AMC 10 · 2024 · #21

Grade 8 algebra

Archetype Lattice / Grid Pattern from Small Cases

sequences-arithmeticsystems-of-equationslinear-equations-two-var convert-to-algebraidentify-subproblems ↑ Prerequisites: sequences-arithmeticsystems-of-equations

📏 Long solution 💡 4 insights

Problem

The numbers, in order, of each row and the numbers, in order, of each column of a $5 \times 5$ array of integers form an arithmetic progression of length $5{.}$ The numbers in positions $(5, 5), \,(2,4),\,(4,3),$ and $(3, 1)$ are $0, 48, 16,$ and $12{,}$ respectively. What number is in position $(1, 2)?$
$\begin{bmatrix} . & ? &.&.&. \\ .&.&.&48&.\\ 12&.&.&.&.\\ .&.&16&.&.\\ .&.&.&.&0\end{bmatrix}$
$\textbf{(A) } 19 \qquad \textbf{(B) } 24 \qquad \textbf{(C) } 29 \qquad \textbf{(D) } 34 \qquad \textbf{(E) } 39$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

Toolkit + CCSS Solution

Understand

Restated: Fill a $5 \times 5$ grid of integers so every row and every column is an arithmetic progression (AP). Four entries are pinned down: $A_{5,5} = 0$, $A_{2,4} = 48$, $A_{4,3} = 16$, $A_{3,1} = 12$. Find $A_{1,2}$.

Givens: Each row of the $5 \times 5$ array is an AP of length $5$; Each column of the $5 \times 5$ array is an AP of length $5$; $A_{5,5} = 0,\; A_{2,4} = 48,\; A_{4,3} = 16,\; A_{3,1} = 12$; Answer choices: (A) $19$, (B) $24$, (C) $29$, (D) $34$, (E) $39$

Unknowns: The entry $A_{1,2}$ in row $1$, column $2$

Understand

Plan

Primary tool: #13 Convert to Algebra

Secondary: #5 Look for a Pattern, #7 Identify Subproblems

The grid has $25$ entries but only two "free" numbers control everything once a corner is fixed. Anchor at $A_{5,5} = 0$: call $a$ the common difference of column $5$ going up, and $b$ the common difference of row $5$ going left. That alone uses Tool #13 (Convert to Algebra) to label the bottom row and right column with multiples of $a$ and $b$. Tool #5 (Look for a Pattern) supplies the bridge: in any AP of length $5$, the middle term is the mean of the endpoints. Applying that to row $3$ (where $A_{3,1} = 12$ pins one endpoint) and to row $2$ (where $A_{2,4} = 48$ is the mean of its two neighbors) gives a clean $2 \times 2$ linear system in $a, b$. Tool #7 (Identify Subproblems) keeps the work in order: first solve for $a, b$; then find row $1$'s common difference from two known row-$1$ entries; then read off $A_{1,2}$. We do not need a $25$-variable assault — two variables suffice.

Execute — Answer: C

#13 Convert to Algebra 6.EE.A.2 Step 1

Anchor at the known corner and introduce two variables.
With $A_{5,5} = 0$ fixed, let $a$ be the column-$5$ common difference going upward and $b$ the row-$5$ common difference going leftward.
Then column $5$ from bottom to top is $0, a, 2a, 3a, 4a$ and row $5$ from right to left is $0, b, 2b, 3b, 4b$.

$$A_{i,5} = (5-i)\,a,\quad A_{5,j} = (5-j)\,b$$

💡 Two letters control an entire row and an entire column — the Grade 6 "variable stands for a number" move.

#5 Look for a Pattern 6.SP.B.5 Step 2

Apply the AP middle-term identity to row $3$.
Row $3$ is an AP of length $5$, so the middle entry $A_{3,3}$ is the mean of the endpoints $A_{3,1} = 12$ and $A_{3,5} = 2a$.
This gives one expression for $A_{3,3}$ in terms of $a$.

$$A_{3,3} = \dfrac{A_{3,1} + A_{3,5}}{2} = \dfrac{12 + 2a}{2} = 6 + a$$

💡 In a length-$5$ AP, the middle term is the average of the two ends — same as the Grade 6 "mean of two numbers" rule.

#13 Convert to Algebra 6.EE.A.2 Step 3

Get a second expression for $A_{3,3}$ from column $3$.
Column $3$ is also an AP, with known entries $A_{4,3} = 16$ and $A_{5,3} = 2b$.
The column's common difference going up is $A_{4,3} - A_{5,3} = 16 - 2b$, so $A_{3,3}$ is one more step up.

$$d_{\text{col }3} = 16 - 2b \;\Rightarrow\; A_{3,3} = 16 + (16 - 2b) = 32 - 2b$$

💡 An AP is determined by any two of its terms — subtract to find the common difference, then add it to step further.

#7 Identify Subproblems 6.EE.B.7 Step 4

Set the two expressions for $A_{3,3}$ equal to get the first linear equation.
The same cell, written two different ways, must be the same number.

$$6 + a = 32 - 2b \;\Longrightarrow\; a + 2b = 26 \quad \text{(Eq.\ 1)}$$

💡 Two routes to the same cell give an equation — a Grade 6 "solve $px + q = r$" setup.

#5 Look for a Pattern 7.EE.B.4 Step 5

Build the second equation from row $2$.
Use the middle-term identity again: $A_{2,4}$ sits between $A_{2,3}$ and $A_{2,5}$ in row $2$'s AP, so $A_{2,4}$ is their mean.
We know $A_{2,4} = 48$ and $A_{2,5} = 3a$.
To get $A_{2,3}$, step up column $3$ once more: $A_{2,3} = A_{3,3} + d_{\text{col }3} = (32 - 2b) + (16 - 2b) = 48 - 4b$.

$$48 = \dfrac{(48 - 4b) + 3a}{2} \;\Longrightarrow\; 96 = 48 - 4b + 3a \;\Longrightarrow\; 3a - 4b = 48 \quad \text{(Eq.\ 2)}$$

💡 Same middle-term trick, applied in row $2$ — three consecutive AP terms have the middle equal to the average of the outer two.

#13 Convert to Algebra 8.EE.C.8 Step 6

Solve the $2 \times 2$ system.
Multiply Eq.\ 1 by $2$ to align the $b$-coefficients, then add to Eq.\ 2.

$$2a + 4b = 52, \;\; 3a - 4b = 48 \;\Longrightarrow\; 5a = 100 \;\Longrightarrow\; a = 20, \;\; b = 3$$

💡 Elimination on a two-by-two linear system — exactly the Grade 8 systems-of-equations skill.

#7 Identify Subproblems 6.EE.B.7 Step 7

Find row $1$'s common difference.
With $a = 20, b = 3$, we know $A_{1,5} = 4a = 80$.
We also know $A_{2,3} = 48 - 4b = 36$ and $d_{\text{col }3} = 16 - 2b = 10$, so $A_{1,3} = A_{2,3} + d_{\text{col }3} = 36 + 10 = 46$.
Row $1$ has $A_{1,3} = 46$ and $A_{1,5} = 80$, two steps apart in the AP.

$$A_{1,5} = A_{1,3} + 2\,d_{\text{row }1} \;\Longrightarrow\; 80 = 46 + 2\,d_{\text{row }1} \;\Longrightarrow\; d_{\text{row }1} = 17$$

💡 Once two terms of an AP are known, one subtraction (and a divide) gives the common difference.

#7 Identify Subproblems 6.EE.A.2 Step 8

Step back one column to read off $A_{1,2}$.

$$A_{1,2} = A_{1,3} - d_{\text{row }1} = 46 - 17 = 29 \;\Rightarrow\; \textbf{(C)}$$

💡 One subtraction inside row $1$'s AP finishes the problem.

[1] #13 6.EE.A.2 Anchor at the known corner and introduce two variables. With $A_{5,5} = 0$ fixed

[2] #5 6.SP.B.5 Apply the AP middle-term identity to row $3$. Row $3$ is an AP of length $5$, so

[3] #13 6.EE.A.2 Get a second expression for $A_{3,3}$ from column $3$. Column $3$ is also an AP,

[4] #7 6.EE.B.7 Set the two expressions for $A_{3,3}$ equal to get the first linear equation. Th

[5] #5 7.EE.B.4 Build the second equation from row $2$. Use the middle-term identity again: $A_{

[6] #13 8.EE.C.8 Solve the $2 \times 2$ system. Multiply Eq.\ 1 by $2$ to align the $b$-coefficie

[7] #7 6.EE.B.7 Find row $1$'s common difference. With $a = 20, b = 3$, we know $A_{1,5} = 4a =

[8] #7 6.EE.A.2 Step back one column to read off $A_{1,2}$.

Review

Reasonableness: Sanity-check by reconstructing row $1$ fully. With $A_{1,3} = 46$ and $d_{\text{row }1} = 17$, row $1$ reads $12, 29, 46, 63, 80$ — and $80 = 4a = A_{1,5}$ matches. Check column $2$ for consistency: $A_{5,2} = 3b = 9$ and the column-$2$ common difference going up is $(A_{1,2} - A_{5,2})/4 = (29 - 9)/4 = 5$, so column $2$ is $9, 14, 19, 24, 29$ from bottom to top. Now check $A_{3,2} = 19$ against row $3$: row $3$ has $A_{3,1} = 12$ and $A_{3,5} = 2a = 40$, so its common difference is $(40 - 12)/4 = 7$, giving row $3$ as $12, 19, 26, 33, 40$. The $A_{3,2} = 19$ matches both ways. All four given entries are recovered, and (C) $29$ is the unique answer.

Alternative: Tool #11 (Work Backwards) plus a structural fact: in any grid whose rows and columns are APs, $A_{i,j}$ is a bilinear function $A_{i,j} = p + qi + rj + s\,ij$ for some constants $p, q, r, s$. Plugging in the four given entries produces four linear equations in $p, q, r, s$; solving yields the same $A_{1,2} = 29$. This is the same algebra repackaged, but it removes the row-by-row bookkeeping.

CCSS standards used (min grade 8)

6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Introducing $a$ and $b$ as the column-$5$ and row-$5$ common differences, and writing every entry in the boundary row/column as a multiple of $a$ or $b$.)
6.SP.B.5 Summarize numerical data sets, including measures of center (mean) (Using the AP middle-term identity (the middle term of three or five consecutive AP terms equals the mean of the endpoints) to relate $A_{3,3}$ to $A_{3,1}, A_{3,5}$ and $A_{2,4}$ to $A_{2,3}, A_{2,5}$.)
6.EE.B.7 Solve real-world and mathematical problems by writing and solving equations of the form $x + p = q$ and $px = q$ (Equating two expressions for the same cell ($6 + a = 32 - 2b$), and solving $80 = 46 + 2\,d_{\text{row }1}$ for the row-$1$ common difference.)
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities to solve problems (Forming the second equation $3a - 4b = 48$ from row $2$'s mean condition combined with column-$3$ stepping.)
8.EE.C.8 Analyze and solve pairs of simultaneous linear equations (Solving the system $\{a + 2b = 26,\; 3a - 4b = 48\}$ by elimination to get $a = 20, b = 3$.)

⭐ When a grid is made of AP rows and AP columns, two letters at one corner control everything. Use "middle term equals the mean of the endpoints" to turn the given entries into a $2 \times 2$ linear system, then back-fill the row you need.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 8)

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