AMC 10 · 2022 · #16

Grade 9 number-theory

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

triangular-numbersperfect-squarespell-equationcompleting-the-squaredigit-sum convert-to-algebracompleting-the-squarepattern-recognition ↑ Prerequisites: triangular-numbersperfect-squares

📏 Long solution 💡 3 insights

Problem

The n-th triangular number adds up one through n. Some triangular numbers are also perfect squares; the three smallest such are 1, 36, and 1225. Find the fourth smallest triangular number that is also a perfect square, then add up its digits.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Convert to Algebra

Searching for "triangular and square" by hand is hopeless — the answers thin out fast (1, 36, 1225, then a jump). So turn the condition into an equation: (n(n+1))/2 = k². Clearing the fraction and completing the square converts it into (2n+1)² - 8k² = 1, a Pell equation. That reframing is the whole game, because Pell equations have a rigid generating structure: every solution is a power of the smallest one. The three examples the problem hands us are not decoration — they are the first three powers, which lets us confirm the pattern before using it to produce the fourth. Tool #13 does the reframing, Tool #5 reads off the generating pattern, and Tool #15 is what makes the pattern visible (rewriting the given n values as 2n+1 values).

1STEP 1

Close the triangular sum

Write it in closed form.

t_n = 1+2+3+…+n = (n(n+1))/2

2STEP 2

Write the condition as an equation

Write the condition as an equation.

(n(n+1))/2 = k² ⟹ n(n+1) = 2k²

3STEP 3

Complete the square

Complete the square.

4n² + 4n = 8k² ⟹ 4n² + 4n + 1 = 8k² + 1 ⟹ (2n+1)² - 8k² = 1

4STEP 4

Recast the known cases

Rewrite the three known cases.

x² - 8k² = 1, with known solutions (x,k) = (3,1), (17,6), (99,35)

5STEP 5

Factor over the radical

Factor over the radical.

x² - 8k² = (x - 2√(2) k)(x + 2√(2) k) = 1

6STEP 6

Spot the pattern in the powers

The solutions are powers of one number.

(3+2√(2))¹ = 3 + 2√(2) · 1 → (3,1), (3+2√(2))² = 17 + 12√(2) = 17 + 2√(2) · 6 → (17,6), (3+2√(2))³ = 99 + 70√(2) = 99 + 2√(2) · 35 → (99,35)

7STEP 7

Take the fourth power

The fourth power gives the next solution.

(3+2√(2))⁴ = (17+12√(2))² = 289 + 408√(2) + 288 = 577 + 2√(2) · 204 → x = 577, k = 204, n = (577-1)/2 = 288

8STEP 8

Build it and add the digits

Adding the digits gives 18.

t₂₈₈ = (288 · 289)/2 = 144 · 289 = 41616 = 204², 4+1+6+1+6 = 18

9STEP 9

Match the choice

It matches a choice exactly.

18 → (D)

Answer

Three independent checks agree. First, 41616 really is triangular: (288 · 289)/2 = 144 · 289 = 41616. Second, it really is square: 204² = 40000 + 1600 + 16 = 41616. Third, it really is the fourth smallest, not just some later one — the powers of 3+2√(2) are strictly increasing and every solution of x²-8k²=1 is one of them, and the first three powers reproduced exactly the three examples the problem stated, so the fourth power must give the fourth smallest. The size is plausible too: the square roots go 1, 6, 35, 204, each roughly 6 times the last, so a jump from 1225 to about 4 × 10⁴ is exactly the growth rate to expect. One more consistency check: 41616 = 144 · 289 and 144 is a multiple of 9, so 41616 is divisible by 9 and its digit sum must be too — which independently eliminates (A) 6 and (C) 12, leaving only 9, 18, and 27 in play.

💡Key takeaway

"Triangular and square at the same time" becomes the single equation (2n+1)² - 8k² = 1, and its solutions are not scattered — they are the powers of 3+2√(2). The three examples in the problem are the first three powers, so the fourth power, 577 + 2√(2) · 204, hands over 204² = 41616 and a digit sum of 18 — choice (D).

Close the triangular sum
Write the condition as an equation
Complete the square
Recast the three known cases
Factor over √(2)
Spot the pattern in the powers
Take the fourth power
Build the number, add the digits
Match the answer choice