https://r2.erweima.ai/mathbot/image/1726810264500hmpp6yf6.jpg

To solve the problem:

We need to prove that \( 7^n - 1 \) is divisible by 6 for all \( n \geq 1 \).

### Approach:

We need to show that \( 7^n - 1 \) is divisible by both 2 and 3, as 6 is the least common multiple of these two numbers.

#### 1. Divisibility by 2:
We begin by showing that \( 7^n - 1 \) is divisible by 2.

Since 7 is an odd number, \( 7^n \) is also odd for all \( n \geq 1 \). Therefore, \( 7^n - 1 \) is the difference between an odd number and 1, which is always even. Hence, \( 7^n - 1 \) is divisible by 2 for all \( n \geq 1 \).

#### 2. Divisibility by 3:
Next, we check divisibility by 3 using **modular arithmetic**:

- For \( n = 1 \): \( 7^1 = 7 \equiv 1 \ (\text{mod}\ 3) \). Thus, \( 7^1 - 1 = 6 \equiv 0 \ (\text{mod}\ 3) \).
- Assume for some \( n \geq 1 \), \( 7^n \equiv 1 \ (\text{mod}\ 3) \). Then by induction:
  \[
  7^{n+1} = 7 \cdot 7^n \equiv 1 \cdot 1 = 1 \ (\text{mod}\ 3)
  \]
Thus, for all \( n \geq 1 \), \( 7^n \equiv 1 \ (\text{mod}\ 3) \), which implies that \( 7^n - 1 \equiv 0 \ (\text{mod}\ 3) \).

#### Conclusion:
Since \( 7^n - 1 \) is divisible by both 2 and 3 for all \( n \geq 1 \), it is divisible by 6 for all \( n \geq 1 \).

Would you like a detailed breakdown of any part of the solution?

### Related Questions:
1. How can modular arithmetic help in proving divisibility problems?
2. Can this proof be extended to numbers other than 7?
3. How does induction work in mathematical proofs?
4. Can the binomial theorem be used to analyze powers of numbers in divisibility problems?
5. How do divisibility rules for larger numbers relate to their prime factors?

### Tip:
When working with divisibility problems, modular arithmetic is a powerful tool to reduce complex calculations to simpler remainders.

Demostrar que 7^n - 1 es divisible entre 6, para toda n ≥ 1.

This problem demonstrates how to prove that 7^n - 1 is divisible by 6 for all n ≥ 1. Key concepts include modular arithmetic and divisibility by 2 and 3. It is ideal for advanced high school students learning number theory.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation