What is the number of decimal places in a cube root of a number having 27 decimal places ?

Options

A. 3
B. 6
C. 9
D. 27

Answer

To find the number of decimal places in the cube root of a number that has 27 decimal places, we can follow these steps:

### Step 1: Understand the number with decimal places
A number with 27 decimal places can be represented as:
\[ N = x \times 10^{-27} \]
where \( x \) is a number that does not have any decimal places (an integer).

### Step 2: Calculate the cube root
We need to find the cube root of \( N \):
\[ \sqrt[3]{N} = \sqrt[3]{x \times 10^{-27}} \]

### Step 3: Separate the components
Using the property of cube roots, we can separate the components:
\[ \sqrt[3]{N} = \sqrt[3]{x} \times \sqrt[3]{10^{-27}} \]

### Step 4: Simplify the cube root of the power of ten
The cube root of \( 10^{-27} \) can be simplified as follows:
\[ \sqrt[3]{10^{-27}} = 10^{-27/3} = 10^{-9} \]

### Step 5: Combine the results
Thus, we have:
\[ \sqrt[3]{N} = \sqrt[3]{x} \times 10^{-9} \]

### Step 6: Determine the number of decimal places
The term \( 10^{-9} \) indicates that the result will have 9 decimal places. The cube root of \( x \) (if \( x \) is a whole number) will not contribute any additional decimal places, as it will be an integer or a number with fewer decimal places than 9.

### Conclusion
Therefore, the number of decimal places in the cube root of a number with 27 decimal places is **9**.

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