If n is an integer, how many values of n will give an integral value of `(51n^2 + 17n + 6)/n` ?

Options

A. 4
B. 3
C. 2
D. none of these

Answer

To solve the problem, we need to determine how many integer values of \( n \) will make the expression

\[ \frac{51n^2 + 17n + 6}{n} \]

an integer.

### Step 1: Simplify the Expression
We can simplify the expression by dividing each term in the numerator by \( n \):

\[ \frac{51n^2}{n} + \frac{17n}{n} + \frac{6}{n} = 51n + 17 + \frac{6}{n} \]

### Step 2: Identify the Condition for Integrality
For the entire expression to be an integer, the term \( \frac{6}{n} \) must also be an integer. This means that \( n \) must be a divisor of 6.

### Step 3: Find the Divisors of 6
The divisors of 6 are the integers that can divide 6 without leaving a remainder. The divisors of 6 are:

- \( 1 \)
- \( 2 \)
- \( 3 \)
- \( 6 \)
- \( -1 \)
- \( -2 \)
- \( -3 \)
- \( -6 \)

### Step 4: Count the Divisors
Now, we count the total number of divisors:

- Positive divisors: \( 1, 2, 3, 6 \) (4 values)
- Negative divisors: \( -1, -2, -3, -6 \) (4 values)

In total, there are \( 4 + 4 = 8 \) divisors.

### Conclusion
Thus, there are **8 integer values of \( n \)** that will make the expression an integer.

### Final Answer
The answer is **8 values of \( n \)**.

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