Question

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.

Answer 1

Pigeon hole principle:

There are $100$ possible addresses and $51$ houses. In order for there to be no houses with consecutive addresses, each house must have at least one address in between it. This can be done by only assigning even numbers to houses (leaving odd addresses as the buffer address)

Assigning even numbers to houses (leaving odd addresses as the buffer address) We now have $\frac{100}{2}=50$ useable addresses.

By the pigeon hole principle, If $N$ objects are placed into $k$ boxes, then there is at least one box containing at least $\left[\frac{N}{k}\right]$ object.

Hence it can’t assigned $50$ unique addresses to $51$ different houses without using any one consecutive integer.

Hence, there must be at least one instance of houses having consecutive integers, meaning there are at least two houses that have addresses that are consecutive integers.