Find the smallest number which when divided by 28 and 32 leaves remainders 8 and 12 respectively.

Explanation

### Step-by-Step Solution:

To find the smallest number which when divided by 28 and 32 leaves remainders 8 and 12 respectively, follow these steps:

1. **Understand the Problem:**
- We need a number \( N \) such that:
- \( N \equiv 8 \pmod{28} \)
- \( N \equiv 12 \pmod{32} \)

2. **Formulate Equations:**
- From the given conditions, we can write:
- \( N = 28k + 8 \) for some integer \( k \)
- \( N = 32m + 12 \) for some integer \( m \)

3. **Equate the Two Expressions:**
- Since both expressions represent \( N \), we equate them:
- \( 28k + 8 = 32m + 12 \)

4. **Simplify the Equation:**
- Rearrange the equation to isolate \( k \) and \( m \):
- \( 28k - 32m = 4 \)
- Divide the entire equation by 4 to simplify:
- \( 7k - 8m = 1 \)

5. **Solve the Diophantine Equation:**
- We need to find integer solutions for \( k \) and \( m \) that satisfy \( 7k - 8m = 1 \).
- Using the Extended Euclidean Algorithm, we find particular solutions:
- \( k = 1 \) and \( m = 1 \) is a particular solution because \( 7(1) - 8(1) = 7 - 8 = -1 \).
- To adjust for the correct sign, we can use \( k = 1 \) and \( m = 0 \) as a starting point.

6. **General Solution:**
- The general solution for \( k \) and \( m \) can be written as:
- \( k = 1 + 8t \)
- \( m = 1 + 7t \)
- where \( t \) is an integer.

7. **Find the Smallest Positive \( N \):**
- Substitute \( k = 1 \) into \( N = 28k + 8 \):
- \( N = 28(1) + 8 = 36 \)
- Check if \( N = 36 \) satisfies both conditions:
- \( 36 \div 28 = 1 \) remainder \( 8 \)
- \( 36 \div 32 = 1 \) remainder \( 4 \)
- Since \( N = 36 \) does not satisfy the second condition, we need to find the correct solution by adjusting \( t \).

8. **Adjust \( t \) to Find the Correct \( N \):**
- Substitute \( k = 1 + 8t \) into \( N = 28k + 8 \):
- \( N = 28(1 + 8t) + 8 = 28 + 224t + 8 = 36 + 224t \)
- We need \( 36 + 224t \equiv 12 \pmod{32} \):
- \( 36 \equiv 4 \pmod{32} \)
- \( 4 + 224t \equiv 12 \pmod{32} \)
- \( 224t \equiv 8 \pmod{32} \)
- \( 224 \equiv 0 \pmod{32} \)
- Thus, \( t \) must be adjusted to satisfy the condition.

9. **Find the Correct \( t \):**
- Since \( 224 \equiv 0 \pmod{32} \), we need \( t \) such that:
- \( 4 + 224t \equiv 12 \pmod{32} \)
- \( 224t \equiv 8 \pmod{32} \)
- \( t = 1 \) works because \( 224 \times 1 \equiv 0 \pmod{32} \)
- Therefore, \( t = 1 \) gives us:
- \( N = 36 + 224 \times 1 = 260 \)

10. **Verify the Solution:**
- Check \( 260 \div 28 \):
- \( 260 = 28 \times 9 + 8 \)
- Check \( 260 \div 32 \):
- \( 260 = 32 \times 8 + 12 \)

Thus, the smallest number \( N \) which when divided by 28 and 32 leaves remainders 8 and 12 respectively is \( \boxed{204} \).