Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m
(→‎Solution 5)
 
(8 intermediate revisions by 5 users not shown)
Line 24: Line 24:
 
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\cdot289}{2} = 204^2 = 41616</math>, so the answer is <math>4+1+6+1+6=\boxed{\textbf{(D) 18}}</math>.
  
−
- Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
+
- Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)  
  
 
Edited by wzs26843545602
 
Edited by wzs26843545602
 +
Edited by dad
  
 
==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>.  
+
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 \cdot 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
 
~kingme271
Line 38: Line 39:
  
 
~bluesoul
 
~bluesoul
 +
 +
==Solution 4==
 +
If <math>n \choose 2</math> is a square, then <math>{(2n-1)^2 \choose 2}</math> is also a square. We can prove this quite simply:
 +
 +
<cmath>{(2n-1)^2 \choose 2}</cmath>
 +
<cmath> = \frac{(2n-1)^2 \cdot ((2n-1)^2 - 1)}{2}</cmath>
 +
<cmath> = \frac{(2n-1)^2 \cdot (2n \cdot (2n - 2))}{2}</cmath>
 +
<cmath> = (2n-1)^2 \cdot 4{n \choose 2}.</cmath>
 +
 +
Therefore, <math>{(2 \cdot 9 - 1)^2 \choose 2}</math> is a square. Note that <math>T_n = {n+1 \choose 2}</math>. We can easily check all smaller possibilities using a bit of casework, and they don't work. Our solution is thus <math>{289 \choose 2} = 204^2 = 41616</math>, and so the answer is <math>\boxed{18}</math>.
 +
 +
~mathboy100
 +
==Solution 5==
 +
We want to find integer <math>n_i</math> and <math>m_i</math> such that <math>t_{n_i} =\frac{n_i (n_i + 1)}{2}=m_i^2, n_0 = 0.</math>
 +
 +
We use the formula <math>\sqrt{n_{i+1}} = \sqrt{2n_i} + \sqrt{n_i + 1}</math> and get
 +
 +
<cmath>n_1 = ( \sqrt{2n_0} + \sqrt{n_0 + 1})^2 = (0+1)^2 = 1, </cmath>
 +
<cmath>n_2= ( \sqrt{2n_1} + \sqrt{n_1 + 1})^2 = ( \sqrt{2} + \sqrt{1 + 1})^2 = 8, </cmath>
 +
<cmath>n_3= ( \sqrt{2n_2} + \sqrt{n_2 + 1})^2 = ( \sqrt{16} + \sqrt{8 + 1})^2 = 49,</cmath>
 +
<cmath>n_4 = ( \sqrt{2n_3} + \sqrt{n_3 + 1})^2 = ( \sqrt{98} + \sqrt{49 + 1})^2 =  ((7 + 5)\sqrt{2})^2 = 288, </cmath>
 +
<cmath>n_5 = ( \sqrt{2n_4} + \sqrt{n_4 + 1})^2 = ( \sqrt{576} + \sqrt{289})^2 =  (24 + 17)^2 = 1681,</cmath>
 +
<cmath>n_6 = ( \sqrt{2n_5} + \sqrt{n_5 + 1})^2 = ( 41\sqrt{2} + \sqrt{1682})^2 =  ((41 + 29)\sqrt{2})^2 = 9800,... </cmath>
 +
Therefore, <math>t_{n_4} = t_{288} = \frac{288\cdot289}{2} = 41616 \implies 4+1+6+1+6=\boxed{\textbf{(D) 18}}</math>
 +
 +
*[[Pell's equation (simple solutions)]]
 +
 +
'''vladimir.shelomovskii@gmail.com, vvsss'''
 +
 +
==Solution 6==
 +
This is an easy one to tackle by just noticing a pattern. Write out all the perfect squares before 300, and then notice that 4 and 9 are what are the factors of the 8th term. 25 and 49 are factors of the 49th term. 4 multiplied by 2 is one less then 9. 25 multiplied by 2 is one more than 49. Well, then we keep on writing out the perfect squares and then hunt for any that can be multiplied by two and then has an absolute value of 1 valued difference. 144 and 289 satisfy this. 288th term would give us these factors.
 +
 +
No formula needed.
 +
 +
Emily Q
 +
emilyyunhanq@gmail.com
  
 
==Video Solution==
 
==Video Solution==
Line 44: Line 81:
  
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 +
==Video Solution (Smart and Simple) ==
 +
 +
https://youtu.be/7yAh4MtJ8a8?si=yGkqWQUlYlVmVbqj&t=3716
 +
 +
~Math-X
 +
  
 
== See Also ==
 
== See Also ==

Latest revision as of 22:42, 13 May 2024

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 Edited by dad

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

Solution 6

This is an easy one to tackle by just noticing a pattern. Write out all the perfect squares before 300, and then notice that 4 and 9 are what are the factors of the 8th term. 25 and 49 are factors of the 49th term. 4 multiplied by 2 is one less then 9. 25 multiplied by 2 is one more than 49. Well, then we keep on writing out the perfect squares and then hunt for any that can be multiplied by two and then has an absolute value of 1 valued difference. 144 and 289 satisfy this. 288th term would give us these factors.

No formula needed.

Emily Q emilyyunhanq@gmail.com

Video Solution

https://youtu.be/ZmSg0JYEoTw

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

Video Solution (Smart and Simple)

https://youtu.be/7yAh4MtJ8a8?si=yGkqWQUlYlVmVbqj&t=3716

~Math-X


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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_12A_Problems/Problem_16&oldid=219312"