9's...

This one again can be solved by induction.

$\begin {matrix} 9^2 & = 81 \\ 99^2& = 9801 \\ 999^2 & = 998001 \\ 9999^2 & = 99980001 \\ 99999^2 & = 9999800001 \end {matrix}$

It can be seen that the number of zeros in a square of a $n$ -digit string of $9$ is $n-1$ . Therefore, the answer is $30-1 = \boxed{29}$ .

$999999999999999999999999999998000000000000000000000000000001$ = $999999999999999999999999999999^{2}$ (wrote it manually)

this comes from a beautiful sequence.

i.e let their be n 9s in the number . then their will be exactly ( n - 1) 9s and 0s in it's square.

In this case it is 30-1 =29

We rewrite the string of 30 9's as $10^{30} - 1.$ Then $(10^{30}-1)^{2} = 10^{60} - 2*10^{30} + 1,$ and we can see that the 29 digits preceding the very last digit are all 0's, while everything else is not. Thus, the answer is 29.

simply n - 1 = 30 -1 = 29