$\large A = \underbrace{9999\cdots9999}_{\text{Number of 9's} = n}5$

Number $A$ has $n$ $9$ 's followed by a $5$ . How many nines and zeros are there in $A^2$ ?

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The square of an odd multiple of $5$ always have the form $\overline{m5}^2 = \overline{p25}$ , where $p=m(m+1)$ . For example, $\blue 15^2 = \red 225, \blue 25^2 = \red 625, \blue 35^2 = \red {12}25, \cdots$ , where $\blue 1 \times 2 = \red 2, \blue 2\times 3 = \red 6, \blue 3\times 4 = \red {12}, \cdots$ . We can prove this as follows:

$\begin{aligned} \overline{m5}^2 & = (10m + 5)^2 = 100m^2 + 100m + 25 = 100m(m+1) + 25 = \overline{p25} \\ \implies p & = m(m+1) \end{aligned}$

For $m = \underbrace{999\cdots9}_{n 9's}$ , $p=\underbrace{999\cdots9}_{n 9's} \times 10^n = \underbrace{999\cdots9}_{n \ 9's}\underbrace{000\cdots0}_{n\ 0's}$ . That is

$n$ nines and $n$ zeros.