Let n be a positive integer. Prove that there is a number of the form
1 · · · 10 · · · 0 (this is a number whose first digits, from the left, are 1’s and whose
last digits are 0’s; numbers with a zero number of 0’s are considered as well) that
is divisible by n.


To prove that for any positive integer \( n \), there is a number of the form \( 1 \cdots 10 \cdots 0 \) (a number whose first digits are 1’s followed by zero or more 0’s) that is divisible by \( n \), we can apply **Dirichlet's Box (Pigeonhole) Principle**.

### Steps for the Proof

1. **Definition of the Sequence**:
   Consider a sequence of numbers of the form:
   \[
   1, 11, 111, 1111, \ldots
   \]
   These are numbers made entirely of 1's. Mathematically, we can express these numbers as:
   \[
   a_k = \underbrace{111\cdots1}_{k \text{ ones}} = \frac{10^k - 1}{9}, \quad \text{for } k = 1, 2, 3, \dots
   \]
   This gives a sequence of numbers \( a_k \) where each \( a_k \) is a number consisting of \( k \) ones.

2. **Consider the Modulo \( n \) Residues**:
   We are interested in the divisibility of such numbers by \( n \). So, consider the sequence \( a_k \mod n \) for different values of \( k \). Since there are only \( n \) possible remainders when dividing by \( n \) (i.e., \( 0, 1, 2, \dots, n-1 \)), by the **Pigeonhole Principle**, in any sequence of \( n+1 \) numbers, at least two of them must have the same remainder when divided by \( n \).

3. **Two Numbers with the Same Remainder**:
   Let \( a_i \equiv a_j \mod n \) for some \( i < j \). This means:
   \[
   a_j - a_i \equiv 0 \mod n
   \]
   The difference \( a_j - a_i \) is a number consisting of \( (j-i) \) ones followed by \( i \) zeros. Explicitly:
   \[
   a_j - a_i = \underbrace{111\cdots1}_{(j-i) \text{ ones}} \underbrace{000\cdots0}_{i \text{ zeros}}
   \]
   Hence, \( a_j - a_i \) is a number of the form \( 1 \cdots 10 \cdots 0 \) (with some ones followed by zeros), and it is divisible by \( n \).

4. **Conclusion**:
   We have found that there is a number of the form \( 1 \cdots 10 \cdots 0 \) that is divisible by \( n \). Thus, we have proven the existence of such a number for any positive integer \( n \).

This concludes the proof.

---

Would you like more details on any step, or have any questions?

Here are five related questions to expand this concept:
1. How can we generalize this result to numbers with alternating digits?
2. What are the implications of this proof for modular arithmetic?
3. Can this result be extended to prove divisibility by any integer combination, like numbers of form \( a \cdots b \)?
4. How does the Pigeonhole Principle work in other divisibility problems?
5. Can you find a number divisible by 7 of the form \( 1 \cdots 10 \cdots 0 \)?

**Tip**: The Pigeonhole Principle is extremely useful in proving existence theorems in number theory, especially when dealing with modular arithmetic!

Let n be a positive integer. Prove that there is a number of the form 1 · · · 10 · · · 0 (this is a number whose first digits, from the left, are 1’s and whose last digits are 0’s; numbers with a zero number of 0’s are considered as well) that is divisible by n.

Explore a detailed proof using Dirichlet's Pigeonhole Principle to show that for any positive integer n, there exists a number of the form 1...10...0 that is divisible by n. This number theory problem involves key concepts like modular arithmetic and divisibility, making it a suitable challenge for undergraduate students.

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