2022 AMC 12A Problems/Problem 16

Problem

A \emph{triangular number} is a positive integer that can be expressed in the form $t_n = 1+2+3+\cdots+n$ , for some positive integer $n$ . The three smallest triangular numbers that are also perfect squares are $t_1 = 1 = 1^2$ , $t_8 = 36 = 6^2$ , and $t_{49} = 1225 = 35^2$ . What is the sum of the digits of the fourth smallest triangular number that is also a perfect square?

Solution

We have $t_n = \frac{n (n+1)}{2}$ . If $t_n$ is a perfect square, then it can be written as $\frac{n (n+1)}{2} = k^2$ , where $k$ is a positive integer.

Thus, $n (n+1) = 2 k^2$ .

Because $n$ and $n+1$ are relatively prime, the solution must be in the form of $n = u^2$ and $n+1 = 2 v^2$ , or $n = 2 v^2$ and $n+1 = u^2$ , where in both forms, $u$ and $v$ are relatively prime and $u$ is odd.

The four smallest feasible $n$ in either of these forms are $n = 1, 8, 49, 288$ .

Therefore, $t_{288} = \frac{288 \cdot 289}{2} = 41616$ .

Therefore, the answer is $4+1+6+1+6=\boxed{\textbf{(D) 18}}$ .

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution

https://youtu.be/ZmSg0JYEoTw

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

2022 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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All AMC 12 Problems and Solutions

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2022 AMC 12A Problems/Problem 16

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Problem

Solution

Video Solution

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