Least number which when divided by 12,18,16,36 leaving remainder 5 in each case is:

Options

A. 60
B. 149
C. 139
D. 180

Explanation

To find the least number which when divided by 12, 18, 16, and 36 leaves a remainder of 5 in each case, we can follow these steps:

### Step 1: Understand the Problem
We need to find a number \( N \) such that:
- \( N \mod 12 = 5 \)
- \( N \mod 18 = 5 \)
- \( N \mod 16 = 5 \)
- \( N \mod 36 = 5 \)

This means that if we subtract 5 from \( N \), the resulting number must be divisible by 12, 18, 16, and 36.

### Step 2: Set Up the Equation
Let \( M = N - 5 \). Then we need to find \( M \) such that:
- \( M \mod 12 = 0 \)
- \( M \mod 18 = 0 \)
- \( M \mod 16 = 0 \)
- \( M \mod 36 = 0 \)

This means \( M \) must be a common multiple of 12, 18, 16, and 36.

### Step 3: Calculate the LCM
To find \( M \), we need to calculate the Least Common Multiple (LCM) of the numbers 12, 18, 16, and 36.

1. **Prime Factorization**:
- \( 12 = 2^2 \times 3^1 \)
- \( 18 = 2^1 \times 3^2 \)
- \( 16 = 2^4 \)
- \( 36 = 2^2 \times 3^2 \)

2. **Determine the Highest Powers**:
- For \( 2 \): The highest power is \( 2^4 \) (from 16).
- For \( 3 \): The highest power is \( 3^2 \) (from 18 and 36).

3. **Calculate the LCM**:
\[ \text{LCM} = 2^4 \times 3^2 = 16 \times 9 = 144 \]

### Step 4: Find the Least Number \( N \)
Since \( M = 144 \), we can find \( N \):
\[ N = M + 5 = 144 + 5 = 149 \]

### Final Answer
The least number which when divided by 12, 18, 16, and 36 leaves a remainder of 5 is **149**.

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