AMC 10 · 2024 · #14

Grade 8 algebra
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)
19
(B)
24
(C)
29
(D)
34
(E)
39
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

Start from the zero

The last row and column each ride a single difference.

A_i,5 = (5-i) a, A₅,j = (5-j) b
2STEP 2

Read the middle cell along its row

Cell three-three is the average of its ends.

A₃,3 = (A₃,1 + A₃,5)/2 = (12 + 2a)/2 = 6 + a
3STEP 3

Read the same cell along its column

Down its column the same cell gets a second expression.

d_col 3 = 16 - 2b → A₃,3 = 16 + (16 - 2b) = 32 - 2b
4STEP 4

The first equation

Equating them gives a plus twice b equals 26.

6 + a = 32 - 2b ⟹ a + 2b = 26 (Eq. 1)
5STEP 5

The second equation from 48

The 48 gives three a minus four b equals 48.

48 = ((48 - 4b) + 3a)/2 ⟹ 96 = 48 - 4b + 3a ⟹ 3a - 4b = 48 (Eq. 2)
6STEP 6

Solve the system

The two differences are 20 and 3.

2a + 4b = 52, 3a - 4b = 48 ⟹ 5a = 100 ⟹ a = 20, b = 3
7STEP 7

Find the first row's difference

Row one steps by 17.

A₁,5 = A₁,3 + 2 d_row 1 ⟹ 80 = 46 + 2 d_row 1 ⟹ d_row 1 = 17
8STEP 8

Read the target cell

46 minus 17 is 29.

A₁,2 = A₁,3 - d_row 1 = 46 - 17 = 29 → (C)
Answer
29
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 (A₁,2 - A₅,2)/4 = (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 (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.