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.
Best Answer
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.