AMC 10 · 2022 · #10

Grade 6 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Factor / Divisibility Constraint

mean-median-mode-rangeconvert-to-algebralinear-equations-one-varsystematic-enumeration convert-to-algebraguess-and-checkidentify-subproblems ↑ Prerequisites: mean-median-mode-rangelinear-equations-one-var

📏 Long solution 💡 3 insights

Problem

Camila writes down five positive integers. The unique mode of these integers is $2$ greater than their median, and the median is $2$ greater than their arithmetic mean. What is the least possible value for the mode?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2022 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: Camila writes down five positive integers. Their unique mode is $2$ more than their median, and their median is $2$ more than their mean. What is the smallest possible value of the mode?

Givens: Five positive integers (repeats allowed); Unique mode (one value appears more than any other); Mode $=$ median $+ 2$; Median $=$ mean $+ 2$; Answer choices: (A) $5$, (B) $7$, (C) $9$, (D) $11$, (E) $13$

Unknowns: The smallest possible value of the mode

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #13 Convert to Algebra, #6 Guess and Check

The problem packages three statistics (mode, median, mean) at once. Tool #7 (Subproblems) — break it into 'what does the sorted list look like?', 'what equation does the mean give?', and 'how small can the unknowns be?'. Tool #13 (Algebra) — once the shape is fixed, one equation in one unknown $M$ tells the whole story. Tool #6 (Guess and Check) — try the smallest legal $x_1, x_2$ first; if parity fails, bump up. We avoid heavier inequality machinery — small directional guesses on positive integers find the floor cleanly.

Execute — Answer: D

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

Sort the five integers: $x_1 \le x_2 \le x_3 \le x_4 \le x_5$.
The median is the middle one, $x_3$.
Call it $M$.
So the mode is $M + 2$.

$$\text{median} = x_3 = M,\;\text{mode} = M + 2$$

💡 In a sorted list of five, the median is just the third number — Grade 6 statistics.

#7 Identify Subproblems 6.SP.A.3 Step 2

Figure out where $M + 2$ sits in the sorted list.
It is bigger than the median $M$, so it must be $x_4$ or $x_5$.
Since the mode appears at least twice (and uniquely most), and the two largest slots are the only places $> M$, force $x_4 = x_5 = M + 2$.
For the mode to be *unique*, no other value can repeat — so $x_1, x_2, x_3 = M$ are all different: $x_1 < x_2 < M$.

$$\text{sorted}:\;x_1, x_2, M, M+2, M+2,\;\;1 \le x_1 < x_2 < M$$

💡 The mode must show up twice in the top slots; the bottom three must all be different so no other value ties the mode.

#13 Convert to Algebra 6.EE.B.7 Step 3

Use the second condition: median = mean + $2$, i.e.
$M = \text{mean} + 2$.
The sum of the five integers is $x_1 + x_2 + M + (M+2) + (M+2) = x_1 + x_2 + 3M + 4$, so the mean is $\frac{x_1 + x_2 + 3M + 4}{5}$.
Then $M = \frac{x_1 + x_2 + 3M + 4}{5} + 2$, i.e.
$5M = x_1 + x_2 + 3M + 4 + 10$, which simplifies to $2M = x_1 + x_2 + 14$.

$$2M = x_1 + x_2 + 14$$

💡 Turning the mean condition into one clean linear equation in $M, x_1, x_2$.

#13 Convert to Algebra 6.EE.B.5 Step 4

Minimize the mode $M + 2$ by minimizing $M$.
From $2M = x_1 + x_2 + 14$, smaller $x_1 + x_2$ means smaller $M$.
The left side $2M$ is even, and $14$ is even, so $x_1 + x_2$ must be even — both even or both odd.

$$x_1 + x_2 \text{ must be even}$$

💡 Parity of both sides forces $x_1 + x_2$ to be even.

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

Try smallest distinct positive integers with even sum.
Smallest $x_1 = 1$.
Then $x_2$ must be odd (to make sum even) and $> 1$, so try $x_2 = 3$.
Sum $= 1 + 3 = 4$.
Then $2M = 4 + 14 = 18$, so $M = 9$.
Check $x_2 < M$: $3 < 9$ ✓.

$$x_1 = 1,\;x_2 = 3 \;\Rightarrow\; 2M = 18 \;\Rightarrow\; M = 9$$

💡 Smallest legal $x_1, x_2$ with even sum makes $M$ smallest — guess-and-check on positive integers.

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

Verify the whole five-number set $\{1, 3, 9, 11, 11\}$.
Sum $1 + 3 + 9 + 11 + 11 = 35$, mean $= 35/5 = 7$.
Median $= 9$.
Mode $= 11$ (appears twice, uniquely).
Check conditions — mode $- $ median $= 11 - 9 = 2$ ✓; median $-$ mean $= 9 - 7 = 2$ ✓.

$$\{1,3,9,11,11\}:\;\text{mean}=7,\;\text{median}=9,\;\text{mode}=11$$

💡 All three conditions hold — the construction is valid.

#7 Identify Subproblems 4.NBT.B.4 Step 7

Smallest possible mode $= M + 2 = 9 + 2 = 11$.
Answer (D).

$$\text{mode}_{\min} = 11 \;\Rightarrow\; \textbf{(D)}$$

💡 Add $2$ to the minimum median to read off the minimum mode.

[1] #7 6.SP.B.5 Sort the five integers: $x_1 \le x_2 \le x_3 \le x_4 \le x_5$. The median is the

[2] #7 6.SP.A.3 Figure out where $M + 2$ sits in the sorted list. It is bigger than the median $

[3] #13 6.EE.B.7 Use the second condition: median = mean + $2$, i.e. $M = \text{mean} + 2$. The s

[4] #13 6.EE.B.5 Minimize the mode $M + 2$ by minimizing $M$. From $2M = x_1 + x_2 + 14$, smaller

[5] #6 5.NBT.B.5 Try smallest distinct positive integers with even sum. Smallest $x_1 = 1$. Then

[6] #6 6.SP.B.5 Verify the whole five-number set $\{1, 3, 9, 11, 11\}$. Sum $1 + 3 + 9 + 11 + 11

[7] #7 4.NBT.B.4 Smallest possible mode $= M + 2 = 9 + 2 = 11$. Answer (D).

Review

Reasonableness: Try a smaller mode and show it fails. Can mode $= 9$? Then $M = 7$ and $2M = 14$, forcing $x_1 + x_2 = 0$ — impossible with positive integers. Can mode $= 7$? Then $M = 5$ and $2M = 10$, forcing $x_1 + x_2 = -4$ — impossible. Can mode $= 5$? Then $M = 3$ and $2M = 6$, forcing $x_1 + x_2 = -8$ — impossible. So choices (A), (B), (C) are all ruled out. (E) $13$ would correspond to $M = 11$ and $x_1 + x_2 = 8$, which works but is not minimal. The smallest valid mode is indeed $11$, matching (D).

Alternative: Tool #2 (Systematic List) over candidate mode values: for each choice $5, 7, 9, 11, 13$, work backwards via $M = \text{mode} - 2$ and $x_1 + x_2 = 2M - 14$. The first one that admits positive distinct $x_1 < x_2 < M$ wins. The list approach matches answer (D) by elimination.

CCSS standards used (min grade 6)

6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures (Computing the mean of the five-element set and reading off the median as the middle term.)
6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number (Using the unique-mode condition to force $x_4 = x_5 = M + 2$ and $x_1 < x_2 < M$.)
6.EE.B.7 Solve real-world problems by writing and solving equations of the form px = q (Turning the mean condition into the equation $2M = x_1 + x_2 + 14$.)
6.EE.B.5 Understand solving an equation or inequality as a process of finding values (Using parity (both sides even) to constrain $x_1 + x_2$ to be even.)
5.NBT.B.5 Fluently multiply multi-digit whole numbers (Computing $2M = 18$ from $x_1 + x_2 = 4$ and reading off $M = 9$.)
4.NBT.B.4 Fluently add and subtract multi-digit whole numbers (Final $M + 2 = 9 + 2 = 11$ for the minimum mode.)

⭐ This AMC 10 problem only needs Grade 6 mean-median-mode and one-step equations you already know — sort the list as $x_1, x_2, M, M+2, M+2$, set up $2M = x_1 + x_2 + 14$, pick the smallest distinct positive $x_1, x_2$ with even sum ($1$ and $3$), and the mode lands at $M + 2 = 11$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 6)

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