Two numbers a and b are chosen at random from the set {1,2,3,..,3n}. The probability that `a^(3)+b^(3)` is divisible by 3, is

Options

A. `(1)/(2)`
B. `(1)(4)`
C. `(1)/(6)`
D. `(1)/(3)`

Explanation

To solve the problem of finding the probability that \( a^3 + b^3 \) is divisible by 3 when two numbers \( a \) and \( b \) are chosen at random from the set \( \{1, 2, 3, \ldots, 3n\} \), we can follow these steps:

### Step 1: Understand the Problem
We need to find the probability that \( a^3 + b^3 \) is divisible by 3. We know from number theory that \( a^3 + b^3 \) can be expressed as \( (a + b)(a^2 - ab + b^2) \). For \( a^3 + b^3 \) to be divisible by 3, \( a + b \) must be divisible by 3.

### Step 2: Group the Numbers
We can categorize the numbers in the set \( \{1, 2, 3, \ldots, 3n\} \) based on their remainders when divided by 3:
- Group \( G_1 \): Numbers that are \( 0 \mod 3 \) (i.e., divisible by 3): \( \{3, 6, 9, \ldots, 3n\} \)
- Group \( G_2 \): Numbers that are \( 1 \mod 3 \): \( \{1, 4, 7, \ldots, 3n - 2\} \)
- Group \( G_3 \): Numbers that are \( 2 \mod 3 \): \( \{2, 5, 8, \ldots, 3n - 1\} \)

Each group contains \( n \) elements.

### Step 3: Identify Favorable Outcomes
Now, we need to find the combinations of \( a \) and \( b \) such that \( a + b \) is divisible by 3. This can happen in the following cases:
1. Both \( a \) and \( b \) are from \( G_1 \).
2. One number is from \( G_2 \) and the other from \( G_3 \).

#### Case 1: Both from \( G_1 \)
The number of ways to choose 2 numbers from \( G_1 \) (which has \( n \) elements) is given by:
\[ \binom{n}{2} \]

#### Case 2: One from \( G_2 \) and one from \( G_3 \)
The number of ways to choose 1 number from \( G_2 \) and 1 number from \( G_3 \) is:
\[ \binom{n}{1} \times \binom{n}{1} = n \times n = n^2 \]

### Step 4: Total Favorable Outcomes
The total number of favorable outcomes is:
\[ \text{Total Favorable Outcomes} = \binom{n}{2} + n^2 \]

### Step 5: Total Outcomes
The total number of ways to choose 2 numbers from the set of \( 3n \) numbers is:
\[ \binom{3n}{2} \]

### Step 6: Calculate the Probability
The probability \( P \) that \( a^3 + b^3 \) is divisible by 3 is given by:
\[ P = \frac{\text{Total Favorable Outcomes}}{\text{Total Outcomes}} = \frac{\binom{n}{2} + n^2}{\binom{3n}{2}} \]

### Step 7: Simplify the Expression
Calculating the binomial coefficients:
\[ \binom{n}{2} = \frac{n(n-1)}{2}, \quad \binom{3n}{2} = \frac{3n(3n-1)}{2} \]
Substituting these into the probability expression:
\[ P = \frac{\frac{n(n-1)}{2} + n^2}{\frac{3n(3n-1)}{2}} = \frac{n(n-1) + 2n^2}{3n(3n-1)} = \frac{3n^2 - n}{3n(3n-1)} \]

### Step 8: Final Calculation
Simplifying further:
\[ P = \frac{n(3n - 1)}{3n(3n - 1)} = \frac{1}{3} \]

Thus, the final probability that \( a^3 + b^3 \) is divisible by 3 is:
\[ \boxed{\frac{1}{3}} \]