AMC 8 · 2008 · #15

Grade 6 arithmeticnumber-theory

Archetype Small-Domain Constrained Search

divisibility-rulesmodular-arithmeticmean-median-mode-rangesystematic-enumeration modular-arithmeticsystematic-enumeration ↑ Prerequisites: divisibility-rulesmean-median-mode-range

📏 Medium solution 💡 3 insights

Problem

In Theresa's first $8$ basketball games, she scored $7, 4, 3, 6, 8, 3, 1$ and $5$ points. In her ninth game, she scored fewer than $10$ points and her points-per-game average for the nine games was an integer. Similarly in her tenth game, she scored fewer than $10$ points and her points-per-game average for the $10$ games was also an integer. What is the product of the number of points she scored in the ninth and tenth games?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: Theresa scored $7, 4, 3, 6, 8, 3, 1, 5$ in her first $8$ games. In game $9$ she scored fewer than $10$ points and her $9$-game average came out to an integer. In game $10$ she also scored fewer than $10$ points and her $10$-game average was an integer too. Find the product of her game-$9$ and game-$10$ scores.

Givens: First $8$ scores: $7, 4, 3, 6, 8, 3, 1, 5$; Game $9$ score is a whole number from $0$ to $9$; Game $10$ score is a whole number from $0$ to $9$; The average over $9$ games is an integer; The average over $10$ games is an integer; Answer choices: (A) $35$, (B) $40$, (C) $48$, (D) $56$, (E) $72$

Unknowns: The product (game $9$ score) $\times$ (game $10$ score)

Understand

Plan

Primary tool: #12 Use Modular Arithmetic

Secondary: #7 Break into Subproblems

"Average is an integer" is just a divisibility statement in disguise: the running total has to be a multiple of the number of games. That makes Tool #12 (divisibility/modular arithmetic) the right lens. Tool #7 splits the work into two clean subproblems — first pin down the game-$9$ score using divisibility by $9$, then use that result to pin down the game-$10$ score using divisibility by $10$. The "fewer than $10$" cap is what forces a unique answer in each step.

Execute — Answer: B

#7 Break into Subproblems 4.OA.A.3 Step 1

Add up the first $8$ scores to get the starting total.

$$7+4+3+6+8+3+1+5 = 37$$

💡 Before any divisibility thinking, just compute the running total — Grade 4 multi-step arithmetic.

#12 Use Modular Arithmetic 6.SP.B.5 Step 2

Translate "the $9$-game average is an integer" into a divisibility statement.
The average is total $\div 9$, so the $9$-game total must be a multiple of $9$.

$$37 + (\text{game }9) \equiv 0 \pmod{9}$$

💡 Mean $=$ sum $\div$ count, so an integer mean forces the sum to be divisible by the count.

#12 Use Modular Arithmetic 4.OA.B.4 Step 3

Find the smallest multiple of $9$ that is at least $37$.
The multiples of $9$ near $37$ are $36$ and $45$.
Since the game-$9$ score is non-negative, the new total must be $\geq 37$, and the cap (game $9$ $< 10$) means the total can rise by at most $9$, so it must be $45$.

$$\text{multiples of }9: \ldots, 36, 45, 54, \ldots \;\Rightarrow\; \text{new total} = 45$$

💡 Grade 4 "find multiples in a range": only $45$ lies in $[37,\,37+9]$.

#7 Break into Subproblems 4.OA.A.3 Step 4

Subtract to read off the game-$9$ score.

$$\text{game }9 = 45 - 37 = 8 \;\;(\text{and } 8 < 10 \checkmark)$$

💡 First subproblem closed: she scored $8$ in game $9$.

#12 Use Modular Arithmetic 4.OA.B.4 Step 5

Now repeat the same idea for game $10$.
The $10$-game total must be a multiple of $10$, and the new total is $45 + (\text{game }10)$ with game $10$ between $0$ and $9$.
The only multiple of $10$ in the interval $[45,\,54]$ is $50$.

$$45 + (\text{game }10) \equiv 0 \pmod{10} \;\Rightarrow\; \text{new total} = 50$$

💡 Multiples of $10$ end in $0$, so once the total passes $45$ the next stop is $50$.

#7 Break into Subproblems 4.OA.A.3 Step 6

Subtract again to read off the game-$10$ score, then multiply the two game scores to answer the question.

$$\text{game }10 = 50 - 45 = 5,\quad 8 \times 5 = 40 \;\Rightarrow\; \textbf{(B)}$$

💡 Second subproblem closed, and the final product is the answer.

[1] #7 4.OA.A.3 Add up the first $8$ scores to get the starting total.

[2] #12 6.SP.B.5 Translate "the $9$-game average is an integer" into a divisibility statement. Th

[3] #12 4.OA.B.4 Find the smallest multiple of $9$ that is at least $37$. The multiples of $9$ ne

[4] #7 4.OA.A.3 Subtract to read off the game-$9$ score.

[5] #12 4.OA.B.4 Now repeat the same idea for game $10$. The $10$-game total must be a multiple o

[6] #7 4.OA.A.3 Subtract again to read off the game-$10$ score, then multiply the two game score

Review

Reasonableness: Check both averages. After $9$ games the total is $45$, so the average is $45 \div 9 = 5$, an integer. After $10$ games the total is $45 + 5 = 50$, so the average is $50 \div 10 = 5$, also an integer. Both game-$9$ and game-$10$ scores ($8$ and $5$) are below $10$, as required. The product $8 \times 5 = 40$ matches choice (B). Sanity check the uniqueness: any larger multiple of $9$ above $37$ would be $54$, requiring a game-$9$ score of $17$, which breaks the "$<10$" cap — so $45$ really is forced.

Alternative: Tool #6 (Guess and Check): try each game-$9$ score $0,1,2,\ldots,9$. The totals are $37,38,\ldots,46$, and only $45$ is divisible by $9$, giving game $9 = 8$. Then try each game-$10$ score $0,1,\ldots,9$. The totals are $45,46,\ldots,54$, and only $50$ is divisible by $10$, giving game $10 = 5$. Product $= 40$. Slower than the divisibility argument but the same answer.

CCSS standards used (min grade 6)

4.OA.A.3 Solve multistep word problems with whole numbers using the four operations (Adding the first $8$ scores to get $37$, and subtracting to recover each game's score from the running total.)
4.OA.B.4 Find all factor pairs; recognize multiples (Locating the next multiple of $9$ above $37$ (namely $45$) and the next multiple of $10$ above $45$ (namely $50$).)
6.SP.B.5 Summarize numerical data sets, including reporting the number of observations and measures of center (Translating "average over $n$ games is an integer" into the divisibility condition "running total is a multiple of $n$".)

⭐ An integer average is just a multiple in hiding. Hop to the next multiple of $9$, then to the next multiple of $10$ — the two hops give the missing scores.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 6)

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