Prove that number is square of natural We have number $a=246810...98100 $. Number $a$ consists of the following even numbers from $2 -100$ . Decide whether number is a square. 
I was given a hint to this task to sum all digits. 
Also I would be grateful if someone could explain whether this rule is true: If the sum of digits is divided by $k^2$ then this number is square of natural
 A: Add up all the digits in your number. What do you get? Is the total divisible by three? Is it divisible by nine? The answers to these two questions will tell you what you want to know. (This is not a general method, but it works in this particular case.)
And no, your last sentence is certainly not true. Take 13, for instance.
A: Hint $\ $ Cast out nines to show the sum is not a square mod $\,9,\,$ hence not the square of an integer. To compute the sum very simply we can use Gauss's grade-school reflection method: pair up each term $\,n\,$ with its negation(inverse) $\:-n\equiv 108\!-\!n\pmod{108}$ so they sum to $\,108\equiv 0\pmod 9\,$ hence cancel out of the sum. In detail:
$\begin{eqnarray}{\rm mod}\ 9\!:\ \color{#c00}{10\equiv 1}\,&\Rightarrow&\quad 2\cdot \color{#c00}{10}^i + 4\cdot \color{#c00}{10}^j +\cdots + 100\cdot\color{#c00}{10}^k\\
&\equiv&\quad 2\cdot \color{#c00}{\ 1}^i\ + 4\cdot \color{#c00}{\ 1}^j\ +\cdots + 100\cdot\color{#c00}{\ 1}^k\\
&\equiv&\quad \quad2 + \quad 4 +\quad 6 +\quad 8 + \cdots + 50 + 52 +\!\smash[t]{\overbrace{ 54}^{\equiv\, 0}}\\
&      &+ \   \color{#0a0}{106 + 104 + 102}+ 100+\cdots + 58+56\ \ -\ \ (\color{#0a0}{106 + 104 - 102})\\
[{\rm by}\ \ 108\equiv 0]\ \  &\equiv&\qquad0 +\quad 0+\quad 0+\quad 0+\cdots +\,\ 0 +\,\ 0\ \ -\ \ \smash[b]{(\underbrace{\quad 7 +\quad 5 +\quad 3}_{\large \color{#c0d}{\equiv\ 15\ \equiv\ -3}})}  \\
&\equiv&\qquad 0-(\color{#c0d}{-3})\,\equiv\, 3\\
\end{eqnarray}
$
Therefore the sum has form $\,\color{#c0d}3\!+\!9n = 3(1\!+\!3n)\neq k^2,\,$ being divisible by $3$ but not by $\,3^2.$
Remark $\ $ This modular nonsquare-test works because squares remain squares mod $\,m,\,$ i.e. if $\,\bar a = (a\ {\rm mod}\ m)\,$ then $\,\overline{a^2} = \bar a^2,\,$ being a special case of $\,\overline{ac} = \bar a \bar c,\,$ which is a special case of the  Congruence product Rule. Contrapositively, if $\,a\,$ is nonsquare mod $\,m\,$ it is a nonsquare integer. 
A: We can say that
$$a = \sum_{n = 1}^{50}{b_n\cdot10^{c_n}}$$
where $b_n$ is the $n$th even number greater than $0$, and $c_n$ is a non-negative integer. Then,
$$\begin{align}a &\equiv \sum_{n=1}^{50}b_n\cdot 1 \pmod 3\\
&= \sum_{n=1}^{50}b_n\pmod 3\\
&= 2550 \pmod 3\\
&\equiv 0 \pmod 3\end{align}$$
So $3$ divides into $a$. Similarly,
$$\begin{align}a &\equiv \sum_{n=1}^{50}b_n\cdot 1 \pmod 9\\
&= \sum_{n=1}^{50}b_n\pmod 9\\
&= 2550 \pmod 9\\
&\equiv 3 \pmod p\end{align}$$
So $3^2$ does not divide into $a$ (and therefore no higher powers of $3$ can divide into $a$). If the exponent of $3$ in the prime factorization of $a$ is equal to $1$, then $a$ cannot be a perfect square.
