Proving Divisibility: Math Problems Explained

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Hey math enthusiasts! Today, we're diving into some cool divisibility proofs. We'll be using mathematical induction to crack these problems. These are classic types of questions that you'll often see in math competitions or even in your regular math classes. Let's break down each part step-by-step. Get ready to flex those math muscles! We'll tackle three problems involving proving divisibility, making sure to show every step so that you understand the process. The first part involves showing that an expression is divisible by 23, the second focuses on divisibility by 11, and the third involves a more complex expression. It's time to put on your thinking caps, and let's get started!

Part a) Proving Divisibility by 23

Let's tackle the first problem: Prove that 2*3^n + 3^(n+1) + 2*3^(n+2) is divisible by 23 for all natural numbers n. This is where things get interesting, so stick with me! The core idea behind proving divisibility is to show that no matter what value of n (a natural number – which means 0, 1, 2, 3, and so on), the given expression will always be a multiple of 23. Let's write out the steps.

Step 1: Analyze the problem

Here, the main focus is to determine whether the given expression is divisible by 23 for all natural numbers 'n'. This requires understanding the properties of exponents and divisibility rules. The given expression involves powers of 3, and our goal is to show that the entire expression is always a multiple of 23, regardless of the value of 'n'. This can be done by simplifying and factoring the expression to demonstrate a clear multiple of 23.

Step 2: Simplify the Expression

First, let's simplify the expression: 2*3^n + 3^(n+1) + 2*3^(n+2). We can rewrite 3^(n+1) as 3*3^n and 3^(n+2) as 9*3^n. So, the expression becomes:

2*3^n + 3*3^n + 2*9*3^n.

Now, let's factor out 3^n:

3^n * (2 + 3 + 18).

Then simplify inside the parentheses:

3^n * 23.

Step 3: Conclusion

Since 3^n * 23 is a product of 23 and some integer (3^n), the expression is always divisible by 23. This holds true for all natural numbers n because 3^n is always an integer. We've proven it! That's it, the expression is always divisible by 23, cool, right? This is the power of simplifying the expressions to find the pattern.

Part b) Proving Divisibility by 11

Alright, let's shift gears and tackle another problem! This time we need to prove that 5^n + 3*5^(n+2) - 2*5^(n+1) is divisible by 11 for all natural numbers n. The process is quite similar to the first part – we aim to simplify the expression and reveal a clear multiple of 11. Are you ready? Let's break it down again step by step. We'll follow the same procedure.

Step 1: Understand the Problem

Here, the objective is to demonstrate that the given expression is divisible by 11 for all natural numbers 'n'. This involves leveraging the properties of exponents and divisibility rules to prove our case. The expression includes powers of 5, and we want to prove that it always results in a multiple of 11, regardless of the value of 'n'. To accomplish this, we'll simplify and factor the expression so that we can clearly see the multiple of 11.

Step 2: Simplify the Expression

First, let's simplify the expression: 5^n + 3*5^(n+2) - 2*5^(n+1). We can rewrite 5^(n+1) as 5*5^n and 5^(n+2) as 25*5^n. So the expression becomes:

5^n + 3*25*5^n - 2*5*5^n.

Now, let's simplify the multiplication:

5^n + 75*5^n - 10*5^n.

Let's factor out 5^n:

5^n * (1 + 75 - 10).

Simplify the terms inside the parentheses:

5^n * 66.

Which can also be written as:

5^n * 6 * 11.

Step 3: Conclusion

Since 5^n * 66 (or 5^n * 6 * 11) is a product of 11 and some integer (5^n * 6), the expression is always divisible by 11. Because 5^n is always an integer for any natural number, it is correct. We have successfully proven that 5^n + 3*5^(n+2) - 2*5^(n+1) is divisible by 11 for all n in N. You're doing great! Keep up the good work!

Part c) Complex Expression Divisibility

Okay, guys, for the final part, we are dealing with a more complex expression: A = 7*12^n - 3^(n+1) + 6*4^(n+1)*9^(n+2) + 18^(n+1)*2^(n+1). Our challenge now is to show that it is divisible by something. Let's get right into it! Remember the process: simplify, simplify, simplify, and look for patterns. Here we go!

Step 1: Simplify the Expression

This expression is much more complicated, so we'll need to break it down carefully. First, let's rewrite some terms to make them easier to work with:

  • 12^n = (3*4)^n = 3^n * 4^n
  • 4^(n+1) = 4 * 4^n
  • 9^(n+2) = 9^2 * 9^n = 81 * 9^n
  • 18^(n+1) = (2*9)^(n+1) = 2^(n+1) * 9^(n+1)

Now, substitute these back into the original expression:

A = 7 * 3^n * 4^n - 3^(n+1) + 6 * 4 * 4^n * 81 * 9^n + 2^(n+1) * 9^(n+1) * 2^(n+1)

Which equals:

A = 7 * 3^n * 4^n - 3 * 3^n + 6 * 4 * 4^n * 81 * 9^n + 2^(n+1) * 9 * 9^n * 2^(n+1)

This simplifies to:

A = 7 * 3^n * 4^n - 3 * 3^n + 1944 * 4^n * 9^n + 9 * 2^(2n+2) * 9^n

We can combine like terms with the same exponents:

A = 7 * 3^n * 4^n - 3 * 3^n + 1944 * 36^n + 9 * 4^(n+1) * 9^n

Which then converts to:

A = 7 * 12^n - 3 * 3^n + 1944 * 36^n + 9 * 36^n

Simplify:

A = 7 * 12^n - 3 * 3^n + (1944 + 9) * 36^n

So it becomes:

A = 7 * 12^n - 3 * 3^n + 1953 * 36^n

Step 2: Further Simplification and Pattern Recognition

Now, we'll try to find a common factor or a pattern that helps us determine divisibility. Notice that the expression involves powers of 3, 4, 9, 12, 18, and 36, which are all related to prime factors 2 and 3. The presence of these numbers and their powers should give us a clue. Remember, our goal is to factor and simplify the expression to hopefully reveal the number it is divisible by.

At first glance, it may seem hard to identify a common factor that ties everything together. The key here is to realize that 36^n can be expressed as (3 * 12)^n or (4 * 9)^n or even (2^2 * 32)n, we may notice that we have 7 * 12^n. By factoring the expression, we can identify a pattern.

Step 3: Conclusion

This expression doesn't seem to have a readily apparent number it is divisible by, so we can't definitively determine its divisibility based on the current form. In this case, additional information or a clearer objective might be needed to determine what A is divisible by. Sometimes, math problems can be a bit tricky, and it might require additional manipulation or different strategies to reach a conclusion. But don't worry, the important thing is that we have explored the expression, tried to simplify it, and look for patterns. Keep practicing and keep up the great work!