 There are 51 houses on a street. Their addresses are between 1000 and 1099 (both inclusive).
Show that there are at least two houses with consecutive addresses. State clearly what the “pi-
geons” and “pigeonholes” are.
1

The **pigeonhole principle** states that if more objects (pigeons) are placed into fewer containers (pigeonholes) than the number of objects, then at least one container must contain more than one object.

### Given Problem:
- There are 51 houses.
- Their addresses are between 1000 and 1099, inclusive.

### Step-by-step explanation:
1. **Total possible house addresses**:  
   The addresses range from 1000 to 1099, so there are 100 addresses in total.

2. **Houses and their possible differences**:
   We want to show that at least two houses have consecutive addresses.

### Applying the pigeonhole principle:
- **Pigeons**: The houses (there are 51 houses).
- **Pigeonholes**: The **gaps** between house addresses where consecutive houses can be placed. There are 50 gaps between the 51 numbers.

The number of available addresses is larger than the number of houses, but we want to show that there are consecutive houses. If we think of addresses as gaps between possible numbers, there are 50 gaps between consecutive numbers.

Now, if we try to assign the 51 houses to 50 gaps (pigeonholes), the pigeonhole principle guarantees that at least one of these gaps must contain more than one house, meaning there are at least two houses with consecutive addresses.

### Conclusion:
Since there are more houses (51 pigeons) than there are gaps between consecutive addresses (50 pigeonholes), by the pigeonhole principle, at least two houses must have consecutive addresses.

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Would you like further clarification or additional details?

Here are 5 related questions to deepen understanding:
1. What is the general form of the pigeonhole principle, and how is it applied in other problems?
2. Can you provide an example where houses have non-consecutive addresses?
3. How many total pairs of consecutive addresses exist between 1000 and 1099?
4. What if there were 60 houses instead of 51? How does the result change?
5. How does the pigeonhole principle apply to real-life scheduling problems?

**Tip**: Always try to think of "pigeons" as items you're placing and "pigeonholes" as categories or slots where those items go. If there are more items than slots, at least one slot will hold more than one item.

There are 51 houses on a street. Their addresses are between 1000 and 1099 (both inclusive). Show that there are at least two houses with consecutive addresses. State clearly what the 'pigeons' and 'pigeonholes' are.

Learn how to apply the Pigeonhole Principle to show that among 51 houses with addresses between 1000 and 1099, at least two houses must have consecutive addresses. Explore a step-by-step breakdown of this fundamental concept in discrete mathematics, suitable for high school students.

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