The smallest number which when divided by `28` and `32 ` leaves remainders `8` and `12` respectively

Answer

To solve the problem of finding the smallest number which when divided by 28 and 32 leaves remainders of 8 and 12 respectively, we can follow these steps:

### Step-by-Step Solution:

1. **Set Up the Equations**:
We need to express the conditions given in the problem in terms of equations.
- Let the smallest number be \( x \).
- From the problem, we have:
- \( x \equiv 8 \, (\text{mod} \, 28) \)
- \( x \equiv 12 \, (\text{mod} \, 32) \)

2. **Rewrite the Equations**:
We can rewrite these congruences in a more useful form:
- From \( x \equiv 8 \, (\text{mod} \, 28) \), we can express \( x \) as:
\[ x = 28k + 8 \quad \text{for some integer } k \]
- From \( x \equiv 12 \, (\text{mod} \, 32) \), we can express \( x \) as:
\[ x = 32m + 12 \quad \text{for some integer } m \]

3. **Equate the Two Expressions**:
Now we have two expressions for \( x \):
\[ 28k + 8 = 32m + 12 \]
Rearranging gives:
\[ 28k - 32m = 4 \]

4. **Simplify the Equation**:
We can simplify this equation by dividing everything by 4:
\[ 7k - 8m = 1 \]

5. **Find Integer Solutions**:
We need to find integer solutions for \( k \) and \( m \). We can try different integer values for \( m \) to find a corresponding \( k \):
- If \( m = 1 \):
\[ 7k - 8(1) = 1 \implies 7k = 9 \implies k = \frac{9}{7} \quad (\text{not an integer}) \]
- If \( m = 2 \):
\[ 7k - 8(2) = 1 \implies 7k = 17 \implies k = \frac{17}{7} \quad (\text{not an integer}) \]
- If \( m = 3 \):
\[ 7k - 8(3) = 1 \implies 7k = 25 \implies k = \frac{25}{7} \quad (\text{not an integer}) \]
- If \( m = 4 \):
\[ 7k - 8(4) = 1 \implies 7k = 33 \implies k = \frac{33}{7} \quad (\text{not an integer}) \]
- If \( m = 5 \):
\[ 7k - 8(5) = 1 \implies 7k = 41 \implies k = 5 \quad (\text{integer solution}) \]

6. **Substitute Back to Find \( x \)**:
Now we have \( k = 5 \) and \( m = 5 \). We can substitute \( k \) back into the equation for \( x \):
\[ x = 28(5) + 8 = 140 + 8 = 148 \]
However, we need to check if \( m = 5 \) gives a valid \( x \) as well:
\[ x = 32(5) + 12 = 160 + 12 = 172 \quad (\text{not valid}) \]

7. **Find the LCM**:
Since we need the smallest number, we can find the least common multiple (LCM) of 28 and 32:
- Prime factorization:
- \( 28 = 2^2 \times 7 \)
- \( 32 = 2^5 \)
- LCM is:
\[ LCM = 2^5 \times 7 = 224 \]

8. **Adjust for Remainders**:
Now, we adjust for the remainders:
\[ x = LCM - \text{(sum of remainders)} = 224 - (8 + 12) = 224 - 20 = 204 \]

### Final Answer:
The smallest number which when divided by 28 and 32 leaves remainders of 8 and 12 respectively is **204**.