Divisibility Rules: Proving 6^n - 1 And 111^n - 6 By 5

Hey guys! Today, we're diving deep into the awesome world of algebra, specifically tackling divisibility rules. We've got a couple of cool expressions to prove: 1) 6^n - 1 and 2) 111^n - 6. The mission, should you choose to accept it, is to prove that no matter what natural number n you throw at these expressions, the result will always be perfectly divisible by 5. Sounds like a challenge? Don't worry, we'll break it down step-by-step. Get ready to flex those mathematical muscles!

Understanding Divisibility and Natural Numbers

Before we jump into the proofs, let's make sure we're all on the same page. When we talk about a number being divisible by 5, it simply means that when you divide it by 5, you get a whole number with no remainder. Think about numbers like 10, 25, 100 – these are all divisible by 5. Now, natural numbers are the positive whole numbers: 1, 2, 3, 4, and so on. Sometimes, people include 0 in the set of natural numbers, but for these kinds of proofs, we usually stick to the positive integers starting from 1.

Why is this important? Because our proofs need to hold true for every single possible value of n that's a natural number. This is where a powerful tool in mathematics called mathematical induction comes in handy. It's like a domino effect: if you can show the first domino falls, and then show that if any domino falls, the next one will too, then you know all the dominos will fall.

We'll be using this inductive reasoning to conquer our problems. We'll first check if our expressions are divisible by 5 for the smallest natural number (which is 1). Then, we'll assume the expression is divisible by 5 for some arbitrary natural number k, and use that assumption to prove it's also divisible by 5 for the next number, k+1. If we can do both, we've essentially proven it's true for all natural numbers n.

Let's get to the exciting part: the actual proofs! Grab your favorite beverage, settle in, and let's explore these algebraic beasts.

Proof 1: Showing 6^n - 1 is Divisible by 5

Alright team, let's tackle the first expression: 6^n - 1. Our goal is to prove that for any natural number n, this expression is always divisible by 5. We're going to use the principle of mathematical induction here, which is a super reliable way to prove statements about natural numbers. It's like a two-step rocket launch: Step 1 is the ignition, and Step 2 is achieving orbit.

Step 1: The Base Case (Ignition)

First, we need to show that the statement is true for the smallest natural number, which is n = 1. Let's plug in n = 1 into our expression:

$6^1 - 1 = 6 - 1 = 5$

Boom! We got 5. Is 5 divisible by 5? Absolutely! So, our base case holds true. This is our first domino successfully toppled.

Step 2: The Inductive Step (Achieving Orbit)

Now for the more challenging part. We need to assume that the statement is true for some arbitrary natural number k. This means we assume that 6^k - 1 is divisible by 5. If it's divisible by 5, we can write it as $6^k - 1 = 5m$ for some integer m. Rearranging this, we get $6^k = 5m + 1$.

Our next mission is to prove that the statement is also true for the next natural number, which is k + 1. That is, we need to show that $6^(k+1) - 1$ is divisible by 5. Let's start by expanding this expression:

$6^(k+1) - 1 = 6^k * 6^1 - 1$

Remember our assumption? We know that $6^k = 5m + 1$. Let's substitute that into our expanded expression:

$= (5m + 1) * 6 - 1$

Now, let's distribute the 6:

$= 30m + 6 - 1$

$= 30m + 5$

We're almost there, guys! To show this is divisible by 5, we can factor out a 5:

$= 5(6m + 1)$

Since m is an integer, $6m$ is also an integer, and $6m + 1$ is definitely an integer. Let's call this new integer p (so, $p = 6m + 1$). Our expression now looks like $5p$.

And there you have it! $5p$ is clearly divisible by 5. We've successfully shown that if the statement is true for k, it's also true for k + 1.

Conclusion for Proof 1

By the principle of mathematical induction, since the base case ($n=1$) is true, and the inductive step (if it's true for k, it's true for k+1) holds, we can confidently conclude that $6^n - 1$ is divisible by 5 for all natural numbers n. Mission accomplished on the first one!

Proof 2: Showing 111^n - 6 is Divisible by 5

Alright, let's move on to our second algebraic challenge: proving that $111^n - 6$ is divisible by 5 for all natural numbers n. We'll stick with our trusty friend, mathematical induction. It's been working well for us, so let's keep using it!

Step 1: The Base Case (Ignition Again)

As always, we start with the smallest natural number, n = 1. Let's substitute n = 1 into the expression:

$111^1 - 6 = 111 - 6 = 105$

Now, we need to ask ourselves: is 105 divisible by 5? Yes, it is! $105 / 5 = 21$. So, our base case is solid. The first domino has fallen successfully.

Step 2: The Inductive Step (Orbiting the Proof)

Next, we make our assumption. We assume that for some arbitrary natural number k, the expression $111^k - 6$ is divisible by 5. This means we can write $111^k - 6 = 5q$ for some integer q. Rearranging this gives us $111^k = 5q + 6$.

Now, our mission is to prove that the statement holds for the next natural number, k + 1. We need to show that $111^(k+1) - 6$ is divisible by 5. Let's expand this expression:

$111^(k+1) - 6 = 111^k * 111^1 - 6$

We know from our assumption that $111^k = 5q + 6$. Let's substitute this in:

$= (5q + 6) * 111 - 6$

Now, we distribute the 111:

$= 5q * 111 + 6 * 111 - 6$

$= 555q + 666 - 6$

$= 555q + 660$

We're super close! To demonstrate divisibility by 5, let's factor out a 5 from both terms:

$= 5(111q + 132)$

Since q is an integer, $111q$ is an integer, and adding 132 ($111q + 132$) still results in an integer. Let's call this new integer r (so, $r = 111q + 132$). Our expression is now $5r$.

And voilà! $5r$ is unmistakably divisible by 5.

Conclusion for Proof 2

Since we've successfully proven the base case ($n=1$) and the inductive step (if it's true for k, it's true for k+1), we can confidently conclude, using the principle of mathematical induction, that $111^n - 6$ is divisible by 5 for all natural numbers n. Another one bites the dust!

Alternative Approaches: Thinking Modulo Arithmetic

While mathematical induction is a fantastic and rigorous way to prove these kinds of statements, sometimes you might encounter problems where thinking in terms of modular arithmetic can offer a quicker insight. It's like a shortcut, but still totally valid!

Modular arithmetic deals with remainders. When we say a number a is congruent to b modulo m (written as $a \equiv b \pmod{m}$), it means a and b have the same remainder when divided by m. In our case, we want to show that our expressions are congruent to 0 modulo 5, which is the same as saying they are divisible by 5.

For $6^n - 1$:

Let's look at the base, 6. What's its remainder when divided by 5? It's 1. So, we can write:

$6 \equiv 1 \pmod{5}$

Now, using the properties of modular arithmetic, if $a \equiv b \pmod{m}$, then $a^n \equiv b^n \pmod{m}$. So, we can raise both sides to the power of n:

$6^n \equiv 1^n \pmod{5}$

Since $1^n$ is always 1 (for any natural number n):

$6^n \equiv 1 \pmod{5}$

This means $6^n$ has a remainder of 1 when divided by 5. If we subtract 1 from $6^n$, the remainder will be $1 - 1 = 0$:

$6^n - 1 \equiv 1 - 1 \pmod{5}$

$6^n - 1 \equiv 0 \pmod{5}$

This directly shows that $6^n - 1$ is divisible by 5.

For $111^n - 6$:

Let's examine 111. What's its remainder when divided by 5? The last digit is 1, so the remainder is 1.

$111 \equiv 1 \pmod{5}$

Applying the same property as before, we raise both sides to the power of n:

$111^n \equiv 1^n \pmod{5}$

$111^n \equiv 1 \pmod{5}$

Now, we need to consider the '- 6' part. What's the remainder of -6 when divided by 5? It's -1, or more commonly written as 4 (since $-6 + 5 = -1$, and $-1 + 5 = 4$). Let's use -1 for simplicity here.

So, we have:

$111^n - 6 \equiv 1 - 6 \pmod{5}$

$111^n - 6 \equiv -5 \pmod{5}$

Since -5 is divisible by 5, its remainder is 0.

$111^n - 6 \equiv 0 \pmod{5}$

This modular arithmetic approach is super slick and often faster once you get the hang of it. It confirms our results from mathematical induction.

Final Thoughts

So there you have it, folks! We've successfully proven, using two powerful methods (mathematical induction and modular arithmetic), that both $6^n - 1$ and $111^n - 6$ are always divisible by 5 for any natural number n. Algebra can seem intimidating sometimes, but breaking it down into steps and understanding the underlying principles makes it totally manageable, and dare I say, even fun! Keep practicing, keep exploring, and never shy away from a good mathematical challenge. Until next time, happy solving!