Berkeley Math Circle Monthly Contest 2

Solution:
We consider three cases:

- $x$ is odd. Then $x+1$ and $x-1$ are consecutive even integers, and thus one of them must be divisible by $4$, the other only by $2$. So $x^{2}-1=(x+1)(x-1)$ is divisible by $8$; since $x \geq 3$, $x^{2}-1$ has four or more digits, the last three of which are $0$'s.

- $x$ is even but $\frac{x}{2}$ is odd. By the same reasoning, $\left(\frac{x}{2}\right)^{2}-1=8k$ for some integer $k$, so
$$x^{2}-1=4\left(\frac{x}{2}\right)^{2}-1=4(8k+1)-1=32k+3$$
Since $x \geq 6$, this number has six or more digits, the last five of which are $00011$. Thus we also get three consecutive zeros in this case.

- $x$ and $\frac{x}{2}$ are both even, that is, $x$ is divisible by $4$. Then $x^{2}$ is divisible by $16$. Since $x \geq 4$, $x^{2}-1$ has at least four digits, the last four of which are $1$'s.