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 of numbers is constructed beginning with the numbers $-1$ , $3$ , and $1$ in the top row. Each adjacent pair of numbers is summed to produce a number in the next row. Each row begins and ends with $-1$ and $1,$ respectively.

$\large{-1}\qquad\large{3}\qquad\large{1}$
$\large{-1}\qquad\large{2}\qquad\large{4}\qquad\large{1}$
$\large{-1}\qquad\large{1}\qquad\large{6}\qquad\large{5}\qquad\large{1}$
If the process continues, one of the rows will sum to $12{,}288$ . In that row, what is 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.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: 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.

Givens: Row 1 is -1, 3, 1.; Each new row's inside numbers are the sums of neighboring pairs from the row above.; Every row starts with -1 and ends with 1.; Rows 2 and 3 are shown: -1, 2, 4, 1 and -1, 1, 6, 5, 1.; Some row has all its numbers adding to 12,288.

Unknowns: The third number from the left in the row whose total is 12,288.

Understand

Plan

Primary tool: #5 Look for a Pattern

Secondary: #9 Solve an Easier Related Problem, #4 Introduce a Variable, #7 Identify Subproblems

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.

Execute — Answer: A

#9 Solve an Easier Related Problem 7.NS.A.1 Step 1

Build a few rows to see the rule

Follow the rule on the first rows to make sure you understand it.
Each inside number is the sum of the two numbers just above it, and every row is capped with -1 on the left and 1 on the right.
Row 1 is -1, 3, 1.
Row 2 is -1, then 2 and 4 (from -1+3 and 3+1), then 1.
Row 3 is -1, then 1, 6, 5 (from -1+2, 2+4, 4+1), then 1.
This matches the picture, so the rule is clear.

$$\text{Row 1: }{-1},\ 3,\ 1 \qquad \text{Row 2: }{-1},\ 2,\ 4,\ 1 \qquad \text{Row 3: }{-1},\ 1,\ 6,\ 5,\ 1$$

💡 Trying the rule on small rows first makes the machine that builds the array obvious.

#5 Look for a Pattern 6.EE.A.1 Step 2

Row totals double, so find the row

Add up each row.
Row 1 gives -1+3+1 = 3, row 2 gives -1+2+4+1 = 6, row 3 gives -1+1+6+5+1 = 12.
The totals double each step.
This happens because every number in a row gets used in two of the sums below it (once with its left neighbor, once with its right neighbor), while the fixed -1 and 1 on the ends cancel out.
So the total of row n is 3 times 2 to the power n-1.
Setting that equal to 12,288 gives 2 to the power n-1 equal to 4096, which is 2 to the 12th, so n = 13.
The target is row 13.

$$3,\ 6,\ 12,\ \dots = 3\cdot 2^{\,n-1};\qquad 3\cdot 2^{\,n-1}=12288 \Rightarrow 2^{\,n-1}=4096=2^{12} \Rightarrow n=13$$

💡 Each number feeds two sums below and the ends cancel, so every row total is exactly twice the last.

#4 Introduce a Variable 6.EE.B.6 Step 3

Track only the 2nd and 3rd numbers

You do not need the whole giant row, only its third number.
Give the numbers names: let a be the 2nd number from the left in a row and b be the 3rd number.
Because a number in the new row is the sum of the two above it, the new 2nd number is (old 1st) + (old 2nd), and the new 3rd number is (old 2nd) + (old 3rd).
From the rows above, a starts at 3 and b starts at 1 in row 1.

$$a_n=\text{2nd number of row }n,\quad b_n=\text{3rd number of row }n;\qquad a_1=3,\ b_1=1$$

💡 Only a few numbers near the left edge affect the third number, so name them and ignore the rest.

#5 Look for a Pattern 4.OA.C.5 Step 4

The 2nd number drops by 1 each row

The new 2nd number is (old 1st) + (old 2nd).
The 1st number is always -1, so the 2nd number just loses 1 every row.
Starting from a = 3 in row 1, the 2nd numbers go 3, 2, 1, 0, -1, and so on.
That means the 2nd number in row n is 4 minus n.

$$a_n=-1+a_{n-1}=a_{n-1}-1,\qquad a_1=3 \;\Rightarrow\; a_n=4-n$$

💡 Adding the fixed -1 each time is the same as counting down by one, giving a straight-line pattern.

#4 Introduce a Variable 7.NS.A.1 Step 5

Add up to reach the 3rd number in row 13

The new 3rd number is (old 2nd) + (old 3rd), so each row the 3rd number grows by that row's 2nd number: b_n = a_{n-1} + b_{n-1}.
Starting from b = 1 in row 1, the 3rd number in row 13 is 1 plus all the 2nd numbers from rows 1 through 12, which are 4-k for k from 1 to 12.
Those add to (twelve 4's) minus (1+2+...+12) = 48 - 78 = -30.
So the 3rd number in row 13 is 1 + (-30) = -29.
That is choice (A).

$$b_n=a_{n-1}+b_{n-1}\;\Rightarrow\; b_{13}=b_1+\sum_{k=1}^{12}a_k=1+\sum_{k=1}^{12}(4-k)=1+(48-78)=-29$$

💡 Each step adds the current 2nd number, so stacking those additions from the start lands on the 3rd number far below.

[1] #9 7.NS.A.1 Follow the rule on the first rows to make sure you understand it. Each inside nu

[2] #5 6.EE.A.1 Add up each row. Row 1 gives -1+3+1 = 3, row 2 gives -1+2+4+1 = 6, row 3 gives -

[3] #4 6.EE.B.6 You do not need the whole giant row, only its third number. Give the numbers nam

[4] #5 4.OA.C.5 The new 2nd number is (old 1st) + (old 2nd). The 1st number is always -1, so the

[5] #4 7.NS.A.1 The new 3rd number is (old 2nd) + (old 3rd), so each row the 3rd number grows by

Review

Reasonableness: 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).

Alternative: Instead of a formula, just build the array. Each row is short near the left, so you only ever need to carry the first three numbers of each row: keep the front -1, update the 2nd number as -1 plus the old 2nd, and the 3rd as old 2nd plus old 3rd. Do this 12 times from row 1 to row 13 and read off the 3rd number, which lands on -29. This is slower but needs no algebra.

CCSS standards used (min grade 7)

7.NS.A.1 Add and subtract rational numbers, including negative integers (Summing the rows and adding the run of positive and negative 2nd numbers to reach -29.)
6.EE.A.1 Write and evaluate numerical expressions involving whole-number exponents (Recognizing the row totals as 3 times a power of 2 and matching 12,288 to row 13.)
6.EE.B.6 Use variables to represent numbers and write expressions to solve problems (Naming the 2nd and 3rd numbers a and b and expressing how each new one comes from the row above.)
4.OA.C.5 Generate a number or shape pattern that follows a given rule (Seeing the 2nd numbers count down 3, 2, 1, 0, ... to get the rule 4 minus n.)

⭐ 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.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

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

Review

CCSS standards used (min grade 7)

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