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Difference between revisions of "2022 AMC 12A Problems/Problem 16"
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==Solution 2 (Bash)==
 
==Solution 2 (Bash)==
  
−
As mentioned above, <math>t_n = \frac{n (n+1)}{2}</math>. If <math>t_n</math> is a perfect square, one of two things must occur when the fraction is split into a product. Either <math>\frac{n}{2}</math> and <math>n+1</math> must both be squares, or <math>n</math> and <math>\frac{n+1}{2}</math> must both be squares, and thus the search for the next perfect square triangular number can be narrowed down by testing values of <math>n</math> that are close to or are perfect squares. After some work, we reach <math>n = 288</math>, <math>1</math> less than <math>289</math>, and <math>t_288 = \frac{288\cdot289}{2} = 144 * 289 = 41616</math>. This product is a perfect square, and thus the sum of the digits of the fourth smallest perfect square triangular number is therefore <math>4+1+6+1+6=\boxed{\textbf{(D) 18}}</math>. ~kingme271
+
As mentioned above, <math>t_n = \frac{n (n+1)}{2}</math>. If <math>t_n</math> is a perfect square, one of two things must occur when the fraction is split into a product. Either <math>\frac{n}{2}</math> and <math>n+1</math> must both be squares, or <math>n</math> and <math>\frac{n+1}{2}</math> must both be squares, and thus the search for the next perfect square triangular number can be narrowed down by testing values of <math>n</math> that are close to or are perfect squares. After some work, we reach <math>n = 288</math>, <math>1</math> less than <math>289</math>, and <math>t_{288} = \frac{288\cdot289}{2} = 144 * 289 = 41616</math>. This product is a perfect square, and thus the sum of the digits of the fourth smallest perfect square triangular number is therefore <math>4+1+6+1+6=\boxed{\textbf{(D) 18}}</math>. ~kingme271
  
  

Revision as of 00:30, 13 November 2022

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)


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 $t_{288} = \frac{288\cdot289}{2} = 144 * 289 = 41616$. 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


See Also

2022 AMC 12A (Problems • Answer Key • Resources)
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

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