Figures sum of a number Let $S(n)$ be the digits sum of a number.
Example: $S(4) = 4$, $S(12) = 3$
Then there exist a number such that:
$S(n) = 10$
$S(n^2) = 100$
The first thing I realized is that if such number exist, it is a even number.
Not only, a realized that $n^2$ need to have at least 12 digits long.
So I think $n$ need to be a big number.
In fact, $n$ need to be greater than $10^5 \sqrt{2}$
But, anyway, I have no idea how to answer it.
 A: Observe that:
\begin{align}
11^2 &=121\\
111^2 &= 12321\\
1111^2 &= 1234321\\
&:\\
111111111^2 &= 12345678987654321
\end{align}
The last number has digit sum of $81$. However, for $1, 111, 111, 111$ the pattern breaks as there is a carry.
To fix that, we consider a number:
$$11110000\dots 0000111111 = 1111\times 10^k + 111111$$
where $k$ is sufficiently large. This number has digit sum of $10$. We also have
\begin{align}(1111\times 10^k + 111111)^2 &= 1111^2 \times 10^{2k} + 2\times1111\times 111111\times 10^k+ 111111^2\\
&=1234321000\dots000246888642000\dots 000012345654321
\end{align}
which has a digit sum of $16+48+36 = 100$.
Considering our construction, $k=12$ should be sufficient, as $111111^2$ has $11$ digits (while in reality, we just need one zero to avoid the carry).
A: By playing around a bit, I found for $n=20110002100111$, we have $n^2 = 404412184466468830466212321$.
Here $S(n) = 10$ and $S(n^2) = 100$.
Some heuristics of how I found these. If each individual digits are large, then the number of digits in $n$ and $n^2$ will be small. So I started experimenting with 1,2. And using 0's to pad.
