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?

$\textbf{(A) } 6 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 27$

Solution 1

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

Solution 2 (Bash)

As mentioned above, $t_n = \frac{n (n+1)}{2}$. If $t_n$ is a perfect square, one of two things must occur when the fraction is split into a product. Either $\frac{n}{2}$ and $n+1$ must both be squares, or $n$ and $\frac{n+1}{2}$ must both be squares, and thus the search for the next perfect square triangular number can be narrowed down by testing values of $n$ that are close to or are perfect squares. After some work, we reach $n = 288$, $1$ less than $289$, and $t_{288} = \frac{288\cdot289}{2} = 144 \cdot 289 = 41616$. This product is a perfect square, and thus the sum of the digits of the fourth smallest perfect square triangular number is therefore $4+1+6+1+6=\boxed{\textbf{(D) 18}}$.

~kingme271

Solution 3

According to the problem, we want to find integer $p$ such $\frac{n(n+1)}{2}=p^2$, after expanding, we have $n^2+n=2p^2, 4n^2+4n=8p^2, (2n+1)^2-8p^2=1$, we call $2n+1=q$, the equation becomes $q^2-8p^2=1$, obviously $(q,p)=(3,1)$ is the elementary solution for this pell equation, thus the forth smallest solution set $q_4+2\sqrt{2}p_4=(3+2\sqrt{2})^4=577+408\sqrt{2}$, which indicates $p=204, p^2=41616$ leads to $\boxed{18}$

~bluesoul

Solution 4

If $n \choose 2$ is a square, then ${(2n-1)^2 \choose 2}$ is also a square. We can prove this quite simply:

\[{(2n-1)^2 \choose 2}\] \[= \frac{(2n-1)^2 \cdot ((2n-1)^2 - 1)}{2}\] \[= \frac{(2n-1)^2 \cdot (2n \cdot (2n - 2))}{2}\] \[= (2n-1)^2 \cdot 4{n \choose 2}.\]

Therefore, ${(2 \cdot 9 - 1)^2 \choose 2}$ is a square. Note that $T_n = {n+1 \choose 2}$. We can easily check all smaller possibilities using a bit of casework, and they don't work. Our solution is thus ${289 \choose 2} = 204^2 = 41616$, and so the answer is $\boxed{18}$.

~mathboy100

Solution 5

We want to find integer $n_i$ and $m_i$ such that $t_{n_i} =\frac{n_i (n_i + 1)}{2}=m_i^2, n_0 = 0.$

We use the formula $\sqrt{n_{i+1}} = \sqrt{2n_i} + \sqrt{n_i + 1}$ and get

\[n_1 = ( \sqrt{2n_0} + \sqrt{n_0 + 1})^2 = (0+1)^2 = 1,\] \[n_2= ( \sqrt{2n_1} + \sqrt{n_1 + 1})^2 = ( \sqrt{2} + \sqrt{1 + 1})^2 = 8,\] \[n_3= ( \sqrt{2n_2} + \sqrt{n_2 + 1})^2 = ( \sqrt{16} + \sqrt{8 + 1})^2 = 49,\] \[n_4 = ( \sqrt{2n_3} + \sqrt{n_3 + 1})^2 = ( \sqrt{98} + \sqrt{49 + 1})^2 =  ((7 + 5)\sqrt{2})^2 = 288,\] \[n_5 = ( \sqrt{2n_4} + \sqrt{n_4 + 1})^2 = ( \sqrt{576} + \sqrt{289})^2 =  (24 + 17)^2 = 1681,\] \[n_6 = ( \sqrt{2n_5} + \sqrt{n_5 + 1})^2 = ( 41\sqrt{2} + \sqrt{1682})^2 =  ((41 + 29)\sqrt{2})^2 = 9800,...\] Therefore, $t_{n_4} = t_{288} = \frac{288\cdot289}{2} = 41616 \implies 4+1+6+1+6=\boxed{\textbf{(D) 18}}$

vladimir.shelomovskii@gmail.com, vvsss

Video Solution

https://youtu.be/ZmSg0JYEoTw

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

See Also

2022 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS