Find the least number which when divided by 12, 16, 24 and 36 leaves a remainder 7 in each case.

Options

A. `151`
B. `121`
C. `141`
D. `111`

Explanation

To find the least number which when divided by 12, 16, 24, and 36 leaves a remainder of 7 in each case, we can follow these steps:

### Step 1: Understand the Problem
We need to find a number \( x \) such that:
- \( x \mod 12 = 7 \)
- \( x \mod 16 = 7 \)
- \( x \mod 24 = 7 \)
- \( x \mod 36 = 7 \)

This means that \( x - 7 \) must be divisible by each of these numbers.

### Step 2: Set Up the Equation
Let \( y = x - 7 \). Then, we need to find \( y \) such that:
- \( y \mod 12 = 0 \)
- \( y \mod 16 = 0 \)
- \( y \mod 24 = 0 \)
- \( y \mod 36 = 0 \)

This means \( y \) is a common multiple of 12, 16, 24, and 36.

### Step 3: Find the Least Common Multiple (LCM)
To find \( y \), we need to calculate the LCM of the numbers 12, 16, 24, and 36.

1. **Prime Factorization**:
- \( 12 = 2^2 \times 3^1 \)
- \( 16 = 2^4 \)
- \( 24 = 2^3 \times 3^1 \)
- \( 36 = 2^2 \times 3^2 \)

2. **Determine the highest powers of each prime**:
- For \( 2 \): highest power is \( 2^4 \) (from 16)
- For \( 3 \): highest power is \( 3^2 \) (from 36)

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

### Step 4: Calculate the Required Number
Now that we have \( y = 144 \), we can find \( x \):
\[ x = y + 7 = 144 + 7 = 151 \]

### Final Answer
The least number which when divided by 12, 16, 24, and 36 leaves a remainder of 7 is **151**.

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