https://file.aiquickdraw.com/mathbot/file/1729030830575bpspn6u0.jpg

The image contains the following statement:

**For all \( n \in \mathbb{Z}^+ \), 5 divides \( 7^n - 2^n \)**.

This is a mathematical statement claiming that for all positive integers \( n \), the expression \( 7^n - 2^n \) is divisible by 5, or equivalently:

\[
7^n - 2^n \equiv 0 \pmod{5}
\]

### Proof by Induction:

Let's prove this statement using **mathematical induction**.

**Base Case:**
For \( n = 1 \):

\[
7^1 - 2^1 = 7 - 2 = 5
\]

Clearly, \( 5 \) is divisible by 5, so the base case holds.

**Inductive Step:**
Assume the statement holds for some \( n = k \), i.e.,

\[
7^k - 2^k \equiv 0 \pmod{5}
\]

This means:

\[
7^k \equiv 2^k \pmod{5}
\]

Now we need to prove the statement for \( n = k+1 \), i.e.,

\[
7^{k+1} - 2^{k+1} \equiv 0 \pmod{5}
\]

Using the property of exponents, we can express \( 7^{k+1} \) and \( 2^{k+1} \) as:

\[
7^{k+1} = 7^k \cdot 7
\]
\[
2^{k+1} = 2^k \cdot 2
\]

Thus, we need to show:

\[
7^k \cdot 7 - 2^k \cdot 2 \equiv 0 \pmod{5}
\]

From the inductive hypothesis, we know \( 7^k \equiv 2^k \pmod{5} \), so we replace:

\[
7^k \cdot 7 - 2^k \cdot 2 \equiv 2^k \cdot 7 - 2^k \cdot 2 \pmod{5}
\]
\[
2^k \cdot (7 - 2) \equiv 2^k \cdot 5 \equiv 0 \pmod{5}
\]

Thus, the statement holds for \( n = k+1 \).

By the principle of mathematical induction, the statement is true for all \( n \in \mathbb{Z}^+ \).

---

Would you like more details or have any questions about this proof?

Here are 5 related questions for you to consider:
1. What is modular arithmetic and how is it used in divisibility proofs?
2. Can this statement be proven using another method besides induction?
3. What other numbers besides 5 can be used to generalize this result?
4. How does the property of exponents help in simplifying complex expressions?
5. What are common pitfalls in applying mathematical induction?

**Tip:** When proving divisibility statements, modular arithmetic simplifies handling large powers and remainders, making the logic clearer.

Learn how to prove that for all n ∈ Z+, 5 divides 7^n - 2^n using mathematical induction. This problem explores number theory concepts and modular arithmetic, suitable for Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation