For any integral value of `n , 3^2n + 9n + 5` when divided by 3 will leave the remainder?

Options

A. 1
B. 2
C. 0
D. 5

Answer

To solve the problem of finding the remainder when \( 3^{2n} + 9n + 5 \) is divided by 3, we can follow these steps:

### Step 1: Analyze each term in the expression
The expression we need to evaluate is \( 3^{2n} + 9n + 5 \).

### Step 2: Evaluate \( 3^{2n} \) modulo 3
Since \( 3^{2n} \) is a power of 3, it is divisible by 3. Therefore, when \( 3^{2n} \) is divided by 3, the remainder is:
\[ 3^{2n} \mod 3 = 0 \]

### Step 3: Evaluate \( 9n \) modulo 3
Similarly, \( 9n \) can be rewritten as \( 3 \times 3n \). This means \( 9n \) is also divisible by 3, and thus:
\[ 9n \mod 3 = 0 \]

### Step 4: Evaluate \( 5 \) modulo 3
Now, we need to evaluate the constant term \( 5 \) modulo 3:
\[ 5 \div 3 = 1 \quad \text{(with a remainder of 2)} \]
So,
\[ 5 \mod 3 = 2 \]

### Step 5: Combine the results
Now we can combine the results from the previous steps:
\[ 3^{2n} + 9n + 5 \mod 3 = 0 + 0 + 2 = 2 \]

### Conclusion
Therefore, when \( 3^{2n} + 9n + 5 \) is divided by 3, the remainder is:
\[ \text{Remainder} = 2 \]