AMC 8 · 2022 · #20

Grade 8 arithmeticalgebra

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

systems-of-equationslinear-equations-one-varlogical-deduction convert-to-algebrabound-inequality-then-enumerate ↑ Prerequisites: linear-equations-one-varmulti-digit-arithmetic

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number $x$ in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of $x$ ?

Pick an answer.

(A)

-1

(B)

(C)

(D)

(E)

AMC 8 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: A $3 \times 3$ grid is filled with integers so that every row and every column has the same sum. Five entries are already filled: top row is $-2, 9, 5$; the middle-right cell is $-1$; the bottom-right cell is $8$; and the bottom-left cell is $x$. Four cells are missing. The number $x$ in the lower-left corner must be strictly larger than each of the other three missing numbers. Find the smallest possible integer value of $x$.

Givens: Top row: $-2,\ 9,\ 5$; Right column has $5$ on top, $-1$ in the middle, $8$ on the bottom; Bottom-left cell is $x$; Four cells are unknown (mid-left, center, bottom-middle, and $x$ itself); All entries are integers; Every row and every column share the same sum; $x$ is strictly greater than each of the other three missing numbers; Answer choices: (A) $-1$, (B) $5$, (C) $6$, (D) $8$, (E) $9$

Unknowns: The smallest integer value of $x$ that satisfies all conditions

Understand

Plan

Primary tool: #13 Convert to Algebra

Secondary: #1 Draw a Diagram, #3 Eliminate Possibilities

The top row is fully filled, so we instantly read off the magic sum $-2 + 9 + 5 = 12$. After that, each missing cell sits in a row or column with only one unknown, so naming the missing cells $a, b, c$ and writing $\text{row sum} = 12$ / $\text{column sum} = 12$ turns the puzzle into a tiny system of linear equations — Tool #13 (Algebra) is the cleanest path. Tool #1 (Diagram) is the labelled $3 \times 3$ grid we keep in front of us so we don't lose track of which cell is which. Once we express $a, b, c$ in terms of $x$, the condition "$x$ beats the others" becomes three simple inequalities, and Tool #3 (Eliminate) lets us check the answer choices $\{-1, 5, 6, 8, 9\}$ against $x > 7$ to land on (D).

Execute — Answer: D

#1 Draw a Diagram 6.NS.C.5 Step 1

Find the common row/column sum.
The top row is the only fully filled row, so its sum determines the magic constant that every row and every column must hit.

$$-2 + 9 + 5 = 12$$

💡 Adding signed integers like $-2$ to a positive sum is exactly the Grade 6 "positive and negative numbers describe quantities" skill.

#13 Convert to Algebra 6.EE.B.6 Step 2

Label the three other missing cells.
Let $a$ be the mid-left cell, $b$ the center cell, and $c$ the bottom-middle cell.
The grid now reads $\begin{pmatrix} -2 & 9 & 5 \\ a & b & -1 \\ x & c & 8 \end{pmatrix}$.

$$a = \text{mid-left},\ b = \text{center},\ c = \text{bottom-middle}$$

💡 Naming unknown cells with letters is exactly Grade 6 "use variables to represent numbers and write expressions."

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

Write each remaining row/column sum as an equation equal to $12$.
Row 2: $a + b + (-1) = 12$.
Row 3: $x + c + 8 = 12$.
Column 1: $-2 + a + x = 12$.
Column 2: $9 + b + c = 12$.

$$a + b = 13,\quad x + c = 4,\quad a + x = 14,\quad b + c = 3$$

💡 Translating each row/column constraint into an equation is Grade 6 equation writing.

#13 Convert to Algebra 8.EE.C.8 Step 4

Solve the system in terms of $x$.
From Column 1: $a = 14 - x$.
From Row 3: $c = 4 - x$.
Substituting into Row 2: $(14 - x) + b = 13$ gives $b = x - 1$.
(Column 2 check: $(x-1) + (4-x) = 3$, confirmed.)

$$a = 14 - x,\quad b = x - 1,\quad c = 4 - x$$

💡 Solving four linked linear equations to express $a, b, c$ in terms of $x$ is the Grade 8 "solve pairs of simultaneous linear equations" idea pushed to a small system.

#13 Convert to Algebra 7.EE.B.4 Step 5

Turn the "$x$ is the largest missing number" condition into inequalities.
We need $x > a$, $x > b$, $x > c$, i.e.
$x > 14 - x$, $x > x - 1$, $x > 4 - x$.
The middle one simplifies to $0 > -1$ (always true).
The first gives $2x > 14 \Rightarrow x > 7$, and the third gives $2x > 4 \Rightarrow x > 2$.
The binding constraint is $x > 7$.

$$x > 7$$

💡 Setting up and solving inequalities like $x > 14 - x$ is Grade 7 "construct simple equations and inequalities to solve problems."

#3 Eliminate Possibilities 7.EE.B.4 Step 6

Pick the smallest integer $x$ with $x > 7$, namely $x = 8$, and check it works against the answer choices.
(A) $-1$, (B) $5$, (C) $6$ all fail $x > 7$.
(D) $8$ gives $a = 6,\ b = 7,\ c = -4$, and indeed $8 > 6,\ 8 > 7,\ 8 > -4$.
(E) $9$ also works but is not the smallest.

$$x = 8 \Rightarrow a = 6,\ b = 7,\ c = -4 \;\Rightarrow\; \textbf{(D)}$$

💡 Eliminating choices that violate $x > 7$ and testing the smallest survivor is the multiple-choice version of solving an inequality for its least integer solution.

[1] #1 6.NS.C.5 Find the common row/column sum. The top row is the only fully filled row, so its

[2] #13 6.EE.B.6 Label the three other missing cells. Let $a$ be the mid-left cell, $b$ the cente

[3] #13 6.EE.B.7 Write each remaining row/column sum as an equation equal to $12$. Row 2: $a + b

[4] #13 8.EE.C.8 Solve the system in terms of $x$. From Column 1: $a = 14 - x$. From Row 3: $c =

[5] #13 7.EE.B.4 Turn the "$x$ is the largest missing number" condition into inequalities. We nee

[6] #3 7.EE.B.4 Pick the smallest integer $x$ with $x > 7$, namely $x = 8$, and check it works a

Review

Reasonableness: Plug $x = 8$ back into the full grid: row 1 $= -2 + 9 + 5 = 12$, row 2 $= 6 + 7 + (-1) = 12$, row 3 $= 8 + (-4) + 8 = 12$; columns are $-2 + 6 + 8 = 12$, $9 + 7 + (-4) = 12$, $5 + (-1) + 8 = 12$. All six sums match, and $8$ is strictly greater than $6, 7, -4$. The smaller answer choices $-1, 5, 6$ all fail $x > 7$, so $x = 8$ is genuinely the minimum.

Alternative: Tool #3 (Eliminate Possibilities) without algebra: scan the choices in increasing order and, for each candidate, fill the grid using the magic sum $12$ to see whether $x$ is the largest of the four missing entries. (A) $x = -1$: bottom row forces $c = 4 - (-1) = 5$, already bigger than $x$, fail. (B) $x = 5$: column 1 forces $a = 14 - 5 = 9$, bigger than $x$, fail. (C) $x = 6$: $a = 14 - 6 = 8 > 6$, fail. (D) $x = 8$: $a = 6,\ b = 7,\ c = -4$, all smaller than $8$ — success on the first try. This bypasses algebraic manipulation entirely and is exactly how a time-pressed AMC 8 solver should attack it.

CCSS standards used (min grade 8)

6.NS.C.5 Understand that positive and negative numbers describe quantities (Combining the signed integers $-2, 9, 5$ (and later $-1, 8, -4$) to verify the constant row/column sum of $12$.)
6.EE.B.6 Use variables to represent numbers and write expressions to solve problems (Introducing the variables $a, b, c$ for the three missing cells besides $x$.)
6.EE.B.7 Solve real-world problems by writing and solving equations of the form px = q (Writing each row/column constraint as a one-step equation that the unknown cell must satisfy ($a + b = 13$, $x + c = 4$, $a + x = 14$, $b + c = 3$).)
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities (Translating "$x$ is larger than $a, b, c$" into the inequalities $x > 14 - x$, $x > x - 1$, $x > 4 - x$ and solving them to get $x > 7$.)
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations (Solving the linked system $\{a+b=13,\ x+c=4,\ a+x=14,\ b+c=3\}$ to express $a, b, c$ as $14-x,\ x-1,\ 4-x$.)

⭐ This AMC 8 problem only needs Grade 8 "solve a small system of linear equations" you already know — once the magic sum $12$ is in hand, the four cells fall out as expressions in $x$ and the inequality $x > 7$ gives $x = 8$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 8)

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