Divisibility Proofs: 5^n + 5^(n+1) By 30, 24*3^n + 2^(n+1) - 3^n By 150

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Divisibility Proofs: 5^n + 5^(n+1) by 30, 24*3^n + 2^(n+1) - 3^n by 150

Hey guys! Let's dive into some cool divisibility problems. We've got two interesting proofs to tackle today, both involving demonstrating that certain expressions are divisible by specific numbers for any natural number 'n'. This kind of problem often involves algebraic manipulation and modular arithmetic, so get ready to put on your math hats! We'll break down each proof step-by-step to make sure everything's crystal clear. Let's get started!

Problem 1: Proving 5^n + 5^(n+1) is Divisible by 30

Our first task is to prove that the expression a = 5^n + 5^(n+1) is divisible by 30 for any natural number 'n' (n ∈ N). To show that a number is divisible by 30, we need to demonstrate that it is divisible by both 5 and 6 (since 5 and 6 are coprime, and 5 * 6 = 30). Let's break it down:

Step 1: Simplify the Expression

First, let's simplify the expression a = 5^n + 5^(n+1). We can factor out 5^n from both terms:

a = 5^n + 5^(n+1) = 5^n + 5^n * 5^1 = 5^n * (1 + 5) = 5^n * 6

So, we have a = 5^n * 6. This already shows that 'a' is divisible by 6, because it's 6 multiplied by some integer (5^n).

Step 2: Prove Divisibility by 5

Now, we need to show that 'a' is divisible by 5. Looking at our simplified expression, a = 5^n * 6, it's pretty obvious. Since 5^n is a power of 5, it's definitely divisible by 5 for any n ≥ 1. If n = 0, then 5^0 = 1, and a = 1 * 6 = 6, which is not divisible by 5. However, the problem states n ∈ N, which usually implies n ≥ 1. If N includes 0, we need to consider that case separately. Assuming N = {1, 2, 3, ...}, then 5^n is always divisible by 5.

For n ≥ 1, 5^n will always have a factor of 5. Therefore, the entire expression 5^n * 6 will also be divisible by 5.

Step 3: Conclusion

Since 'a' is divisible by both 5 and 6 (and 5 and 6 are coprime), it follows that 'a' is divisible by 30. To summarize:

  • a = 5^n * 6
  • 5^n is divisible by 5 for n ≥ 1
  • Therefore, 5^n * 6 is divisible by 5
  • Since 'a' is divisible by both 5 and 6, 'a' is divisible by 30.

Thus, we've proven that the number a = 5^n + 5^(n+1) is divisible by 30 for any n ∈ N (where N = {1, 2, 3, ...}). If N includes 0, the statement is not strictly true, as 5^0 + 5^(0+1) = 1 + 5 = 6, which is not divisible by 30. However, for n ≥ 1, the proof holds.

Problem 2: Proving 24 * 3^n + 2^(n+1) - 3^n is Divisible by 150

Next up, we need to demonstrate that the expression a = 24 * 3^n + 2^(n+1) - 3^n is divisible by 150 for any natural number 'n' (n ∈ N). This is a bit trickier, but we can handle it! Remember that 150 = 2 * 3 * 5^2, so we need to show divisibility by 2, 3, and 25 (since 25 is 5 squared).

Step 1: Simplify the Expression

First, let's simplify the expression a = 24 * 3^n + 2^(n+1) - 3^n:

a = 24 * 3^n + 2^(n+1) - 3^n = 24 * 3^n - 3^n + 2^(n+1) = 23 * 3^n + 2^(n+1) = 23 * 3^n + 2 * 2^n

So, we have a = 23 * 3^n + 2 * 2^n.

Step 2: Analyze Divisibility by 2

Let's consider the term 2 * 2^n. This can be rewritten as 2^(n+1). For any n ≥ 1, 2^(n+1) will always be divisible by 2. Now we need to examine the term 23 * 3^n. Since 23 is odd and 3^n is always odd for any n, their product will also be odd. Therefore, 23 * 3^n is not divisible by 2.

However, a = (23 * 3^n) + (2 * 2^n). Since 2 * 2^n is always even, let's see if the whole expression 'a' is even. An odd number plus an even number is always odd. Therefore, 'a' is not divisible by 2 for any n. This means the original statement that 'a' is divisible by 150 for any n is incorrect. We need to find a specific condition or error in the initial problem statement.

Important Note: Since we've shown that 'a' is not divisible by 2 for any n, it cannot be divisible by 150. Thus, there is something wrong with the original problem statement or the assumption that it holds for all n ∈ N. Let's explore if there are specific values of 'n' for which it might hold true, or if there's a typo in the expression.

Step 3: Investigating Specific Values of 'n'

Let's test a few values of 'n' to see if we can spot a pattern:

  • If n = 1: a = 24 * 3^1 + 2^(1+1) - 3^1 = 24 * 3 + 2^2 - 3 = 72 + 4 - 3 = 73. This is not divisible by 150.
  • If n = 2: a = 24 * 3^2 + 2^(2+1) - 3^2 = 24 * 9 + 2^3 - 9 = 216 + 8 - 9 = 215. This is not divisible by 150.
  • If n = 3: a = 24 * 3^3 + 2^(3+1) - 3^3 = 24 * 27 + 2^4 - 27 = 648 + 16 - 27 = 637. This is not divisible by 150.

These examples reinforce our earlier conclusion that the expression is likely never divisible by 150 for any n.

Step 4: Re-evaluating the Problem Statement

It's possible there's a typo in the problem statement. Let's consider what changes would be needed to make it divisible by 150. Since 150 = 2 * 3 * 5^2, the expression needs to be even and divisible by 3 and 25.

Hypothetical Correction: Let's imagine the problem meant to say:

a = 25 * 3^n + 2^(n+1) - 3^n. In this case, a = 24 * 3^n + 2^(n+1). It's still unlikely to be divisible by 150 for all n, but at least it's closer to potentially being even for some n.

Step 5: Conclusion and Discussion

Based on our analysis, the original statement that a = 24 * 3^n + 2^(n+1) - 3^n is divisible by 150 for any n ∈ N is incorrect. The expression is not divisible by 2, and therefore cannot be divisible by 150. The examples for n = 1, 2, and 3 further confirm this conclusion.

Guys, it's crucial to carefully examine the problem statements and assumptions. Sometimes, there might be errors or typos that lead to incorrect conclusions. In this case, the expression as given simply does not satisfy the divisibility requirement.

It might be worthwhile to double-check the original source of the problem to see if there's a correction available, or to consider alternative interpretations if the problem is intended to be a more open-ended challenge.