AMC 8 · 2019 · #23

Grade 6 number-theoryalgebra

Archetype Multiplicative Ratio with Integrality Constraints └ Sub-archetype Integrality / Divisibility Screen

lcmmultiplesfraction-arithmeticlinear-equations-one-var convert-to-algebrabound-inequality-then-enumerate ↑ Prerequisites: lcmfraction-arithmetic

📏 Long solution 💡 3 insights

Problem

After Euclid High School's last basketball game, it was determined that $\frac{1}{4}$ of the team's points were scored by Alexa and $\frac{2}{7}$ were scored by Brittany. Chelsea scored $15$ points. None of the other $7$ team members scored more than $2$ points. What was the total number of points scored by the other $7$ team members?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2019 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: Euclid High School's team has a total score $T$. Alexa scored $\tfrac{1}{4}T$, Brittany scored $\tfrac{2}{7}T$, and Chelsea scored $15$. The remaining $7$ teammates each scored at most $2$ points. Find the total $X$ scored by those $7$ teammates.

Givens: Alexa's points $= \tfrac{1}{4}T$; Brittany's points $= \tfrac{2}{7}T$; Chelsea's points $= 15$; $7$ other players, each scoring at most $2$ points; Answer choices: (A) $10$, (B) $11$, (C) $12$, (D) $13$, (E) $14$

Unknowns: $X$ = total points scored by the other $7$ players

Understand

Plan

Primary tool: #6 Guess and Check

Secondary: #3 Eliminate Possibilities, #7 Identify Subproblems

The condition that $\tfrac{1}{4}T$ and $\tfrac{2}{7}T$ must both be whole numbers forces $T$ to be a multiple of $\text{lcm}(4,7)=28$, so the candidate totals are just $T=28,56,84,\dots$ — a tiny list. Tool #6 (Guess and Check) on those candidates is the fastest path: for each candidate compute Alexa + Brittany + Chelsea, subtract from $T$, and keep the one whose leftover $X$ obeys $0 \le X \le 14$. Tool #3 (Eliminate) supports it by ruling out totals where $X$ would be negative or above $14$. Tool #7 (Identify Subproblems) sets up the bookkeeping: "score budget", "three named players", "seven others".

Execute — Answer: B

#7 Identify Subproblems 5.NF.B.6 Step 1

Split the team into named scorers and the seven others, then write the score budget.
The four pieces must add up to $T$.

$$\underbrace{\tfrac{1}{4}T}_{\text{Alexa}} + \underbrace{\tfrac{2}{7}T}_{\text{Brittany}} + \underbrace{15}_{\text{Chelsea}} + \underbrace{X}_{\text{7 others}} = T$$

💡 Taking a fraction of a whole quantity ($\tfrac{1}{4}$ of $T$, $\tfrac{2}{7}$ of $T$) is the Grade 5 "fraction of a whole" word-problem move.

#3 Eliminate Possibilities 6.NS.B.4 Step 2

Since every player's score is a whole number, $\tfrac{1}{4}T$ must be a whole number (so $4 \mid T$) and $\tfrac{2}{7}T$ must be a whole number (so $7 \mid T$).
The smallest $T$ that works for both is $\text{lcm}(4,7)=28$, and $T$ must be a multiple of $28$.

$$T \in \{28,\,56,\,84,\,112,\dots\}$$

💡 Finding the least common multiple of $4$ and $7$ to combine the divisibility requirements is Grade 6 LCM reasoning.

#3 Eliminate Possibilities 6.EE.B.8 Step 3

Bound how large $T$ can be.
The three named players alone contribute $\tfrac{1}{4}T + \tfrac{2}{7}T + 15 = \tfrac{15}{28}T + 15$, and the remaining $\tfrac{13}{28}T - 15$ points belong to the $7$ others.
Since each of those $7$ scores at most $2$, the others contribute at most $14$.
So we need $\tfrac{13}{28}T - 15 \le 14$, i.e.
$T \le \tfrac{28 \cdot 29}{13} \approx 62.5$.

$$\tfrac{13}{28}T - 15 \le 14 \;\Longrightarrow\; T \le 62.46\ldots$$

💡 Translating "each of $7$ scores at most $2$" into an inequality on $T$ is exactly the Grade 6 inequality-from-a-real-world-constraint standard.

#6 Guess and Check 6.NS.B.4 Step 4

Combine the two filters: $T$ is a multiple of $28$ and $T \le 62$.
Only $T=28$ and $T=56$ survive — a list small enough to check by hand.

Candidates for $T$: $\{28,\,56\}$

💡 Intersecting "multiple of $28$" with the upper bound leaves a tiny candidate set — a Grade 6 multiples-and-bounds check.

#6 Guess and Check 5.NF.B.6 Step 5

Test $T=28$: Alexa $= 7$, Brittany $= 8$, Chelsea $= 15$.
Together that is $7+8+15=30$, already $2$ more than the team total of $28$ — impossible.
Eliminate.

$T=28 \Rightarrow X = 28 - (7+8+15) = -2 < 0$ — reject.

💡 Computing $\tfrac{1}{4}$ and $\tfrac{2}{7}$ of a whole number is a Grade 5 fraction-of-a-whole calculation.

#6 Guess and Check 4.OA.A.3 Step 6

Test $T=56$: Alexa $= 14$, Brittany $= 16$, Chelsea $= 15$.
Together that is $14+16+15=45$, so the $7$ others scored $X = 56 - 45 = 11$.
Since $0 \le 11 \le 14$, this passes every constraint, so $X = \mathbf{11}$.

$$T=56 \Rightarrow X = 56 - (14+16+15) = \mathbf{11} \;\Rightarrow\; \textbf{(B)}$$

💡 Adding and subtracting whole numbers to compute the leftover score is a Grade 4 multi-step word-problem skill.

[1] #7 5.NF.B.6 Split the team into named scorers and the seven others, then write the score bud

[2] #3 6.NS.B.4 Since every player's score is a whole number, $\tfrac{1}{4}T$ must be a whole nu

[3] #3 6.EE.B.8 Bound how large $T$ can be. The three named players alone contribute $\tfrac{1}{

[4] #6 6.NS.B.4 Combine the two filters: $T$ is a multiple of $28$ and $T \le 62$. Only $T=28$ a

[5] #6 5.NF.B.6 Test $T=28$: Alexa $= 7$, Brittany $= 8$, Chelsea $= 15$. Together that is $7+8+

[6] #6 4.OA.A.3 Test $T=56$: Alexa $= 14$, Brittany $= 16$, Chelsea $= 15$. Together that is $14

Review

Reasonableness: Cross-check that $X=11$ can actually happen: $7$ players each scoring at most $2$ can hit $11$ total, e.g. four players score $2$ and three players score $1$ ($4 \times 2 + 3 \times 1 = 11$). The final tallies $14+16+15+11=56=T$ match perfectly. The next candidate, $T=84$, would force the $7$ others to score $\tfrac{13}{28}(84)-15 = 39-15 = 24 > 14$, which is impossible — confirming $T=56$ is unique.

Alternative: Tool #13 (Convert to Algebra) gives the same answer: write $\tfrac{1}{4}T + \tfrac{2}{7}T + 15 + X = T$, clear denominators to get $\tfrac{13}{28}T = 15 + X$, set $T = 28k$, and solve $X = 13k - 15$ subject to $0 \le X \le 14$. Only $k=2$ works, giving $X = 11$. The guess-and-check route avoids the variable manipulation but lands at the same conclusion.

CCSS standards used (min grade 6)

4.OA.A.3 Solve multi-step word problems using four operations with whole numbers (Adding Alexa, Brittany, and Chelsea's scores and subtracting from $T$ to get the leftover $X$ (e.g. $56-(14+16+15)=11$).)
5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers (Reading "$\tfrac{1}{4}$ of $T$" and "$\tfrac{2}{7}$ of $T$" as the fractions-of-a-whole that give each player's score (e.g. $\tfrac{1}{4} \times 56 = 14$).)
6.NS.B.4 Find greatest common factor and least common multiple of two numbers (Combining the divisibility requirements ($4 \mid T$ and $7 \mid T$) into $\text{lcm}(4,7)=28 \mid T$, which restricts $T$ to multiples of $28$.)
6.EE.B.8 Write an inequality of the form x > c or x < c and graph on a number line (Turning "each of $7$ players scores at most $2$" into the inequality $X \le 14$, then $\tfrac{13}{28}T - 15 \le 14$, which bounds $T$.)

⭐ This AMC 8 problem only needs Grade 6 least-common-multiple reasoning you already know — $T$ has to be a multiple of both $4$ and $7$, so it jumps in steps of $28$!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 6)

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