2022 AMC 12A Problems/Problem 16
Contents
Problem
A is a positive integer that can be expressed in the form , for some positive integer . The three smallest triangular numbers that are also perfect squares are , , and . What is the sum of the digits of the fourth smallest triangular number that is also a perfect square?
Solution 1
We have . If is a perfect square, then it can be written as , where is a positive integer.
Thus, . Rearranging, we get , a Pell equation. So must be a truncation of the continued fraction for :
Therefore, , so the answer is .
- Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Edited by wzs26843545602 Edited by dad
Solution 2 (Bash)
As mentioned above, . If is a perfect square, one of two things must occur when the fraction is split into a product. Either and must both be squares, or and must both be squares, and thus the search for the next perfect square triangular number can be narrowed down by testing values of that are close to or are perfect squares. After some work, we reach , less than , and . This product is a perfect square, and thus the sum of the digits of the fourth smallest perfect square triangular number is therefore .
~kingme271
Solution 3
According to the problem, we want to find integer such , after expanding, we have , we call , the equation becomes , obviously is the elementary solution for this pell equation, thus the forth smallest solution set , which indicates leads to
~bluesoul
Solution 4
If is a square, then is also a square. We can prove this quite simply:
Therefore, is a square. Note that . We can easily check all smaller possibilities using a bit of casework, and they don't work. Our solution is thus , and so the answer is .
~mathboy100
Solution 5
We want to find integer and such that
We use the formula and get
Therefore,
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. By the way, this isn't very rigorous.
No formula needed.
Emily Q emilyyunhanq@gmail.com
Video Solution
~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 |
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