What is the least number which when divided by 4,6,8 and 9 leaves zero remainder in each case but when divided by 13 leaves a remainder of 7?

Answer

To find the least number that meets the given conditions, we can follow these steps:

### Step 1: Find the LCM of 4, 6, 8, and 9
To find the least number that leaves a remainder of 0 when divided by 4, 6, 8, and 9, we first need to calculate the least common multiple (LCM) of these numbers.

- **Prime factorization**:
- 4 = 2²
- 6 = 2 × 3
- 8 = 2³
- 9 = 3²

- **Taking the highest power of each prime**:
- For 2: highest power is 2³ (from 8)
- For 3: highest power is 3² (from 9)

- **Calculating LCM**:
\[ \text{LCM} = 2³ \times 3² = 8 \times 9 = 72 \]

### Step 2: Verify that 72 leaves a remainder of 0 when divided by 4, 6, 8, and 9
- **Dividing 72 by 4**:
\[ 72 \div 4 = 18 \quad (\text{remainder } 0) \]
- **Dividing 72 by 6**:
\[ 72 \div 6 = 12 \quad (\text{remainder } 0) \]
- **Dividing 72 by 8**:
\[ 72 \div 8 = 9 \quad (\text{remainder } 0) \]
- **Dividing 72 by 9**:
\[ 72 \div 9 = 8 \quad (\text{remainder } 0) \]

### Step 3: Check the condition for division by 13
Now, we need to check if 72 leaves a remainder of 7 when divided by 13.

- **Dividing 72 by 13**:
\[ 72 \div 13 = 5 \quad (\text{remainder } 7) \]

### Conclusion
Since 72 meets both conditions (leaving a remainder of 0 when divided by 4, 6, 8, and 9, and leaving a remainder of 7 when divided by 13), the least number we are looking for is **72**.

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