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.

Short answer

The resultant answer is that we can't assign 50 unique addresses to 51 different houses. Therefore, 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.

Step-by-step solution

Given data

There are 51 houses on a street is the given data.

Concept of Pigeonhole principle

If the number of pigeons exceeds the number of pigeonholes, at least one hole will hold at least two pigeons, according to the pigeon hole principle.

Contradict and simplify the expression

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 \(100/2 = 50\) useable addresses.

This implies that we can't assign 50 unique addresses to 51 different houses. Therefore, 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.