If the number formed by the last two digits of a three digit integer is an integral multiple of 6, the orginal integer itself will always be divisible by

Options

A. 6
B. 3
C. 2
D. 12

Answer

To solve the problem, we need to analyze the conditions under which a three-digit integer is divisible by certain numbers based on the last two digits being an integral multiple of 6.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- Let the three-digit integer be represented as \( xyz \), where \( x \), \( y \), and \( z \) are the digits of the number.
- The last two digits of this integer are \( yz \), which forms a number.

2. **Condition of Divisibility by 6**:
- A number is divisible by 6 if it is divisible by both 2 and 3.
- Therefore, \( yz \) must be divisible by 2 and 3.

3. **Divisibility by 2**:
- For \( yz \) to be divisible by 2, the last digit \( z \) must be even (i.e., \( z = 0, 2, 4, 6, \) or \( 8 \)).

4. **Divisibility by 3**:
- For \( yz \) to be divisible by 3, the sum of the digits \( y + z \) must be divisible by 3.

5. **Analyzing the Original Integer \( xyz \)**:
- The original integer \( xyz \) can be expressed as \( 100x + 10y + z \).
- We need to determine if this integer is divisible by 6, 3, 2, or 12 based on the conditions we established.

6. **Divisibility by 3**:
- Since \( yz \) is divisible by 3, we know that \( y + z \) is divisible by 3.
- Therefore, \( xyz \) will also be divisible by 3 because \( 100x \) (where \( x \) is a digit) does not affect the divisibility by 3.

7. **Divisibility by 2**:
- Since \( z \) is even, \( xyz \) will also be divisible by 2.

8. **Conclusion**:
- Since \( xyz \) is divisible by both 2 and 3, it follows that \( xyz \) is divisible by \( 6 \).
- Therefore, if the last two digits of a three-digit integer form an integral multiple of 6, the original integer itself will always be divisible by \( 6 \).

### Final Answer:
The original integer will always be divisible by **6**.