Question

Show that any integer consisting of \(3^{n}\) identical digits is divisible by \(3^{n}\). Verify this for \(222 ; 777 ; 222,222,222 ;\) and \(555,555.555 .\) Prove the general statement for all \(n \in \mathbb{N}\) by induction.

Short answer

Any integer with \(3^n\) identical digits is divisible by \(3^n.\)

Step-by-step solution

Rewrite Given Terms

Let's first understand what it means for numbers such as 222, 777, etc., to be expressed as integers with 3^n identical digits. Consider 222, where each digit is 2, and the number has three digits: 2 repeated. Similarly, 777 is composed of the digit 7 repeated in three places.

Express the Numbers Mathematically

For a number consisting of single digit \(d\) repeated \(k\) times (e.g., 222 is three times the digit 2), it can be written as:\[ d + 10d + 10^2d + \ldots + 10^{k-1}d = d \times \frac{10^k - 1}{9}\]This is because the number itself can be seen as a geometric series.

Check Divisibility by Powers of 3

For any integer expressed as above, every single digit number \(d \times \frac{10^k - 1}{9}\) is divisible by \(3^n\) when \(10^k - 1\) is divisible by \(9\cdot3^n\).For k identical digits, k must equal \(3^n\) for some n. Hence, \(10^k - 1\) is always divisible by \(9\cdot3^n\).

Verify Specific Cases

- For 222: \(222 = 2 \times \frac{10^3 - 1}{9} = 2 \times 111\). Since 111 is divisible by 3, 222 is divisible by \(3^2=9\).- For 777: \(777 = 7 \times \frac{10^3 - 1}{9} = 7 \times 111\). Again, 111 is divisible by 3, so 777 is divisible by \(9\).- For 222,222,222, it is \(2 \times \frac{10^9 - 1}{9} = 2 \times (111...111)\) (with nine 1's). \[(10^9 - 1) = 999,999,999\text{ is divisible by } 27 = 3^3.\]- 555,555,555: Same logic applies; each block of \(10^3 - 1\) ensures divisibility.

Prove by Mathematical Induction

_Base Case:_ For \(n=1\), any integer of the form 111 (3 repeated digits) divides by \(3^1=3\)._Inductive Step:_ Assume true for \(n=k\) that any integer of form \(3^k\) repeated digits is divisible by \(3^k\). We prove for \(n=k+1\).Assume number \(N\) has \(3^{k+1}\) digits:\[ N = d \times \frac{10^{3^{k+1}} - 1}{9} \]Since \(10^{3^{k+1}} - 1\) is divisible by \(3^{k+1}\) by its construction in terms of geometric progression, each repetition follows the rule.Hence by induction, any integer with \(3^n\) identical digits is divisible by \(3^n.\)

Key concepts

Divisibility

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.

Geometric Series

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

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.