2022 AMC 12A Problem 16

Question

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

Rewrite

\frac{n(n+1)}{2}=y^2

as the Pell equation

(2n+1)^2-8y^2=1

Generate successive positive solutions by multiplying

(2n+1)+y\sqrt8

by

3+\sqrt8

If

t_n=y^2,

then

\frac{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.

Problem 16 in other years