`(x-1)(x-2)(x-3)=6y`
How many integer solutions exist for the given equation?

Options

A. 0
B. 1
C. 2
D. More than 2

Answer

To solve the equation \((x-1)(x-2)(x-3) = 6y\) for integer solutions, we will analyze the left-hand side and find integer values for \(x\) and \(y\).

### Step 1: Rewrite the equation
We start with the equation:
\[ (x-1)(x-2)(x-3) = 6y \]
This means that the product of the three consecutive integers \((x-1)\), \((x-2)\), and \((x-3)\) must be a multiple of 6.

### Step 2: Analyze the product
The product of three consecutive integers is always divisible by 6 because:
- Among any three consecutive integers, at least one is even (ensuring divisibility by 2).
- Among any three consecutive integers, at least one is divisible by 3.

Thus, the left-hand side \((x-1)(x-2)(x-3)\) is divisible by 6 for any integer \(x\).

### Step 3: Set up the equation for \(y\)
Since \((x-1)(x-2)(x-3)\) is divisible by 6, we can express \(y\) as:
\[ y = \frac{(x-1)(x-2)(x-3)}{6} \]
We need to find integer values for \(y\), which means \((x-1)(x-2)(x-3)\) must be divisible by 6.

### Step 4: Find integer solutions
To find integer solutions, we can calculate \((x-1)(x-2)(x-3)\) for various integer values of \(x\):

1. **For \(x = 1\)**:
\[ (1-1)(1-2)(1-3) = 0 \quad \Rightarrow \quad y = 0 \]

2. **For \(x = 2\)**:
\[ (2-1)(2-2)(2-3) = 0 \quad \Rightarrow \quad y = 0 \]

3. **For \(x = 3\)**:
\[ (3-1)(3-2)(3-3) = 0 \quad \Rightarrow \quad y = 0 \]

4. **For \(x = 4\)**:
\[ (4-1)(4-2)(4-3) = 3 \quad \Rightarrow \quad y = \frac{3}{6} = \frac{1}{2} \quad \text{(not an integer)} \]

5. **For \(x = 5\)**:
\[ (5-1)(5-2)(5-3) = 8 \quad \Rightarrow \quad y = \frac{8}{6} = \frac{4}{3} \quad \text{(not an integer)} \]

6. **For \(x = 6\)**:
\[ (6-1)(6-2)(6-3) = 15 \quad \Rightarrow \quad y = \frac{15}{6} = \frac{5}{2} \quad \text{(not an integer)} \]

7. **For \(x = 7\)**:
\[ (7-1)(7-2)(7-3) = 24 \quad \Rightarrow \quad y = \frac{24}{6} = 4 \quad \text{(integer)} \]

8. **For \(x = 8\)**:
\[ (8-1)(8-2)(8-3) = 42 \quad \Rightarrow \quad y = \frac{42}{6} = 7 \quad \text{(integer)} \]

9. **For \(x = 9\)**:
\[ (9-1)(9-2)(9-3) = 56 \quad \Rightarrow \quad y = \frac{56}{6} = \frac{28}{3} \quad \text{(not an integer)} \]

10. **For \(x = 10\)**:
\[ (10-1)(10-2)(10-3) = 72 \quad \Rightarrow \quad y = \frac{72}{6} = 12 \quad \text{(integer)} \]

Continuing this process, we can find that there are integer solutions for various values of \(x\).

### Conclusion
The integer solutions for \(y\) occur at \(x = 1, 2, 3, 7, 8, 10\) and potentially more as we continue checking higher values of \(x\).

### Final Count of Integer Solutions
We can summarize that there are multiple integer solutions for the equation, specifically more than 2.