The decimal expansion of the rational number `327/(2^(3)xx5)` will terminate after

Options

A. one decimal place
B. Two decimal place
C. Three decimal place
D. More than three decimal place

Answer

To determine how many decimal places the decimal expansion of the rational number \( \frac{327}{2^3 \times 5} \) will terminate after, we can follow these steps:

### Step 1: Identify the denominator
The denominator of the given rational number is \( 2^3 \times 5 \).

### Step 2: Check the form of the denominator
For a decimal expansion to terminate, the denominator (after simplification) must be of the form \( 2^m \times 5^n \), where \( m \) and \( n \) are non-negative integers.

### Step 3: Analyze the denominator
In our case, the denominator is \( 2^3 \times 5^1 \). This is already in the required form, as it consists of powers of 2 and 5.

### Step 4: Determine the maximum power of 10
The maximum power of 10 that can be formed from the denominator is determined by the minimum of the powers of 2 and 5 in the denominator:
- The power of 2 is 3 (from \( 2^3 \)).
- The power of 5 is 1 (from \( 5^1 \)).

### Step 5: Calculate the minimum power
The minimum of these two values is:
\[ \min(3, 1) = 1 \]

### Step 6: Determine the number of decimal places
The number of decimal places in the decimal expansion is given by:
\[ \text{Number of decimal places} = \max(3, 1) = 3 \]

Thus, the decimal expansion of the rational number \( \frac{327}{2^3 \times 5} \) will terminate after **3 decimal places**.

### Summary of the solution:
The decimal expansion of \( \frac{327}{2^3 \times 5} \) will terminate after **3 decimal places**.

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