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 2 houses have addresses that are consecutive integers.

Solution

For integers between 1000 and 1099, the set with the most nonconsecutive integers should contain all even or odd integers between 1000 and 1099, inclusive.

In either case, the set has 50 elements. Therefore, the set with the most nonconsecutive integers contain 50 elements.In this case, there are 51 houses, 50 addresses can be assigned to these houses.

According to the generalized pigeonhole principle, there is at least one address assigned to at least houses. The Pigeonhole Principle [ceil (51/ 50)]=2...

Because each house has an unique house address, we can not assign the same address to two different houses. Therefore, even the set with the most nonconsecutive integers are not enough for us to assign the address for the 51 houses. As a result, some houses must have address that are consecutive integers.