Find the smallest number which on dividing separately by 12, 20 and 24, leaves 7 as a remainder.

Answer

To find the smallest number that, when divided by 12, 20, and 24, leaves a remainder of 7, we can follow these steps:

### Step 1: Understand the problem
We need to find a number \( N \) such that:
- \( N \mod 12 = 7 \)
- \( N \mod 20 = 7 \)
- \( N \mod 24 = 7 \)

This means that \( N - 7 \) should be divisible by 12, 20, and 24.

### Step 2: Set up the equation
Let \( M = N - 7 \). We need to find the smallest \( M \) such that:
- \( M \mod 12 = 0 \)
- \( M \mod 20 = 0 \)
- \( M \mod 24 = 0 \)

This means \( M \) should be a common multiple of 12, 20, and 24.

### Step 3: Find the Least Common Multiple (LCM)
To find \( M \), we need to calculate the LCM of 12, 20, and 24.

1. **Prime factorization**:
- \( 12 = 2^2 \times 3^1 \)
- \( 20 = 2^2 \times 5^1 \)
- \( 24 = 2^3 \times 3^1 \)

2. **Take the highest power of each prime**:
- For \( 2 \): highest power is \( 2^3 \) (from 24)
- For \( 3 \): highest power is \( 3^1 \) (from 12 and 24)
- For \( 5 \): highest power is \( 5^1 \) (from 20)

3. **Calculate the LCM**:
\[ \text{LCM} = 2^3 \times 3^1 \times 5^1 = 8 \times 3 \times 5 = 120 \]

### Step 4: Calculate \( N \)
Now that we have \( M = 120 \), we can find \( N \):
\[ N = M + 7 = 120 + 7 = 127 \]

### Step 5: Conclusion
The smallest number which, when divided by 12, 20, and 24, leaves a remainder of 7 is \( \boxed{127} \).

---