Find the least number which when divided by 9, 12 and 15, leaves 5 as the remainder in each case.

Options

A. 180
B. 50
C. 185
D. 77

Answer

To find the least number which, when divided by 9, 12, and 15, leaves a remainder of 5 in each case, we can follow these steps:

### Step 1: Understand the Problem
We need to find a number \( x \) such that:
- \( x \mod 9 = 5 \)
- \( x \mod 12 = 5 \)
- \( x \mod 15 = 5 \)

This means that \( x - 5 \) should be divisible by 9, 12, and 15.

### Step 2: Set Up the Equation
Let’s define a new variable:
\[ y = x - 5 \]
Now, we need to find the least \( y \) such that:
- \( y \mod 9 = 0 \)
- \( y \mod 12 = 0 \)
- \( y \mod 15 = 0 \)

This means \( y \) must be a common multiple of 9, 12, and 15.

### Step 3: Find the Least Common Multiple (LCM)
To find the least number \( y \) that is divisible by 9, 12, and 15, we need to calculate the LCM of these numbers.

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

2. **Take the highest power of each prime**:
- For \( 2 \): \( 2^2 \) (from 12)
- For \( 3 \): \( 3^2 \) (from 9)
- For \( 5 \): \( 5^1 \) (from 15)

3. **Calculate the LCM**:
\[ \text{LCM} = 2^2 \times 3^2 \times 5^1 = 4 \times 9 \times 5 \]
\[ = 36 \times 5 = 180 \]

### Step 4: Calculate \( x \)
Now that we have \( y = 180 \), we can find \( x \):
\[ x = y + 5 = 180 + 5 = 185 \]

### Conclusion
The least number which, when divided by 9, 12, and 15, leaves a remainder of 5 is:
\[ \boxed{185} \]