If the 6 - digit number 479xyz is exacatly divisible,by 7,11 and 3, then the product of the digit x,y and z will be

Options

A. 1001
B. 794
C. 252
D. 479

Answer

To solve the problem, we need to find the digits \( x, y, z \) in the 6-digit number \( 479xyz \) such that the number is divisible by 7, 11, and 3.

### Step-by-Step Solution:

1. **Understand the divisibility requirements**:
- A number is divisible by 7, 11, and 3 if it is divisible by their least common multiple (LCM).
- The LCM of 7, 11, and 3 is \( 7 \times 11 \times 3 = 231 \).

2. **Formulate the number**:
- The number can be expressed as \( 479000 + 100x + 10y + z \).

3. **Check divisibility by 231**:
- We need to find the smallest \( n \) such that \( 479xyz \) is divisible by 231. We can start by calculating \( 479000 \mod 231 \).
- Calculate \( 479000 \div 231 \) to find the remainder:
\[ 479000 \div 231 \approx 2079.1736 \quad \text{(take the integer part 2079)} \]
\[ 2079 \times 231 = 479049 \]
\[ 479000 - 479049 = -49 \quad \text{(which means we need to add 49 to make it divisible)} \]

4. **Adjust the number**:
- To make \( 479000 + 100x + 10y + z \) divisible by 231, we need:
\[ 100x + 10y + z \equiv 49 \mod 231 \]

5. **Find suitable values for \( x, y, z \)**:
- We can try different combinations of \( x, y, z \) (each ranging from 0 to 9) to find values that satisfy the equation \( 100x + 10y + z = 49 \).
- The simplest way is to set \( x = 0 \), \( y = 4 \), \( z = 9 \) which gives:
\[ 100(0) + 10(4) + 9 = 40 + 9 = 49 \]

6. **Calculate the product**:
- Now we have \( x = 0 \), \( y = 4 \), and \( z = 9 \).
- The product \( x \times y \times z = 0 \times 4 \times 9 = 0 \).

### Final Answer:
The product of the digits \( x, y, z \) is \( 0 \).