After training for her entire life, Erica is approaching the pinnacle of her Tennis career. She is about to compete in the Women’s Singles Tennis finals at the Olympics and has a shot at gold. But disaster... while sitting in the locker room, she has started experiencing pre-competition nerves because she cannot f igure out the following problems. (a) If x and y are both divisible by k, prove that gcd(x,y) = k ·gcd(x k, y k), where k > 1.



We are tasked with proving that if \( x \) and \( y \) are both divisible by \( k \) (where \( k > 1 \)), then the greatest common divisor (gcd) of \( x \) and \( y \) can be expressed as:

\[
\gcd(x, y) = k \cdot \gcd\left(\frac{x}{k}, \frac{y}{k}\right)
\]

### Step-by-step proof:

**Given:**
- \( x = k \cdot x_1 \) and \( y = k \cdot y_1 \), where \( x_1 = \frac{x}{k} \) and \( y_1 = \frac{y}{k} \) are integers.
- We are asked to prove that \( \gcd(x, y) = k \cdot \gcd(x_1, y_1) \).

**Step 1: Express \( x \) and \( y \) in terms of \( k \)**
- Since \( x = k \cdot x_1 \) and \( y = k \cdot y_1 \), we see that both \( x \) and \( y \) are divisible by \( k \). Hence, \( x_1 \) and \( y_1 \) are integers.

**Step 2: Use the properties of gcd**
- By definition, \( \gcd(x, y) \) is the greatest integer \( d \) that divides both \( x \) and \( y \).
- Also, \( \gcd(x_1, y_1) \) is the greatest integer that divides both \( x_1 \) and \( y_1 \).

We now explore how the gcd of the scaled numbers \( x \) and \( y \) relates to the gcd of \( x_1 \) and \( y_1 \).

**Step 3: Factor out \( k \)**
- Since \( x = k \cdot x_1 \) and \( y = k \cdot y_1 \), any common divisor of \( x \) and \( y \) must also divide \( k \), as \( k \) is a factor of both \( x \) and \( y \).
- Let \( d = \gcd(x, y) \). Since \( x = k \cdot x_1 \) and \( y = k \cdot y_1 \), the integer \( d \) must be divisible by \( k \). Thus, we can write \( d = k \cdot d_1 \), where \( d_1 \) is the gcd of \( x_1 \) and \( y_1 \).

**Step 4: Conclusion**
- Therefore, we have the relationship \( \gcd(x, y) = k \cdot \gcd(x_1, y_1) \).
- This completes the proof.

\[
\boxed{\gcd(x, y) = k \cdot \gcd\left(\frac{x}{k}, \frac{y}{k}\right)}
\]

Would you like more details or have any questions?

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Here are 5 related questions you might find useful:

1. How do you compute the gcd of two numbers using the Euclidean algorithm?
2. What is the relationship between the lcm and gcd of two numbers?
3. Can you generalize this property to more than two numbers?
4. How can we efficiently calculate gcd for very large numbers?
5. What are the real-world applications of gcd in cryptography and number theory?

**Tip**: When working with gcd, remember that prime factorization can also help reveal insights about common divisors!

After training for her entire life, Erica is approaching the pinnacle of her Tennis career. She is about to compete in the Women’s Singles Tennis finals at the Olympics and has a shot at gold. But disaster... while sitting in the locker room, she has started experiencing pre-competition nerves because she cannot figure out the following problems. (a) If x and y are both divisible by k, prove that gcd(x,y) = k ·gcd(x/k, y/k), where k > 1.

Explore a detailed proof showing that if x and y are divisible by k, then gcd(x, y) = k · gcd(x/k, y/k). This problem covers key number theory concepts, including properties of the greatest common divisor and divisibility rules. Suitable for advanced students in number theory.

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