Question

If “p” is completely divided by the number 17, and \( p = x^2 \cdot y \), where x and y are distinct prime numbers, which of these numbers must be divisible by 289?

• A. \( x^2 \)
• B. \( y^2 \)
• C. \( xy \)
• D. \( x^2y^2 \)
• E. \( x^3y \)

Hint:

When evaluating divisibility by a square of a prime number, ensure that both the prime factor and its square are represented in the factors of the number.

Answer 1

The Correct Option is D

Step 1: Understanding the Problem. 
We know that \( p = x^2 \cdot y \), where x and y are distinct prime numbers. For \( p \) to be divisible by 17, at least one of the prime factors (x or y) must be 17. Therefore, either \( x = 17 \) or \( y = 17 \). 
Step 2: Divisibility by 289. 
Since 289 is \( 17^2 \), for \( p \) to be divisible by 289, it must contain at least two factors of 17. This means that both \( x^2 \) and \( y^2 \) must each contain a factor of 17. Therefore, the correct answer is \( x^2y^2 \), which contains \( 17^2 \). 
Final Answer: \[ \boxed{x^2y^2} \]