There are 51 houses on a street Each house has an address be

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.

Solution

Let us make pigeonhole for two consecutive addresses numbers as (1000, 1001), (1002, 1003) ... (1098, 1099). This way there are 99 pigeonholes for the numbers between 1000 and 1099. But since except 1000 and 1099, all the other numbers are labelled twice, every housing bearing those numbers will occupy two pigeonholes simultaneously. Only two houses numbered 1000 and 1099 occupy only one pigeonhole.

Hence, there are 2 + 49 * 2 = 100 pigeons. Now, with 99 pigeonholes and 100 pigeons, using pigeonhole principle, it is concluded that there would be atleast one pigeonhole, which has two pigeons.

So, from this, we can say that at least two houses have consecutive numbers as addresses.