AMC 8 · 2023 · #20

Grade 6 arithmetic

Archetype Small-Domain Constrained Search

mean-median-mode-rangeoptimization-countingsystematic-enumeration caseworksystematic-enumerationbound-inequality-then-enumerate ↑ Prerequisites: mean-median-mode-rangemental-arithmetic

📏 Long solution 💡 4 insights

Problem

Two integers are inserted into the list $3, 3, 8, 11, 28$ to double its range. The mode and median remain unchanged. What is the maximum possible sum of the two additional numbers?

$\textbf{(A) } 56 \qquad \textbf{(B) } 57 \qquad \textbf{(C) } 58 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 61$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: Insert two integers into the list $3, 3, 8, 11, 28$ so that the new list of seven numbers has DOUBLE the original range while keeping the same mode and the same median. Among all valid pairs, what is the LARGEST possible sum of the two inserted numbers?

Givens: Original list: $3, 3, 8, 11, 28$ (already sorted); Original range = $28 - 3 = 25$; Original mode = $3$ (the only repeated value); Original median = $8$ (the middle value of five); After insertion, the new list has $5+2=7$ numbers; New range must equal $2 \times 25 = 50$; Answer choices: (A) $56$, (B) $57$, (C) $58$, (D) $60$, (E) $61$

Unknowns: The two new integers (call them $x$ and $y$, with $x \le y$); The maximum possible value of $x + y$

Understand

Plan

Primary tool: #2 Make a Systematic List

Secondary: #6 Guess & Check, #3 Eliminate Possibilities

The new range of $50$ can come from three different sources: the original $3$ stays as the minimum and a new large $y$ becomes the maximum, OR the original $28$ stays as the maximum and a new small $x$ becomes the minimum, OR BOTH endpoints are new numbers. Tool #2 (Systematic List) handles this cleanly: list the three cases, solve each one in isolation, then compare. Inside each case the median and mode constraints squeeze $x$ and $y$ into a narrow integer band, and Tool #6 (Guess & Check) picks the extreme integer in that band so the sum is as big as it can be. Tool #3 (Eliminate) finally checks the winning sum against the five answer choices so a typo cannot slip through.

Execute — Answer: D

#2 Make a Systematic List 6.SP.B.5 Step 1

Read off the three statistics that constrain the puzzle.
The list $3,3,8,11,28$ is already sorted, so the range is just last $-$ first, the mode is the only repeated value, and the median (for five numbers) is the third one.

$$\text{range}=28-3=25,\quad \text{mode}=3,\quad \text{median}=8$$

💡 Summarizing a small data set with its range, mode, and median is exactly the Grade 6 statistics standard.

#2 Make a Systematic List 6.EE.B.8 Step 2

Translate the three conditions into algebraic squeezes on $x$ and $y$.
Let $x \le y$ be the two new integers.
The new list has $7$ numbers, so its median is the 4th term: keeping the median at $8$ forces $x \le 8$ and $y \ge 8$.
Keeping the mode at $3$ blocks any new repeats, so $x \ne y$ (otherwise that shared value would tie $3$) and neither value equals $8$, $11$, or $28$.
Combining these we get the strict squeeze $x < 8 < y$ with $y \ne 11, 28$.

$$x \le 8,\; y \ge 8;\quad x \ne 8,\; y \ne 8,11,28;\quad x \ne y\;\Rightarrow\; x < 8 < y$$

💡 Writing each "stays the same" condition as an inequality $x < 8$, $y > 8$ is the Grade 6 inequality standard.

#2 Make a Systematic List 6.SP.B.5 Step 3

Now apply the new-range condition with Tool #2.
The new max minus the new min equals $50$.
Either the min stays at $3$, or the max stays at $28$, or BOTH endpoints become new.
That gives exactly three cases.

$$\text{Case 1: min}=3,\;\text{max}=y;\quad \text{Case 2: min}=x,\;\text{max}=28;\quad \text{Case 3: min}=x,\;\text{max}=y$$

💡 Listing every way the new max and min can be assembled is the Grade 6 "summarize the data" mindset, just applied to a hypothetical list.

#6 Guess & Check 6.EE.B.5 Step 4

Case 1: $\min=3$, $\max=y$.
Then $y-3=50$, so $y=53$.
The min stays at $3$ only if $x\ge 3$, and step 2 already gave $x<8$.
So $x$ is an integer with $3\le x\le 7$, and Tool #6 (Guess \& Check) picks the biggest allowed integer, $x=7$.
Sum $=7+53=60$.
Quick verify: list $\{3,3,7,8,11,28,53\}$ has range $53-3=50$, median (4th term) $=8$, mode $=3$.
All conditions hold.

$$y=53,\; x_{\max}=7\;\Rightarrow\; x+y=7+53=60$$

💡 Solving $y-3=50$ for the unknown $y$ is exactly Grade 6 "find the value that makes the equation true."

#6 Guess & Check 6.NS.C.5 Step 5

Case 2: $\min=x$, $\max=28$.
Then $28-x=50$, so $x=-22$.
The max stays at $28$ only if $y\le 28$, and step 2 gave $8<y\ne 11,28$, so the biggest allowed integer is $y=27$.
Sum $=-22+27=5$.
This is far smaller than $60$, so it cannot beat Case 1.

$$x=-22,\; y_{\max}=27\;\Rightarrow\; x+y=-22+27=5$$

💡 Handling a negative integer like $-22$ as a real value on the number line is the Grade 6 "positives and negatives describe quantities" standard.

#6 Guess & Check 6.EE.B.5 Step 6

Case 3: both endpoints are new, so $x<3$ and $y>28$ with $y-x=50$.
Maximizing $x+y=x+(x+50)=2x+50$ means making $x$ as big as possible.
The biggest integer below $3$ is $x=2$, which gives $y=52$.
Sum $=2+52=54$.
Still less than $60$.

$$x_{\max}=2,\; y=x+50=52\;\Rightarrow\; x+y=2+52=54$$

💡 Picking the largest integer that satisfies $x<3$ is a Grade 6 "solve the inequality" move.

#3 Eliminate Possibilities 6.SP.B.5 Step 7

Compare the three sums and match the choice.
Case 1 wins with $60$, beating Case 2's $5$ and Case 3's $54$.
Tool #3 (Eliminate) confirms $60$ is among the five listed answers — it is choice (D).

$$\max(60,\,5,\,54)=60\;\Rightarrow\;\textbf{(D)}$$

💡 Picking the largest summary value from a short list of candidates is the Grade 6 data-summary skill in action.

[1] #2 6.SP.B.5 Read off the three statistics that constrain the puzzle. The list $3,3,8,11,28$

[2] #2 6.EE.B.8 Translate the three conditions into algebraic squeezes on $x$ and $y$. Let $x \l

[3] #2 6.SP.B.5 Now apply the new-range condition with Tool #2. The new max minus the new min eq

[4] #6 6.EE.B.5 Case 1: $\min=3$, $\max=y$. Then $y-3=50$, so $y=53$. The min stays at $3$ only

[5] #6 6.NS.C.5 Case 2: $\min=x$, $\max=28$. Then $28-x=50$, so $x=-22$. The max stays at $28$ o

[6] #6 6.EE.B.5 Case 3: both endpoints are new, so $x<3$ and $y>28$ with $y-x=50$. Maximizing $x

[7] #3 6.SP.B.5 Compare the three sums and match the choice. Case 1 wins with $60$, beating Case

Review

Reasonableness: Sanity-check the winner $\{3,3,7,8,11,28,53\}$: range $=53-3=50$ (double of $25$, correct), median is the 4th term $=8$ (unchanged), and $3$ still appears twice while every other number appears once, so mode $=3$ (unchanged). The sum $7+53=60$ matches choice (D). The other choices $56,57,58,61$ would require either $y\ne 53$ (breaking the range) or $x>7$ (which forces $x=8$ and ruins the mode/median by giving $\{3,3,8,8,11,28,53\}$, a bimodal list with median $8$ but $3$ no longer the unique mode). Everything lines up.

Alternative: Tool #3 (Eliminate Possibilities) from the choices: in any winning case the new big number must be $y=53$ (otherwise the range is wrong by the Case-1 logic), so the sum is $x+53$ where $x\in\{3,4,5,6,7\}$. The only choice of the form $53+x$ with $x\le 7$ is $53+7=60=$ (D). Choices $56,57,58$ would need $x=3,4,5$ — all valid for range but smaller — and $61$ would need $x=8$, which destroys the mode. So (D) is forced even without listing every case.

CCSS standards used (min grade 6)

6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures (Reading off range, mode, and median of the original five-number list, and re-checking those same measures for each candidate seven-number list.)
6.EE.B.5 Understand solving an equation or inequality as a process of finding values (Solving $y-3=50$ in Case 1, $28-x=50$ in Case 2, and $y-x=50$ in Case 3 to pin down the new endpoints.)
6.EE.B.8 Write an inequality of the form x > c or x < c and graph on a number line (Encoding the median-stays and mode-stays conditions as the integer inequalities $x<8$ and $y>8$ (and the case-specific squeezes like $3\le x\le 7$, $y\le 28$, $x<3$).)
6.NS.C.5 Understand that positive and negative numbers describe quantities (Working with the negative integer $x=-22$ that appears in Case 2 and adding $-22+27=5$ to compute that case's sum.)

⭐ This AMC 8 problem only needs Grade 6 statistics (range, mode, median) and a little inequality reasoning you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 6)

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