Two numbers X and Y are chosen at random from the set {1,2,3...,3n}.Find the probability that X^2-Y^2 is divisible by 3?

To find the probability that \(X^2 - Y^2\) is divisible by 3 when choosing two numbers \(X\) and \(Y\) from the set \(\{1, 2, 3, \ldots, 3n\}\), we can start by using the difference of squares formula. This formula states that \(X^2 - Y^2\) can be factored as \((X - Y)(X + Y)\). For \(X^2 - Y^2\) to be divisible by 3, at least one of these factors must be divisible by 3.

Understanding the Remainders

First, let's analyze the possible remainders when numbers are divided by 3. Any integer can yield one of three remainders: 0, 1, or 2. We can categorize the numbers in our set based on these remainders:

• Numbers congruent to 0 modulo 3: \(3, 6, 9, \ldots, 3n\) (total of \(n\) numbers)
• Numbers congruent to 1 modulo 3: \(1, 4, 7, \ldots, 3n-2\) (total of \(n\) numbers)
• Numbers congruent to 2 modulo 3: \(2, 5, 8, \ldots, 3n-1\) (total of \(n\) numbers)

Calculating the Total Outcomes

The total number of ways to choose two numbers \(X\) and \(Y\) from the set is given by:

Total outcomes = (3n) × (3n) = 9n²

Identifying Favorable Outcomes

Next, we need to determine when \(X^2 - Y^2\) is divisible by 3. This occurs under the following conditions:

• Both \(X\) and \(Y\) are congruent to 0 modulo 3.
• Both \(X\) and \(Y\) are congruent to 1 modulo 3.
• Both \(X\) and \(Y\) are congruent to 2 modulo 3.
• One is congruent to 1 and the other to 2 modulo 3.

Calculating Each Case

Now, let’s calculate the number of favorable outcomes for each case:

• **Case 1**: Both \(X\) and \(Y\) are 0 mod 3: - Choices = \(n \times n = n^2\)
• **Case 2**: Both \(X\) and \(Y\) are 1 mod 3: - Choices = \(n \times n = n^2\)
• **Case 3**: Both \(X\) and \(Y\) are 2 mod 3: - Choices = \(n \times n = n^2\)
• **Case 4**: One is 1 mod 3 and the other is 2 mod 3: - Choices = \(n \times n = n^2\)

Adding these cases together gives us:

Favorable outcomes = n² + n² + n² + n² = 4n²

Finding the Probability

The probability \(P\) that \(X^2 - Y^2\) is divisible by 3 is then calculated as:

P = (Favorable outcomes) / (Total outcomes) = 4n² / 9n² = 4/9

Thus, the probability that \(X^2 - Y^2\) is divisible by 3 when choosing two numbers from the set \(\{1, 2, 3, \ldots, 3n\}\) is \(\frac{4}{9}\).