Question

三角数是可以写成 $t_n=1+2+3+\cdots+n$ 形式的正整数，其中 $n$ 为正整数。最小的三个同时也是完全平方数的三角数为 $t_1=1=1^2$ 、 $t_8=36=6^2$ 和 $t_{49}=1225=35^2$ 。第四小的同时也是完全平方数的三角数的各位数字之和是多少？

A 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?

$6$

$9$

$12$

$18$

$27$

答案：D

知识点：三角形数完全平方数递推

难度评级：1800

解答：

若 $t_n=y^2,$ 则 $n(n+1)/2=y^2,$ 即 $(2n+1)^2-8y^2=1.$ 正的佩尔方程解可依次通过将 $(2n+1)+y\sqrt8$ 乘以 $3+\sqrt8.$ 得到。

从 $(2n+1,y)=(3,1),$ 开始，依次得到 $(17,6),$ $(99,35),$ 然后是 $(577,204).$ 因此第四个值对应 $n=288$ ，并且等于 $204^2=41616.$

它的数位和为 $4+1+6+1+6=18.$

因此，正确答案是 D。

If $t_n=y^2,$ then $n(n+1)/2=y^2,$ or $(2n+1)^2-8y^2=1.$ The positive Pell solutions occur successively by multiplying $(2n+1)+y\sqrt8$ by $3+\sqrt8.$

Starting from $(2n+1,y)=(3,1),$ this gives $(17,6),$ $(99,35),$ and then $(577,204).$ Thus the fourth value has $n=288$ and equals $204^2=41616.$

The sum of its digits is $4+1+6+1+6=18.$

Thus, the correct answer is D.

2022 AMC 12A 第 16 题

答案：D

难度评级：1800

解答：

其他年份的第 16 题