AMC 10 · 2025 · #19

Grade 7 arithmetic
Archetype Lattice / Grid Pattern from Small Cases └ Sub-archetype Figure-N Growth Sequence
recursive-sequencesequences-arithmeticexponents pattern-recognition ↑ Prerequisites: sequences-arithmeticexponents
📏 Medium solution 💡 3 insights
Problem
An array grows one row at a time. The top row is -1, 3, 1. To make the next row, add each side-by-side pair of numbers, then place -1 at the front and 1 at the back. One of the rows adds up to 12,288. In that row, find the third number from the left.

Pick an answer.

(A)
-29
(B)
-21
(C)
-14
(D)
-18
(E)
-3

AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Look for a Pattern

Building the whole array down to a giant row is too much work, so hunt for patterns instead. First, watch the row totals: they follow a clean doubling rule that tells you which row is the target without writing it out. Then track only the two numbers you care about — the 2nd and 3rd from the left — since each depends only on a couple of numbers directly above. Each of those makes its own simple pattern, so a short calculation reaches the far-off row.

1STEP 1

Build a few rows to see the rule

Apply the rule to row 1: -1, 3, 1 gives -1, 2, 4, 1, and then -1, 1, 6, 5, 1 — exactly the picture, so the rule is clear.

Row 1: {-1}, 3, 1 Row 2: {-1}, 2, 4, 1 Row 3: {-1}, 1, 6, 5, 1
2STEP 2

Row totals double, so find the row

Row totals run 3, 6, 12 — doubling, since each number feeds two sums below and the ends cancel. So 12,288 is row 13.

3, 6, 12, … = 3 · 2^ n-1; 3 · 2^ n-1=12288 → 2^ n-1=4096=2¹² → n=13
3STEP 3

Track only the 2nd and 3rd numbers

Name them: a is a row's 2nd number, b its 3rd. New a is old 1st plus old 2nd, new b is old 2nd plus old 3rd; row 1 has a = 3, b = 1.

a_n=2nd number of row n, b_n=3rd number of row n; a₁=3, b₁=1
4STEP 4

The 2nd number drops by 1 each row

The 1st number is always -1, so the 2nd number just loses 1 a row: 3, 2, 1, 0, -1, … In row n it is 4 minus n.

a_n=-1+a_n-1=a_n-1-1, a₁=3 → a_n=4-n
5STEP 5

Add up to reach the 3rd number in row 13

b picks up each row's 2nd number, so row 13 gives 1 plus the sum of 4-k for k = 1 to 12 = 1 - 30 = -29, choice (A).

b_n=a_n-1+b_n-1 → b₁₃=b₁+Σ_k=1¹²a_k=1+Σ_k=1¹²(4-k)=1+(48-78)=-29
Answer
-29
Walk the 3rd number down row by row using b_n = a_n-1 + b_n-1: 1, 4, 6, 7, 7, 6, 4, 1, -3, -8, -14, -21, -29. The 13th value is -29, matching the closed-form sum. The 2nd numbers 3, 2, 1, ... also match 4-n, so the pieces agree. A negative third number makes sense because the 2nd numbers turn negative partway down and drag the running total below zero. -29 is choice (A).
💡Key takeaway

When only one number in a huge pattern matters, track just the few numbers that feed it and let their small patterns carry you there.

  • Build a few rows to see the rule
  • Row totals double, so find the row
  • Track only the 2nd and 3rd numbers
  • The 2nd number drops by 1 each row
  • Add up to reach the 3rd number in row 13

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