2022 AMC 12A Problem 16

Attempt Problem 16 of the 2022 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AMC 12A solutions, or check the answer key.

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16.

A triangular number is a positive integer that can be expressed in the form tn=1+2+3++n,t_n=1+2+3+\cdots+n, for some positive integer n.n. The three smallest triangular numbers that are also perfect squares are t1=1=12,t_1=1=1^2, t8=36=62,t_8=36=6^2, and t49=1225=352.t_{49}=1225=35^2. What is the sum of the digits of the fourth smallest triangular number that is also a perfect square?

66

99

1212

1818

2727

Answer: D
Concepts:triangular numberperfect squarerecursion
Difficulty rating: 1800
Solution:

If tn=y2,t_n=y^2, then n(n+1)/2=y2,n(n+1)/2=y^2, or (2n+1)28y2=1.(2n+1)^2-8y^2=1. The positive Pell solutions occur successively by multiplying (2n+1)+y8(2n+1)+y\sqrt8 by 3+8.3+\sqrt8.

Starting from (2n+1,y)=(3,1),(2n+1,y)=(3,1), this gives (17,6),(17,6), (99,35),(99,35), and then (577,204).(577,204). Thus the fourth value has n=288n=288 and equals 2042=41616.204^2=41616.

The sum of its digits is 4+1+6+1+6=18.4+1+6+1+6=18.

Thus, the correct answer is D.

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