Math Problem Statement

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.

Solution

To prove that for any positive integer nn, there is a number of the form 11001 \cdots 10 \cdots 0 (a number whose first digits are 1’s followed by zero or more 0’s) that is divisible by nn, 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: 1,11,111,1111,1, 11, 111, 1111, \ldots These are numbers made entirely of 1's. Mathematically, we can express these numbers as: ak=1111k ones=10k19,for k=1,2,3,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 aka_k where each aka_k is a number consisting of kk ones.

  2. Consider the Modulo nn Residues: We are interested in the divisibility of such numbers by nn. So, consider the sequence akmodna_k \mod n for different values of kk. Since there are only nn possible remainders when dividing by nn (i.e., 0,1,2,,n10, 1, 2, \dots, n-1), by the Pigeonhole Principle, in any sequence of n+1n+1 numbers, at least two of them must have the same remainder when divided by nn.

  3. Two Numbers with the Same Remainder: Let aiajmodna_i \equiv a_j \mod n for some i<ji < j. This means: ajai0modna_j - a_i \equiv 0 \mod n The difference ajaia_j - a_i is a number consisting of (ji)(j-i) ones followed by ii zeros. Explicitly: ajai=1111(ji) ones0000i zerosa_j - a_i = \underbrace{111\cdots1}_{(j-i) \text{ ones}} \underbrace{000\cdots0}_{i \text{ zeros}} Hence, ajaia_j - a_i is a number of the form 11001 \cdots 10 \cdots 0 (with some ones followed by zeros), and it is divisible by nn.

  4. Conclusion: We have found that there is a number of the form 11001 \cdots 10 \cdots 0 that is divisible by nn. Thus, we have proven the existence of such a number for any positive integer nn.

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 aba \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 11001 \cdots 10 \cdots 0?

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

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Number Theory
Divisibility
Modular Arithmetic

Formulas

a_k = \frac{10^k - 1}{9}, where a_k is a number made up of k ones
a_j - a_i = \underbrace{111\cdots1}_{(j-i) \text{ ones}} \underbrace{000\cdots0}_{i \text{ zeros}}

Theorems

Dirichlet's Box (Pigeonhole) Principle

Suitable Grade Level

Undergraduate