The least number which when divided by 4,6,8,12 and 16 leaves a remainder of 2 in each case is :

Options

A. 46
B. 48
C. 50
D. 56

Answer

To find the least number that leaves a remainder of 2 when divided by 4, 6, 8, 12, and 16, we can follow these steps:

### Step 1: Understand the Problem
We need to find a number \( N \) such that:
- \( N \mod 4 = 2 \)
- \( N \mod 6 = 2 \)
- \( N \mod 8 = 2 \)
- \( N \mod 12 = 2 \)
- \( N \mod 16 = 2 \)

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

### Step 2: Set Up the Equation
Let \( M = N - 2 \). Then, \( M \) must be divisible by 4, 6, 8, 12, and 16. We need to find the least common multiple (LCM) of these numbers.

### Step 3: Find the LCM
To find the LCM, we can use the prime factorization method:
- \( 4 = 2^2 \)
- \( 6 = 2^1 \times 3^1 \)
- \( 8 = 2^3 \)
- \( 12 = 2^2 \times 3^1 \)
- \( 16 = 2^4 \)

Now, we take the highest power of each prime:
- For \( 2 \): the highest power is \( 2^4 \) (from 16)
- For \( 3 \): the highest power is \( 3^1 \) (from 6 and 12)

Thus, the LCM is:
\[ \text{LCM} = 2^4 \times 3^1 = 16 \times 3 = 48 \]

### Step 4: Calculate the Least Number
Since \( M = 48 \), we can find \( N \):
\[ N = M + 2 = 48 + 2 = 50 \]

### Step 5: Conclusion
The least number which when divided by 4, 6, 8, 12, and 16 leaves a remainder of 2 is **50**.