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Difference between revisions of "2022 AMC 12A Problems/Problem 16"
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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== See Also ==
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{{AMC12 box|year=2022|ab=A|num-b=15|num-a=17}}
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Revision as of 21:55, 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)

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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