If the number formed by the last two digits of a three digit Integer is an integral multiple of 6, the original 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 given and derive a conclusion based on them.

### Step-by-Step Solution:

1. **Understanding the Problem**: We have a three-digit integer, and we know that the last two digits of this integer form a number that is an integral multiple of 6. We need to determine what this implies about the original three-digit integer.

2. **General Representation of the Number**: Let's denote the three-digit integer as \( N \). We can express \( N \) in terms of its digits:
\[ N = 100x + 10y + z \]
where \( x \) is the hundreds digit (1 to 9), \( y \) is the tens digit (0 to 9), and \( z \) is the units digit (0 to 9).

3. **Last Two Digits as a Multiple of 6**: The last two digits of \( N \) are represented by \( 10y + z \). According to the problem, this number must be a multiple of 6:
\[ 10y + z \equiv 0 \mod 6 \]

4. **Divisibility by 2**: For a number to be a multiple of 6, it must be divisible by both 2 and 3. Since \( 10y + z \) is a two-digit number, we can analyze its divisibility by 2:
- A number is divisible by 2 if its last digit (in this case, \( z \)) is even. Therefore, \( z \) must be one of the digits: 0, 2, 4, 6, or 8.

5. **Divisibility by 3**: Next, we check the condition for divisibility by 3:
- A number is divisible by 3 if the sum of its digits is divisible by 3. Thus, we need to check if \( 10y + z \) meets this condition.

6. **Conclusion about the Original Integer**: Since \( 10y + z \) is a multiple of 6, it is guaranteed to be divisible by 2. Therefore, the original three-digit integer \( N \) must also be divisible by 2, as the last digit \( z \) being even ensures that \( N \) is even.

7. **Final Answer**: Hence, we conclude that if the last two digits of a three-digit integer form a number that is an integral multiple of 6, then the original integer itself will always be divisible by **2**.