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
NikoIsLife
06.12.2018 17:28
The sequence goes:
$$2019,121,16,49,169,256,169,256,169,...$$It just keeps on repeating $169,256$.
After the $5^{th}$ term, every odd term becomes $169$, and every even term becomes $256$.
The answer is $\boxed{256}$.
ahwawm
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
ahwawm
06.12.2018 17:43
Therefore the correct answer is :
Notice that after 144, it becomes $(9) ^ {2}$ and everything after that will be equal to 81
the_referee
06.12.2018 17:43
The sequence becomes $2019,144,81,81,81,...$ The 81 continues on forever, so the answer should be 81.
ahwawm
06.12.2018 17:44
@ above you are right the_referee wrote:
The sequence becomes $2019,144,81,81,81,...$ The 81 continues on forever, so the answer should be 81.
lion11202
06.12.2018 18:39
Sorry, I had a typo. 2019 should be 2018.
Purple_Planet
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?
Let $S(n)$ be the sum of the digits of $n$.
Notice that the series goes on like $(S(n))^2$.
The first few terms are
$2018,121,16,49,169,256,169,256,169,256,\ldots $
It can be seen that from $5^{th}$ term onwards $169, 256$ are getting repeated. If we consider that the sequence starts from $169$ in the $5^{th}$ term so, we have to find the $2014^{th}$ term now.
Let $a$ be the position of $n$ where $n$ is an integer in the sequence.
Notice that at $a\equiv 0\pmod 2$ we have $256$ and at
$a\equiv 1\pmod 2$ we have $169$. Since, $2014\equiv 0\pmod 2$ we will find $\boxed{256}$ as the $2018^{th}$ term of the series $2018,121,16,49,169,256,169,256,169,256,\ldots$