If each of the three nonzero numbers a, b and c is divisible by 3, then abc must be divisible by which one of the following the number?

Options

A. 8
B. 27
C. 81
D. 121

Answer

To solve the problem, we need to determine the divisibility of the product of three nonzero numbers \( a \), \( b \), and \( c \), given that each of these numbers is divisible by 3.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
We know that \( a \), \( b \), and \( c \) are nonzero numbers and each is divisible by 3. This means:
\[ a = 3x, \quad b = 3y, \quad c = 3z \]
where \( x \), \( y \), and \( z \) are integers.

2. **Calculating the Product \( abc \)**:
We can express the product \( abc \) as follows:
\[ abc = (3x)(3y)(3z) \]

3. **Simplifying the Expression**:
When we multiply these terms, we can factor out the 3s:
\[ abc = 3 \cdot 3 \cdot 3 \cdot xyz = 27xyz \]

4. **Identifying the Divisibility**:
From the expression \( abc = 27xyz \), we can see that \( abc \) is a multiple of 27. Since \( xyz \) is an integer (as \( x \), \( y \), and \( z \) are integers), it follows that \( abc \) is divisible by 27.

5. **Conclusion**:
Therefore, the product \( abc \) must be divisible by 27.

### Final Answer:
The correct answer is **27**.