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Difference between revisions of "2022 AMC 12A Problems/Problem 16"

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where <math>k</math> is a positive integer.
 
where <math>k</math> is a positive integer.
  
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Thus, <math>n (n+1) = 2 k^2</math>.
+
Thus, <math>n (n+1) = 2 k^2</math>. Rearranging, we get <math>(2n+1)^2-2(2k)^2=1</math>, a Pell equation. So <math>\frac{2n+1}{2k}</math> must be a truncation of the continued fraction for <math>\sqrt{2}</math>:
  
−
Because <math>n</math> and <math>n+1</math> are relatively prime, the solution must be in the form of <math>n = u^2</math> and <math>n+1 = 2 v^2</math>, or <math>n = 2 v^2</math> and <math>n+1 = u^2</math>, where in both forms, <math>u</math> and <math>v</math> are relatively prime and <math>u</math> is odd.
+
<cmath>\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*}</cmath>
  
−
The four smallest feasible <math>n</math> in either of these forms are <math>n = 1, 8, 49, 288</math>.
+
Therefore, <math>t_{288} = \frac{288\cdot289}{2} = 204^2 = 41616</math>, so the answer is <math>4+1+6+1+6=\boxed{\textbf{(D) 18}}</math>.
  
−
Therefore, <math>t_{288} = \frac{288 \cdot 289}{2} = 41616</math>.
+
- Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
−
Therefore, the answer is <math>4+1+6+1+6=\boxed{\textbf{(D) 18}}</math>.
+
Edited by wzs26843545602
−
 
 
−
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
  
 
==Video Solution==
 
==Video Solution==

Revision as of 23:41, 11 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)

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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