The least number which when divided by 6, 9, 12, 15, 18 leaves the same remainder 2 in each case is :

Options

A. 180
B. 176
C. 182
D. 178

Answer

To find the least number which, when divided by 6, 9, 12, 15, and 18, leaves the same remainder of 2, we can follow these steps:

### Step 1: Identify the Numbers
We need to work with the numbers 6, 9, 12, 15, and 18.

### Step 2: Find the Least Common Multiple (LCM)
To find the LCM of these numbers, we can use their prime factorization:

- 6 = 2 × 3
- 9 = 3²
- 12 = 2² × 3
- 15 = 3 × 5
- 18 = 2 × 3²

Now, we take the highest power of each prime number that appears in these factorizations:

- For 2: the highest power is 2² (from 12)
- For 3: the highest power is 3² (from 9)
- For 5: the highest power is 5¹ (from 15)

Thus, the LCM is calculated as follows:

\[ \text{LCM} = 2^2 \times 3^2 \times 5^1 = 4 \times 9 \times 5 \]

### Step 3: Calculate the LCM
Now, we calculate the LCM:

\[ 4 \times 9 = 36 \]
\[ 36 \times 5 = 180 \]

So, the LCM of 6, 9, 12, 15, and 18 is 180.

### Step 4: Add the Remainder
Since we need a number that leaves a remainder of 2 when divided by these numbers, we add 2 to the LCM:

\[ \text{Least number} = \text{LCM} + 2 = 180 + 2 = 182 \]

### Conclusion
The least number which when divided by 6, 9, 12, 15, and 18 leaves the same remainder of 2 is **182**.

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