The least number which wehn divided by 4, 6, 8 and 9 leaves zero remainder in each case and when divided by 13 leaves a remainder of 7 is :

Options

A. 144
B. 72
C. 36
D. 85

Answer

To solve the problem, we need to find the least number that meets two conditions:

1. It leaves a remainder of 0 when divided by 4, 6, 8, and 9.
2. It leaves a remainder of 7 when divided by 13.

### 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 these numbers, we need to calculate the Least Common Multiple (LCM).

- **Prime Factorization**:
- 4 = 2²
- 6 = 2¹ × 3¹
- 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 the LCM**:
\[ \text{LCM} = 2^3 \times 3^2 = 8 \times 9 = 72 \]

### Step 2: Check the second condition

Now we have the LCM, which is 72. We need to check if this number satisfies the second condition, which is leaving a remainder of 7 when divided by 13.

- **Dividing 72 by 13**:
\[ 72 \div 13 = 5 \quad \text{(quotient)} \]
\[ 72 - (13 \times 5) = 72 - 65 = 7 \quad \text{(remainder)} \]

Since the remainder is indeed 7, the number 72 satisfies both conditions.

### Final Answer

The least number which when divided by 4, 6, 8, and 9 leaves a remainder of 0 and when divided by 13 leaves a remainder of 7 is **72**.

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