91Ó°ÊÓ

Divisibility is quite simply the ability of one number, called the dividend, to be divided by another number, called the divisor, without leaving a remainder. In mathematical terms, a number \( a \) is divisible by \( b \) if there exists an integer \( c \) such that \( a = b \times c \). This means that when you divide \( a \) by \( b \), you get a whole number.
In our exercise, we're exploring the property of divisibility within numbers that have repeating digits. Specifically, these numbers must be divisible by \(3^n\), where \(n\) is a natural number. This concept becomes interesting because it involves using patterns found within number systems and relates to deeper mathematical properties like geometric sequences. Here, if a number is made up of a set of repeated sequences having \(3^n\) digits, it will consistently follow the rules of divisibility by \(3^n\). You can verify this by breaking down the number into a mathematical expression involving a geometric series. Once expressed in this form, it becomes clear that the structure of the number supports the divisibility by powers of 3.

A geometric series is a series of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. Mathematically, a geometric series with first term \( a \) and common ratio \( r \) can be expressed as:
\[ a + ar + ar^2 + ar^3 + \, \ldots \]
In the context of our exercise, the expression \( d \times \left( 1 + 10 + 10^2 + \ldots + 10^{k-1} \right) \) represents a geometric series where each number is made from a digit \( d \) repeated \( k \) times. This translates into:
\[ d \times \frac{10^k - 1}{9} \]
This formula comes from the sum of a finite geometric series and is key to show the divisibility property. It simplifies checking divisibility because the term \( \frac{10^k - 1}{9} \) is structured such that it aligns with divisibility by \(9\), ensuring easier multiplication checks. The parenthetical term is essentially just a sequence of nines, like 111 which is \( 111 = \frac{1000-1}{9} \). This approach facilitates understanding how numbers with repeated digits behave regarding divisibility.

Natural numbers are a fundamental concept in mathematics. They include all the positive integers starting from 1 (sometimes 0 as well) going upwards: \(1, 2, 3, \ldots\). They are the numbers we naturally count with and form the basis for other number systems.
Their importance in our exercise lies in the range of \( n \) values we consider (\(n \in \mathbb{N}\)). When we state a property or rule about natural numbers, like any integer consisting of \(3^n\) identical digits is divisible by \(3^n\), it's a broad and important claim since it's true for the infinite set of these numbers. Applying mathematical induction to prove such statements provides a robust method by way of handling infinitely many cases. We start with a base case (like \( n = 1 \)) and assume it holds true for \( n = k \), then we prove it has to hold for \( n = k + 1 \). This locked step manner covers all natural numbers without having to check each one individually.