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, $n (n+1) = 2 k^2$ . Rearranging, we get $(2n+1)^2-2(2k)^2=1$ , a Pell equation. So $\frac{2n+1}{2k}$ must be a truncation of the continued fraction for $\sqrt{2}$ :

$\begin{eqnarray*} 1+\frac12&=&\frac{2\cdot1+1}{2\cdot1}\\ 1+\frac1{2+\frac1{2+\frac12}}&=&\frac{2\cdot8+1}{2\cdot6}\\ 1+\frac1{2+\frac1{2+\frac1{2+\frac1{2+\frac12}}}}&=&\frac{2\cdot49+1}{2\cdot35}\\ 1+\frac1{2+\frac1{2+\frac1{2+\frac1{2+\frac1{2+\frac1{2+\frac12}}}}}}&=&\frac{2\cdot288+1}{2\cdot204} \end{eqnarray*}$

Therefore, $t_{288} = \frac{288\cdot289}{2} = 204^2 = 41616$ , so the answer is $4+1+6+1+6=\boxed{\textbf{(D) 18}}$ .

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

Edited by wzs26843545602

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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Video Solution

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