Question

A sequence is constructed according to the following rule: its first term is 7, and each next term is one more than the sum of the digits of the square of its previous term. Which number is on 2018th place?

Answer 1

Answer:

5

Step-by-step explanation:

According to the described rule, we have

$a_1=7\\ \\a_1^2=7^2=49\Rightarrow a_2=4+9+1=14\\ \\a_2^2=14^2=196\Rightarrow a_3=1+9+6+1=17\\ \\a_3^2=17^2=289\Rightarrow a_4=2+8+9+1=20\\ \\a_4^2=20^2=400\Rightarrow a_5=4+0+0+1=5\\ \\a_5^2=5^2=25\Rightarrow a_6=2+5+1=8\\ \\a_6^2=8^2=64\Rightarrow a_7=6+4+1=11\\ \\a_7^2=11^2=121\Rightarrow a_8=1+2+1+1=5\\ \\\text{and so on...}$

We can see the pattern

$a_5=a_8=a_{11}=a_{14}=...=5\\ \\a_6=a_9=a_{12}=a_{15}=...=8\\ \\a_7=a_{10}=a_{13}=a_{16}=...=11$

In other words, for all $k\ge 2$

$a_{3k-1}=5\\ \\a_{3k}=8\\ \\a_{3k+1}=11$

Now,

$a_{2018}=a_{3\cdot 673-1}=5$