Is there a perfect square in the sequence $6, 96, 996, 9996, ... , 99999...996, ...$?

Question

$6, 96, 996, 9996,\dots, 99999\cdots996,\dots$

Consider the above number sequence. Here in the $n$ $^{th}$ term $n-1$ digits are $9$. How can we tell about the existence of a perfect square number in this sequence?

Answer 1

The sum of the digits of the $n^{th}$ term of this sequence is $$\text{Sum}=9(n-1)+6=9n-3=3(3n-1)$$ Thus, every number in the sequence is divisible by $3$ but not by $9$. Hence, no number in the sequence can be a perfect square.

Answer 2

HINT: consider divisibility by a suitable small number $n$ and also by $n^2$