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

(Created page with "==Problem== A \emph{triangular number} is a positive integer that can be expressed in the form <math>t_n = 1+2+3+\cdots+n</math>, for some positive integer <math>n</math>. Th...")
 
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==Solution==
 
==Solution==
  
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https://youtu.be/08YkinzFcCc
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We have <math>t_n = \frac{n (n+1)}{2}</math>.
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If <math>t_n</math> is a perfect square, then it can be written as
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<math>\frac{n (n+1)}{2} = k^2</math>,
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where <math>k</math> is a positive integer.
  
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(Professor Chen Education Palace, www.professorchenedu.com)
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Thus, <math>n (n+1) = 2 k^2</math>.
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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.
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The four smallest feasible <math>n</math> in either of these forms are <math>n = 1, 8, 49, 288</math>.
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Therefore, <math>t_{288} = \frac{288 \cdot 289}{2} = 41616</math>.
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Therefore, the answer is <math>4+1+6+1+6=\boxed{\textbf{(D) 18}}</math>.
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Revision as of 20:30, 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, $n (n+1) = 2 k^2$.

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)

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