick an answer.

(A)

a prime number

(B)

divisible by 2

(C)

divisible by 5

(D)

divisible by 7

(E)

divisible by 11

AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: A single-elimination tournament starts with $100$ players and ends when exactly one player is left unbeaten. Some players get first-round byes and the rest are paired off, but every game still knocks one player out. Find the total number of matches played, then decide which description of that number is correct.

Givens: There are $100$ players to start; Single elimination: losing one match removes a player for good; In round one, $28$ players get a bye and the other $72$ are paired into matches; Play continues until only one unbeaten player remains

Unknowns: The total number of matches played in the whole tournament, and which answer description ((A) prime, or divisible by (B) $2$, (C) $5$, (D) $7$, (E) $11$) it fits

Understand

Plan

Primary tool: #16 Change Focus / Count the Complement

Secondary: #9 Solve an Easier Related Problem, #3 Eliminate Possibilities

Counting matches round by round means tracking byes and how many players survive each round — fiddly and easy to slip on. Tool #16 (Change Focus) swaps the question: instead of counting matches, count the players who get eliminated, because each match removes exactly one player, so matches and eliminations are the same number. Tool #9 (Solve an Easier Related Problem) then makes the count trivial — to leave $1$ champion out of $100$ players, exactly $99$ must be eliminated, and the bye numbers become irrelevant. Finally tool #3 (Eliminate Possibilities) checks $99$ against the five descriptions to pick the one that fits.

Execute — Answer: E

#16 Change Focus / Count the Complement 2.OA.A.1 Step 1

Match each match to one loser

In single elimination every match ends with exactly one loser, and that loser is out of the tournament.
So there is a perfect one-to-one pairing: each match played corresponds to exactly one player eliminated.
That means the number of matches equals the number of players eliminated.
This is the key switch — stop counting matches and count eliminations instead.

💡 One game knocks out exactly one player, so matches and eliminations rise together, one for one.

#9 Solve an Easier Related Problem 2.NBT.B.5 Step 2

Count how many get eliminated

The tournament ends with exactly one unbeaten champion.
Everyone else has lost a match and been eliminated.
Out of $100$ players, all but the single winner are eliminated, so the number eliminated is $100 - 1 = 99$.
By the pairing from the previous step, the total number of matches is therefore $99$.
Notice the $28$ byes and $72$ paired players never entered this count — they were a distraction.

$$100 - 1 = 99 \text{ matches}$$

💡 Exactly one player never loses, so the other $99$ each lost once, meaning $99$ deciding matches.

#3 Eliminate Possibilities 4.OA.B.4 Step 3

Test 99 against the choices

Now check which description $99$ fits.
Factor it: $99 = 9 \times 11 = 3 \times 3 \times 11$.
So $99$ is composite, ruling out (A).
It is odd, so not divisible by $2$, ruling out (B).
It does not end in $0$ or $5$, so not divisible by $5$, ruling out (C).
Dividing by $7$ gives $99 = 7 \times 14 + 1$, not a whole number, ruling out (D).
But $99 = 9 \times 11$, so it is divisible by $11$.
That is choice (E).

$$99 = 3^2 \times 11 \Rightarrow 11 \mid 99 \Rightarrow \textbf{(E)}$$

💡 Factoring $99$ into $9 \times 11$ exposes $11$ as a divisor and shows no other choice fits.

[1] #16 2.OA.A.1 In single elimination every match ends with exactly one loser, and that loser is

[2] #9 2.NBT.B.5 The tournament ends with exactly one unbeaten champion. Everyone else has lost a

[3] #3 4.OA.B.4 Now check which description $99$ fits. Factor it: $99 = 9 \times 11 = 3 \times 3

Review

Reasonableness: A quick sanity check by rounds confirms $99$: round one plays $36$ matches (the $72$ paired players), leaving $28$ byes $+ 36$ winners $= 64$ players; then $32 + 16 + 8 + 4 + 2 + 1 = 63$ more matches to whittle $64$ down to $1$. Total $36 + 63 = 99$, exactly matching the elimination count. Since $99 = 9 \times 11$ is divisible by $11$ and by no other option, (E) is the only description that fits.

Alternative: Add the matches round by round instead of using the elimination idea. From $100$ players: $72$ play ($36$ matches) leaving $64$; then $64 \to 32$ matches, $32 \to 16$, $16 \to 8$, $8 \to 4$, $4 \to 2$, $2 \to 1$. Summing $36 + 32 + 16 + 8 + 4 + 2 + 1 = 99$. Same total, but it takes far more bookkeeping than counting the $99$ eliminations directly.

CCSS standards used (min grade 4)

2.OA.A.1 Solve one- and two-step word problems using addition and subtraction within 100 (Recognizing that each match removes exactly one player, so matches and eliminations pair up one-to-one.)
2.NBT.B.5 Fluently add and subtract within 100 (Computing $100 - 1 = 99$ eliminations to leave a single champion.)
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Factoring $99 = 9 \times 11$ to test each description and confirm divisibility by $11$.)

⭐ To find the number of matches, don't count games — count who gets knocked out: one champion means the other $99$ players each lost once, so $99$ matches.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Match each match to one loser

Count how many get eliminated

Test 99 against the choices

Review

CCSS standards used (min grade 4)

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