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

Options

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

Answer

To solve the problem of finding the last number which, when divided by 6, 9, 12, 15, and 18, leaves the same remainder of 2 in each case, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**: We need to find a number \( N \) such that:
- \( N \mod 6 = 2 \)
- \( N \mod 9 = 2 \)
- \( N \mod 12 = 2 \)
- \( N \mod 15 = 2 \)
- \( N \mod 18 = 2 \)

This means that \( N - 2 \) must be divisible by each of these numbers.

2. **Formulating the Equation**: Let \( M = N - 2 \). Then, we need to find \( M \) such that:
- \( M \mod 6 = 0 \)
- \( M \mod 9 = 0 \)
- \( M \mod 12 = 0 \)
- \( M \mod 15 = 0 \)
- \( M \mod 18 = 0 \)

This implies that \( M \) is a common multiple of 6, 9, 12, 15, and 18.

3. **Finding the Least Common Multiple (LCM)**: We need to calculate the LCM of the numbers 6, 9, 12, 15, and 18.

- **Prime Factorization**:
- \( 6 = 2^1 \times 3^1 \)
- \( 9 = 3^2 \)
- \( 12 = 2^2 \times 3^1 \)
- \( 15 = 3^1 \times 5^1 \)
- \( 18 = 2^1 \times 3^2 \)

- **Taking the highest power of each prime**:
- For \( 2 \): highest power is \( 2^2 \) (from 12)
- For \( 3 \): highest power is \( 3^2 \) (from 9 and 18)
- For \( 5 \): highest power is \( 5^1 \) (from 15)

- Therefore, the LCM is:
\[ \text{LCM} = 2^2 \times 3^2 \times 5^1 = 4 \times 9 \times 5 \]

4. **Calculating the LCM**:
- First calculate \( 4 \times 9 = 36 \)
- Then calculate \( 36 \times 5 = 180 \)

So, \( \text{LCM}(6, 9, 12, 15, 18) = 180 \).

5. **Finding the Required Number**: Since \( M = 180 \), we can find \( N \):
\[ N = M + 2 = 180 + 2 = 182 \]

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

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