Let $m$ be a natural number with digits consisting entirely of $6'$s and $0'$s. Prove that $m$ is not the square of a natural number. 
Question: Let $m$ be a natural number with digits consisting entirely of $6'$s and $0'$s. Prove that $m$ is not the square of a natural number. 

My approach: Given that $m\in\mathbb{N}$ with digits consisting entirely of $6'$s and $0'$s. Let this property be called $P$. 
Now for the sake of contradiction let us assume that $m$ is a perfect square, that is $m=k^2$ for some $k\in\mathbb{N}$. Now since $m$ ends either with $6$ or $0$, implies that $m$ is even, which in turn implies that $2|m$. Therefore, $2|k^2\implies 2|k\implies k=2l$, for some $l\in\mathbb{N}$. 
Thus $k^2=4l^2\implies m=4l^2\implies 4|m.$ 
Also observe that the sum of the digits of $m$ is equal to $6j$, for some $j\in\mathbb{Z}.$ Now since $3|6j$, implies that $3|m$. Proceeding as above we will have $9|m$. 
Now since $\gcd(4,9)=1$ and $4|m, 9|m$, implies that $36|m$. 
Now clearly two cases are possible: 
$1.$ $m$ ends with $6$ and 
$2.$ $m$ ends with $0$. 
Observe that if $(1)$ holds true then $\frac{m}{6}$ ends with $1$, which implies that $\frac{m}{6}$ is odd. But $6|\frac{m}{6}\implies 2|\frac{m}{6},$ which implies that $\frac{m}{6}$ is even. Thus this case leads to a contradiction, which implies that $m$ doesn't ends with $6$. Or in other words this implies that all the natural numbers $m$ (the natural numbers having property $P$) ending with $6$ cannot be a perfect square. 
Now if $(2)$ holds true, then $5|m.$ Now since $m$ is a perfect square, implies that $5^2|m$. Now since $\gcd(5^2,6^2)=1$, implies that $5^2\times 6^2|m$. Now this clearly means that $m$ ends with $00$.  
How to proceed from here?
The problem can be solved by taking$\pmod {100}$ of all the natural numbers and eventually arrive at a contradiction, but that doesn't seem to be efficient enough. 
 A: If $m$ ends with $00$, you can simply start again with $m'=m/100$.
And if $m$ ends in $60$, then $5|m$, therefore $25|m$; but multiples of $25$ must end in $00,25,50,$ or $75$, a contradiction.
A: Okay, here goes my solution. 
Given that $m\in\mathbb{N}$ with digits consisting entirely of $6'$s and $0'$s. Let this property be called $P$. 
Now for the sake of contradiction let us assume that $m$ is a perfect square, that is $m=k^2$ for some $k\in\mathbb{N}$. Now since $m$ ends either with $6$ or $0$, implies that $m$ is even, which in turn implies that $2|m$. Therefore, $2|k^2\implies 2|k\implies k=2l$, for some $l\in\mathbb{N}$. 
Thus $k^2=4l^2\implies m=4l^2\implies 4|m.$ 
Also observe that the sum of the digits of $m$ is equal to $6j$, for some $j\in\mathbb{Z}.$ Now since $3|6j$, implies that $3|m$. Proceeding as above we will have $9|m$. 
Now since $\gcd(4,9)=1$ and $4|m, 9|m$, implies that $36|m$. 
Now clearly two cases are possible: 
$1.$ $m$ ends with $6$ and 
$2.$ $m$ ends with $0$. 
Observe that if $(1)$ holds true then $\frac{m}{6}$ ends with $1$, which implies that $\frac{m}{6}$ is odd. But $6|\frac{m}{6}\implies 2|\frac{m}{6},$ which implies that $\frac{m}{6}$ is even. Thus this case leads to a contradiction, which implies that $m$ doesn't ends with $6$.
Now if $(2)$ holds true, then $5|m.$ Now since $m$ is a perfect square, implies that $5^2|m$. Now since $\gcd(5^2,36)=1$, implies that $5^2\times 6^2|m$. Thus $100|m$, which clearly means that $m$ ends with $00$. Now this also implies that $m'=\frac{m}{100}$ is yet again a perfect square ($\because$ $100=10^2$ is a perfect square) consisting entirely  of digits $6'$s and $0'$s. Then we can yet again conclude that $100|m'$ and $m'$ ends with $00$. Now let $m''=\frac{m'}{100}$ and keep continuing this iteration. After this iteration ends we can certainly conclude that $m=100^j$ for some $j\in\mathbb{N}$. This also implies that $m$ is entirely a combination of $1$ and $2j$ $0'$s, which contradicts the fact that $m$ has property $P$. Hence $m$ does not end with $0$. 
Thus $m$ does not end neither with $0$ nor with $6$, which is a clear contradiction to the property $P$. 
Hence $m$ is not the square of a natural number.  
A: The answer given by @TonyK establishes that $m$ cannot end in a string of one or more $0$s, so it must end in $06$ or $66$. Hence, $m=100k+6$ or $m=100k+66$. Since $m$ is even, its square must be evenly divisible by $4$. But $(100k+6,\ 100k+66)\equiv 2 \bmod 4$, so a contradiction results, and the hypothesis that $m$ can be a square composed entirely of the digits $0,6$ must be false.
A: You don't have to do it for 2).
Let $N = N'\times 10^k$ so that $N$ ends with $k$ zeros but once you remove the $k$ zeros $N'$ will end with $6$.
If $N=m^2$ is a perfect square then $k$ is an even number of zeros and $m^2 = N'\times 10^k$ and $m = \sqrt {N'}\times 10^{\frac k2}$ and $N'$ is a perfect square that ends in $6$ and which contains only sixes and zeros.
......And you are back to case 1).
.....
Or if you grab the tiger from the other end (the one without the point teeth) and say
$n$ is a number that ends with $k$ zeros ($k$ could be zero and $n$ could and with no zeros).  Then $n = m\times 10^k$ where $m$ is a number that does not equal $0$.
Then $m = 10a + b$ where $b = 1,....,9$ and 
$n = (10a + b)\times 10^k$.
And $n^2 = (100a^2 + 20ab + b^2)\times 10^{2k}$.
Now suppose $n^2$ has only $0$s and $6$s.  Then $100a^2 + 20ab + b^2$ has only $0$s and $6$es.
So $b^2$ must end in a six or zero but of $1^2, 2^2, 3^3....., 9^2$ only $4$ and $6$ will end in a $6$ and none of them end in a $0$.  So $b = 4$ or $b = 6$.
If $b = 4$ then $100a^2 + 80a + 16$ contains only zeros and sixes.  So $80a + 16$ must end with either $06$ or $66$.  Which means $8a$ must end in either $9$ or $5$. But that's impossible as $8a$ is even.
If $b = 6$ then $100a^2 + 120a + 36$ contains only zeros and sixes.  So $120a+36$ must end with either $06$ or $66$.  Which means $12a$ must end in either $7$ or $3$.  But that's impossible as $12a$ is even.
So it is impossible for $n^2$ to have only sixes and zeros.
A: Another approach which I think is simpler is to use the following fact:

If $m$ is a perfect square, then for every prime $p$ the power of $p$ occurring in $m$ - that is, the largest $s$ such that $p^s\vert m$ - must be even.

To see how this is relevant, first consider ${m\over 2}$. This has only $3$s and $0$s as digits. In particular, we have ${m\over 2}=10^k\cdot c$ for some $k$ and some $c$ which is not divisible by either $2$ or $5$ (since $c$ ends in "$3$").
But then we have $$m=2\cdot 10^k\cdot c=2^{\color{red}{k+1}}5^kc$$ for some $k$ and some $c$ not divisible by either $2$ or $5$, and this is impossible: if $k$ is odd then $m$ can't be a square since the power of $5$ occurring in $m$ is odd, and if $k$ is even then $m$ can't be a square since the power of $2$ occurring in $m$ is odd.

A bit more snappily: let $Pow_a(b)$ be the largest $n$ such that $a^n\vert b$. Then if $m$ consists only of $6$s and $0$s, we have $Pow_2(m)=Pow_5(m)+1$. This means that one of $Pow_2(m)$ and $Pow_5(m)$ is odd. But if $m$ were a perfect square then $Pow_p(m)$ would be even for every prime $p$.
A: I have another approach
Any number is of the form $10^n a$  where n is a whole number so if we square it we get $10^{2n} a^2$ basically if we remove the trailing zeroes the resulting number must also be a square.
Incase of m it looks like
m=6.....6
or
$m=6(1.....1)=2*3(1....1)$ and we notice that the resulting number is odd  but for m to be square we needed at least another two.
Therefore m is not perfect square
