Find the smallest number which leave remainder 2, when divided by 6, 9,12, 15 and 18? (a) 180 (b) 182 (c) 178 (d) 176

Options

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

Answer

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

### Step 1: Understand 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 all of these numbers.

### Step 2: Calculate the LCM of the divisors
To find \( N - 2 \), we first need to calculate the Least Common Multiple (LCM) of the numbers 6, 9, 12, 15, and 18.

#### Finding the LCM:
1. **Prime factorization**:
- \( 6 = 2^1 \times 3^1 \)
- \( 9 = 3^2 \)
- \( 12 = 2^2 \times 3^1 \)
- \( 15 = 3^1 \times 5^1 \)
- \( 18 = 2^1 \times 3^2 \)

2. **Take 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)

3. **Calculate the LCM**:
\[ \text{LCM} = 2^2 \times 3^2 \times 5^1 = 4 \times 9 \times 5 = 180 \]

### Step 3: Find the smallest number \( N \)
Now that we have the LCM, we can find \( N \):
\[ N - 2 = 180 \implies N = 180 + 2 = 182 \]

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

### Answer
(b) 182