If x7y5 is exactly divisible by 3, then the least value of (x+y) is

Options

A. 6
B. 0
C. 4
D. 3

Answer

To solve the problem, we need to determine the least value of \(x + y\) such that the number \(x7y5\) is divisible by 3.

### Step-by-Step Solution:

1. **Understand the divisibility rule for 3**: A number is divisible by 3 if the sum of its digits is divisible by 3.

2. **Identify the digits of the number**: The number \(x7y5\) has the digits \(x\), \(7\), \(y\), and \(5\).

3. **Calculate the sum of the digits**:
\[ \text{Sum} = x + 7 + y + 5 = x + y + 12 \]

4. **Set up the condition for divisibility by 3**:
We need \(x + y + 12\) to be divisible by 3. This means:
\[ x + y + 12 \equiv 0 \ (\text{mod} \ 3) \]

5. **Simplify the condition**:
Since \(12\) is divisible by \(3\) (as \(12 \div 3 = 4\)), we can simplify the condition to:
\[ x + y \equiv 0 \ (\text{mod} \ 3) \]

6. **Find the least value of \(x + y\)**:
The smallest non-negative integer that satisfies \(x + y \equiv 0 \ (\text{mod} \ 3)\) is \(0\). However, since \(x\) and \(y\) are digits (0 to 9), the next possible values are \(0\), \(3\), \(6\), \(9\), etc.

7. **Check the least value**:
The least value of \(x + y\) that is non-negative and satisfies the condition is \(3\).

### Conclusion:
Thus, the least value of \(x + y\) such that \(x7y5\) is exactly divisible by 3 is **3**.

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