AMC 8 · 2022 · #20

Grade 8 arithmeticalgebra
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 xx in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of xx?

(A) 1(B) 5(C) 6(D) 8(E) 9\textbf{(A) } -1 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 8 \qquad \textbf{(E) } 9 \qquad

Pick an answer.

(A)
-1
(B)
5
(C)
6
(D)
8
(E)
9
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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

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) $9$

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 $x = 8$.

⭐ 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 $x = 8$.