AMC12 2022 A

AMC12 2022 A · Q16

AMC12 2022 A · Q16. It mainly tests Perfect squares & cubes, Number theory misc.

A $\text{triangular number}$ 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?

一个$\text{三角函数}$是能表示为 $t_n = 1+2+3+\cdots+n$ 的正整数，其中 $n$ 是某个正整数。同时是完全平方的三个最小的三角形数是 $t_1 = 1 = 1^2$，$t_8 = 36 = 6^2$，和 $t_{49} = 1225 = 35^2$。第四个最小的同时是三角形数和完全平方的三角形数的各位数字之和是多少？

(A) 6 6

(B) 9 9

(D) 18 18

(E) 27 27

Answer

Correct choice: (D)

正确答案：（D）

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, $n (n+1) = 2 k^2$. Rearranging, we get $(2n+1)^2-2(2k)^2=1$, a Pell equation (see https://artofproblemsolving.com/wiki/index.php/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 $4+1+6+1+6=\boxed{\textbf{(D) 18}}$.

我们有 $t_n = \frac{n (n+1)}{2}$。如果 $t_n$ 是完全平方，则可以写成 $\frac{n (n+1)}{2} = k^2$，其中 $k$ 是正整数。因此，$n (n+1) = 2 k^2$。重新整理，得到 $(2n+1)^2-2(2k)^2=1$，这是一个 Pell 方程（参见 https://artofproblemsolving.com/wiki/index.php/Pell_equation）。所以 $\frac{2n+1}{2k}$ 必须是 $\sqrt{2}$ 的连分数截断： \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*} 因此，$t_{288} = \frac{288\cdot289}{2} = 204^2 = 41616$，所以答案是 $4+1+6+1+6=\boxed{\textbf{(D) 18}}$。

Topics