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

Revision as of 15:21, 18 December 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 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 $t_{288} = \frac{288\cdot289}{2} = 144 * 289 = 41616$ . 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

Video Solution

https://youtu.be/ZmSg0JYEoTw

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

@@ Line 4: / Line 4: @@
 <math>t_1 = 1 = 1^2</math>, <math>t_8 = 36 = 6^2</math>, and <math>t_{49} = 1225 = 35^2</math>. What is the sum of the digits of the fourth smallest triangular number that is also a perfect square?
-==Solution==
+==Solution 1==
 We have <math>t_n = \frac{n (n+1)}{2}</math>.
@@ Line 25: / Line 25: @@
 Edited by wzs26843545602
-==Video Solution==
-https://youtu.be/ZmSg0JYEoTw
-~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 ==Solution 2 (Bash)==
@@ Line 41: / Line 34: @@
 ~bluesoul
+==Video Solution==
+https://youtu.be/ZmSg0JYEoTw
+~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 == See Also ==
 {{AMC12 box|year=2022|ab=A|num-b=15|num-a=17}}
 {{MAA Notice}}

2022 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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