What is the sum of digit of the least number which when divided by 12,15,18,27 leaves the same remainder 9 in each case and is also a multiple of 11 ?

Options

A. 17
B. 19
C. 20
D. 18

Answer

To solve the problem, we need to find the least number that meets the following criteria:

1. When divided by 12, 15, 18, and 27, it leaves a remainder of 9.
2. It is a multiple of 11.

Let's break down the solution step by step.

### Step 1: Find the Least Common Multiple (LCM)
First, we need to find the LCM of the numbers 12, 15, 18, and 27.

- **Prime Factorization**:
- \(12 = 2^2 \times 3^1\)
- \(15 = 3^1 \times 5^1\)
- \(18 = 2^1 \times 3^2\)
- \(27 = 3^3\)

- **LCM Calculation**:
- Take the highest power of each prime factor:
- For \(2\): \(2^2\) (from 12)
- For \(3\): \(3^3\) (from 27)
- For \(5\): \(5^1\) (from 15)

Thus, the LCM is:
\[ LCM = 2^2 \times 3^3 \times 5^1 = 4 \times 27 \times 5 \]
Calculating this step by step:
\[ 4 \times 27 = 108 \]
\[ 108 \times 5 = 540 \]
So, \(LCM(12, 15, 18, 27) = 540\).

### Step 2: Formulate the Number
The number we are looking for can be expressed as:
\[ N = 540k + 9 \]
where \(k\) is a non-negative integer.

### Step 3: Ensure the Number is a Multiple of 11
We need \(N\) to be a multiple of 11:
\[ 540k + 9 \equiv 0 \mod{11} \]
Calculating \(540 \mod 11\):
\[ 540 \div 11 = 49 \quad \text{(remainder 1)} \]
So, \(540 \equiv 1 \mod{11}\). Therefore, we rewrite the equation:
\[ 1k + 9 \equiv 0 \mod{11} \]
This simplifies to:
\[ k + 9 \equiv 0 \mod{11} \implies k \equiv -9 \equiv 2 \mod{11} \]
Thus, \(k\) can be expressed as:
\[ k = 11m + 2 \quad \text{(for some integer } m\text{)} \]

### Step 4: Substitute Back to Find N
Substituting \(k\) back into the equation for \(N\):
\[ N = 540(11m + 2) + 9 = 5940m + 1080 + 9 = 5940m + 1089 \]
The least value occurs when \(m = 0\):
\[ N = 1089 \]

### Step 5: Calculate the Sum of the Digits of N
Now, we find the sum of the digits of \(N = 1089\):
\[ 1 + 0 + 8 + 9 = 18 \]

### Final Answer
The sum of the digits of the least number is **18**.

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