AMC 8 · 2018 · #13

Grade 6 arithmeticalgebra

Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration

mean-median-mode-rangelinear-equations-one-varbound-inequality-then-enumeratesystematic-enumeration bound-inequality-then-enumeratesystematic-enumerationconvert-to-algebra ↑ Prerequisites: linear-equations-one-varmulti-digit-arithmetic

📏 Medium solution 💡 3 insights

Problem

Laila took five math tests, each worth a maximum of 100 points. Laila's score on each test was an integer between 0 and 100, inclusive. Laila received the same score on the first four tests, and she received a higher score on the last test. Her average score on the five tests was 82. How many values are possible for Laila's score on the last test?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2018 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: Laila took $5$ tests, each scored on integers from $0$ to $100$. She got the SAME score on tests $1$ through $4$ (call it $x$) and a STRICTLY HIGHER score on test $5$ (call it $y$, with $y > x$). Her average across the $5$ tests was $82$. How many different integer values are possible for $y$?

Givens: $5$ tests, each integer score from $0$ to $100$ inclusive; Tests $1$-$4$ all received the same score $x$; Test $5$ received a higher score $y$, so $y > x$; The average of the $5$ scores is $82$; Answer choices: (A) $4$, (B) $5$, (C) $9$, (D) $10$, (E) $18$

Unknowns: The number of integer values of $y$ (the last-test score) consistent with all conditions

Understand

Plan

Primary tool: #6 Guess and Check

Secondary: #2 Make a Systematic List, #3 Eliminate Possibilities

Once we turn "average $= 82$" into the total $4x + y = 410$, the unknown $y$ is squeezed into a small window (just above $82$, at most $100$). That is a tiny, bounded set — perfect for Tool #6 (Guess and Check): try each candidate $y$ and see whether the forced $x = (410 - y)/4$ is a whole number that is less than $y$. Tool #2 (Systematic List) keeps the candidate scan in strict order (smallest $y$ first) so nothing is missed or repeated, and Tool #3 (Eliminate Possibilities) is how we drop any $y$ whose forced $x$ is not an integer or violates $y > x$.

Execute — Answer: A

#6 Guess and Check 6.SP.A.3 Step 1

Turn the average condition into a total.
The average of the five scores is $82$, so the sum of the five scores is $5 \times 82 = 410$.
Since the first four scores are all $x$ and the fifth is $y$, this gives $4x + y = 410$.

$$\dfrac{x + x + x + x + y}{5} = 82 \;\Longrightarrow\; 4x + y = 410$$

💡 Average $\times$ count $=$ total — the Grade 6 definition of mean as a single summary number lets us swap the average for an easy sum.

#3 Eliminate Possibilities 6.EE.B.8 Step 2

Find the window for $y$.
Because $y > x$, the last score must beat the average of $82$ (otherwise $x \ge 82$, forcing $4x \ge 328$ and $y \le 82 \le x$, contradiction).
Combined with the cap $y \le 100$, we get $83 \le y \le 100$.

$$y > 82 \text{ and } y \le 100 \;\Longrightarrow\; 83 \le y \le 100$$

💡 Writing inequalities like $y > 82$ and $y \le 100$ to trap a variable is the Grade 6 inequality-on-a-number-line idea.

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

For each candidate $y$ in that window, solve $4x = 410 - y$.
The fifth-test score $y$ is valid only when $410 - y$ is divisible by $4$ (so $x$ is a whole number) AND the resulting $x$ is less than $y$.
Divisibility by $4$ means $y$ must leave the same remainder mod $4$ as $410$.
Since $410 = 4 \cdot 102 + 2$, we need $y$ to leave remainder $2$ when divided by $4$.

$$4x = 410 - y \;\Longrightarrow\; y \text{ leaves remainder } 2 \text{ when divided by } 4$$

💡 Checking which numbers leave a particular remainder when divided by $4$ is just factor / multiple thinking from Grade 4.

#2 Make a Systematic List 4.OA.C.5 Step 4

List every $y$ from $83$ to $100$ that leaves remainder $2$ on division by $4$.
Starting at $86$ (which is $4 \cdot 21 + 2$) and adding $4$ each time: $86, 90, 94, 98$.
The next one, $102$, breaks the $\le 100$ cap.

$$y \in \{86,\; 90,\; 94,\; 98\}$$

💡 Generating numbers by the rule "start at $86$, add $4$" is exactly the Grade 4 shape-and-number pattern skill.

#6 Guess and Check 4.OA.A.3 Step 5

Verify each candidate by computing $x = (410 - y)/4$ and checking $y > x$.
All four pass, so all four are valid.
Count them.

$y = 86 \Rightarrow x = 81$, $y = 90 \Rightarrow x = 80$, $y = 94 \Rightarrow x = 79$, $y = 98 \Rightarrow x = 78$ — all integers with $y > x$. Count $= 4 \;\Rightarrow\; \textbf{(A)}$

💡 Plugging each candidate back into the equation and counting the survivors is multi-step word-problem reasoning from Grade 4.

[1] #6 6.SP.A.3 Turn the average condition into a total. The average of the five scores is $82$,

[2] #3 6.EE.B.8 Find the window for $y$. Because $y > x$, the last score must beat the average o

[3] #6 4.OA.B.4 For each candidate $y$ in that window, solve $4x = 410 - y$. The fifth-test scor

[4] #2 4.OA.C.5 List every $y$ from $83$ to $100$ that leaves remainder $2$ on division by $4$.

[5] #6 4.OA.A.3 Verify each candidate by computing $x = (410 - y)/4$ and checking $y > x$. All f

Review

Reasonableness: Sanity check the extremes. The smallest legal $y$ is $86$ with $x = 81$: average $= (4 \cdot 81 + 86)/5 = 410/5 = 82$ and $86 > 81$. The largest legal $y$ is $98$ with $x = 78$: average $= (4 \cdot 78 + 98)/5 = 410/5 = 82$ and $98 > 78$. Both ends work, and the four candidates step by $4$ (since each unit increase in $y$ must be matched by $\tfrac{1}{4}$ unit drop in $x$, which requires $y$ to change by multiples of $4$ to keep $x$ integer). Four values matches choice (A).

Alternative: Tool #11 (Work Backwards) starting from the boundaries: the largest possible $y$ is $100$, but $410 - 100 = 310$ is not divisible by $4$, so $y = 100$ fails. Step down by $1$ until divisibility works: $y = 98$ gives $x = 78$ — valid. Then jump down by $4$ each time ($94, 90, 86$). The next jump lands on $82$, which violates $y > x$ (would give $x = 82 = y$), so stop. Four values, answer (A).

CCSS standards used (min grade 6)

6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number (Treating the average $82$ as a single summary of all five scores, which lets us replace it with the total $5 \times 82 = 410$ and write $4x + y = 410$.)
6.EE.B.8 Write an inequality of the form x > c or x < c and graph on a number line (Trapping the last-test score with the two inequalities $y > 82$ (from $y > x$ and the average) and $y \le 100$ (problem cap) to bound the candidate set.)
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Recognizing which values of $y$ make $410 - y$ a multiple of $4$, so that $x = (410 - y)/4$ is a whole number.)
4.OA.C.5 Generate a number or shape pattern following a given rule (Listing the valid $y$ values by the rule "start at $86$, add $4$ each time, stop at $100$" to produce $86, 90, 94, 98$.)
4.OA.A.3 Solve multi-step word problems using four operations with whole numbers (Verifying each candidate $y$ by computing the matching $x = (410 - y)/4$ with whole-number arithmetic and confirming $y > x$.)

⭐ This AMC 8 problem only needs Grade 6 mean-as-a-total reasoning plus a quick inequality you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 6)

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