Source: 2018 Canadian Open Math Challenge Part A Problem 4 In the sequence of positive integers, starting with $2018, 121, 16, ...$ each term is the square of the sum of digits of the previous term. What is the $2018^{\text{th}}$ term of the sequence?
Problem
Source:
Tags: Comc, 2018 COMC
06.12.2018 17:28
06.12.2018 17:39
lion11202 wrote: Source: 2018 Canadian Open Math Challenge Part A Problem 4 In the sequence of positive integers, starting with $2019, 121, 16, ...$ each term is the square of the sum of digits of the previous term. What is the $2018^{\text{th}}$ term of the sequence? After 2019, it is supposed to be $(9 + 1 + 2) ^ {2}$ = $(12) ^ {2}$ = 144 and not 121
06.12.2018 17:43
Therefore the correct answer is :
06.12.2018 17:43
06.12.2018 17:44
@ above you are right the_referee wrote:
06.12.2018 18:39
Sorry, I had a typo. 2019 should be 2018.
07.12.2018 04:48
lion11202 wrote: Source: 2018 Canadian Open Math Challenge Part A Problem 4 In the sequence of positive integers, starting with $2018, 121, 16, ...$ each term is the square of the sum of digits of the previous term. What is the $2018^{\text{th}}$ term of the sequence?