 There are 51 houses on a street. Their addresses are between 1000 and 1099 (both inclusive).
Show if it is possible there are at least two houses with consecutive addresses. 




To determine if it is possible that there are at least two houses with consecutive addresses, let's break it down step by step:

1. **Total available addresses:**  
   The addresses are between 1000 and 1099, which includes:
   \[
   1099 - 1000 + 1 = 100 \text{ addresses}.
   \]

2. **Number of houses on the street:**  
   There are 51 houses on the street.

3. **Possible arrangement of houses:**  
   If all houses had unique addresses and no consecutive addresses, then the 51 houses would need to be spread out across 100 available addresses, with no two houses next to each other. 

   To avoid consecutive addresses, each house would need to have at least one "gap" between them. This would mean placing each house with at least one space between, i.e., using a pattern like 1000, 1002, 1004, ..., etc.

4. **Space required without consecutive addresses:**  
   If we want to ensure no two houses have consecutive addresses, 51 houses would need at least 50 gaps between them (since a gap between each pair of houses). The minimum number of addresses required would be:
   \[
   51 \text{ houses} + 50 \text{ gaps} = 101 \text{ addresses}.
   \]

   However, we only have 100 addresses available.

### Conclusion:
Since we only have 100 addresses but would need 101 to ensure no consecutive houses, it is **impossible** to arrange the houses without at least two of them having consecutive addresses.

Would you like further details on any step? Here are 5 follow-up questions for deeper understanding:

1. How can we calculate the number of gaps needed between houses for a given total of addresses?
2. What if the number of addresses was larger, say 200? Would it change the outcome?
3. Can we model this problem using the pigeonhole principle?
4. How does this problem relate to real-life city planning or address assignment?
5. What are some strategies to distribute houses to minimize consecutive addresses?

**Tip:** In problems like this, the pigeonhole principle is a powerful tool to determine whether a specific arrangement is possible!

There are 51 houses on a street. Their addresses are between 1000 and 1099 (both inclusive). Show if it is possible there are at least two houses with consecutive addresses.

Explore whether it's possible to arrange 51 houses with non-consecutive addresses between 1000 and 1099. This problem demonstrates key concepts in combinatorics and the pigeonhole principle, with a step-by-step breakdown of the calculations involved. Suitable for Grades 8-10.

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