The product of the digits of a three digit number which is perfect square as well as a perfect cube is :

Options

A. 126
B. 256
C. 18
D. none of these

Answer

To solve the problem of finding the product of the digits of a three-digit number that is both a perfect square and a perfect cube, we can follow these steps:

### Step 1: Understand the requirements
A number that is both a perfect square and a perfect cube is a perfect sixth power. This means we need to find three-digit numbers of the form \( n^6 \).

### Step 2: Identify the range of \( n \)
We need to find the values of \( n \) such that \( n^6 \) is a three-digit number. The smallest three-digit number is 100, and the largest is 999.

### Step 3: Calculate the sixth roots
- The sixth root of 100 is approximately \( 2.15 \) (since \( 2^6 = 64 \) and \( 3^6 = 729 \)).
- The sixth root of 999 is approximately \( 3.16 \).

Thus, \( n \) can take values of 3 or less, specifically \( n = 3 \) (since \( n \) must be an integer).

### Step 4: Calculate \( n^6 \) for integer values of \( n \)
- For \( n = 3 \):
\[ 3^6 = 729 \]

### Step 5: Find the digits of the number
The digits of 729 are 7, 2, and 9.

### Step 6: Calculate the product of the digits
Now, we calculate the product of the digits:
\[ 7 \times 2 \times 9 = 126 \]

### Conclusion
The product of the digits of the three-digit number (729) that is both a perfect square and a perfect cube is **126**.

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