There are 51 houses on a street. (counting problem) There are 51 houses on a street. Each house has an address between 1000 and 1099, inclusive. Show that at least two houses have addresses that are consecutive integers.

You could prove this by contradiction. Suppose there are no pairs of houses with consecutive integers. Then we can consider the couples $$ (n,n+1) $$ where $n$ is the number of a house and $n+1\leq 1099$. We have $50$ of those couples, and by hypothesis they don't intersect. But this implies there are $$ 50\cdot 2 = 100 $$ different numbers between $1000$ and $1099$, which is absurd.