Let 𝑎 and 𝑏 be integers and 𝑛 be a natural number. Prove by contrapositive:
If 10𝑎 ≢ 10𝑏 (mod 𝑛), then 𝑛 ∤ 10.

To prove the statement by contrapositive, we must first understand the structure of the statement. The original statement is:

"If \(10a \equiv 10b \pmod{n}\), then \(n \mid 10\)."

The contrapositive of this statement is:

"If \(n \nmid 10\), then \(10a \not\equiv 10b \pmod{n}\)."

### Step-by-Step Proof by Contrapositive:

#### 1. **Assume \( n \nmid 10 \):**

This means that \( n \) does **not** divide 10. In other words, 10 is not divisible by \( n \), which implies that the greatest common divisor \( \gcd(n, 10) > 1 \).

#### 2. **We need to prove that \( 10a \not\equiv 10b \pmod{n} \):**

This means we need to show that under this assumption, there are integers \( a \) and \( b \) such that \( 10a \not\equiv 10b \pmod{n} \). In other words, the difference \( 10a - 10b \) is **not** divisible by \( n \).

#### 3. **Derive the equation from the assumption \( n \nmid 10 \):**

Since \( n \nmid 10 \), it follows that 10 is not divisible by \( n \). Therefore, if we multiply any integers \( a \) and \( b \) by 10, the products \( 10a \) and \( 10b \) will still not behave in a way that makes \(

Let 𝑎 and 𝑏 be integers and 𝑛 be a natural number. Prove by contrapositive: If 10𝑎 ≢ 10𝑏 (mod 𝑛), then 𝑛 ∤ 10.

Explore a detailed step-by-step proof using contrapositive reasoning in number theory. The problem involves proving that if 10a is not congruent to 10b modulo n, then n does not divide 10, covering modular arithmetic and divisibility rules. Suitable for advanced high school and undergraduate students.

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