Surprisingly I did not get any correct answers to the puzzle.

The solution to the puzzle lies in divisibility rules – any number with 100 0’s, 1’s and 2’s will be divisible by 3 since the digits add up to 300. However, the number will not be divisible by 9, no matter what the sequence of these digits is. Therefore, the number can never be a perfect square. Argument remains the same with 10 0’s, 1’s and 2’s!

I got a couple of answers that argued that for any large square, there necessarily needs to be digits other than 0, 1 and 2. That is not necessarily true, for example square of 110 is 12100.

Hope you enjoyed this!