AMC 10 · 2024 · #21

Grade 8 algebra

Archetype Lattice / Grid Pattern from Small Cases └ Sub-archetype Lattice Constraint Analysis

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

📏 Long solution 💡 4 insights

Problem

Fill a 5 × 5 grid of integers so every row and every column is an arithmetic progression (AP). Four entries are pinned down: A₅,5 = 0, A₂,4 = 48, A₄,3 = 16, A₃,1 = 12. Find A₁,2.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Convert to Algebra

The grid has 25 entries but only two "free" numbers control everything once a corner is fixed. Anchor at A₅,5 = 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 A₃,1 = 12 pins one endpoint) and to row 2 (where A₂,4 = 48 is the mean of its two neighbors) gives a clean 2 × 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 A₁,2. We do not need a 25-variable assault — two variables suffice.

1STEP 1

Anchor at A₅,5 = 0; let a step up column 5 and b step left along row 5, so the last column and bottom row are multiples of a and b.

A_i,5 = (5-i) a, A₅,j = (5-j) b

2STEP 2

Row 3 is a length-5 AP, so its middle A₃,3 is the mean of the ends A₃,1 = 12 and A₃,5 = 2a: A₃,3 = 6 + a.

A₃,3 =

\frac{A₃,1 + A₃,5}{2}

\frac{12 + 2a}{2}

= 6 + a

3STEP 3

Column 3 is an AP too, with A₄,3 = 16 and A₅,3 = 2b; its upward step is 16 - 2b, so A₃,3 = 32 - 2b.

d_col 3 = 16 - 2b → A₃,3 = 16 + (16 - 2b) = 32 - 2b

4STEP 4

The same cell written two ways must match: 6 + a = 32 - 2b, giving a + 2b = 26 (Eq. 1).

6 + a = 32 - 2b ⟹ a + 2b = 26 (Eq. 1)

5STEP 5

In row 2, A₂,4 = 48 is the mean of A₂,3 = 48 - 4b and A₂,5 = 3a, which gives 3a - 4b = 48 (Eq. 2).

48 =

\frac{(48 - 4b) + 3a}{2}

⟹ 96 = 48 - 4b + 3a ⟹ 3a - 4b = 48 (Eq. 2)

6STEP 6

Eliminate b: double Eq. 1 and add Eq. 2 to get 5a = 100, so a = 20 and then b = 3.

2a + 4b = 52, 3a - 4b = 48 ⟹ 5a = 100 ⟹ a = 20, b = 3

7STEP 7

With a and b known, A₁,3 = 46 and A₁,5 = 80 sit two steps apart in row 1, so its common difference is 17.

A₁,5 = A₁,3 + 2 d_row 1 ⟹ 80 = 46 + 2 d_row 1 ⟹ d_row 1 = 17

8STEP 8

Step back one column to read off A₁,2.

A₁,2 = A₁,3 - d_row 1 = 46 - 17 = 29 → (C)

Answer

Sanity-check by reconstructing row 1 fully. With A₁,3 = 46 and d_row 1 = 17, row 1 reads 12, 29, 46, 63, 80 — and 80 = 4a = A₁,5 matches. Check column 2 for consistency: A₅,2 = 3b = 9 and the column-2 common difference going up is

\frac{A₁,2 - A₅,2}{4}

\frac{29 - 9}{4}

= 5, so column 2 is 9, 14, 19, 24, 29 from bottom to top. Now check A₃,2 = 19 against row 3: row 3 has A₃,1 = 12 and A₃,5 = 2a = 40, so its common difference is

\frac{40 - 12}{4}

= 7, giving row 3 as 12, 19, 26, 33, 40. The A₃,2 = 19 matches both ways. All four given entries are recovered, and (C) 29 is the unique answer.

💡Key takeaway

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 × 2 linear system, then back-fill the row you need.

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