Find out the smallest number of five digits which is completely divisible by 12, 15 and 18 respectively.

Options

A. 10010
B. 10015
C. 10020
D. 10080

Answer

To find the smallest five-digit number that is completely divisible by 12, 15, and 18, we will follow these steps:

### Step 1: Find the LCM of 12, 15, and 18
To find the least common multiple (LCM), we first need to find the prime factorization of each number.

- **12**:
- \(12 = 3^1 \times 2^2\)

- **15**:
- \(15 = 3^1 \times 5^1\)

- **18**:
- \(18 = 3^2 \times 2^1\)

Next, we take the highest power of each prime number that appears in the factorizations:

- For \(2\): The highest power is \(2^2\) (from 12).
- For \(3\): The highest power is \(3^2\) (from 18).
- For \(5\): The highest power is \(5^1\) (from 15).

Now, we can calculate the LCM:

\[ \text{LCM} = 2^2 \times 3^2 \times 5^1 = 4 \times 9 \times 5 \]

Calculating this step-by-step:

\[ 4 \times 9 = 36 \]
\[ 36 \times 5 = 180 \]

So, the LCM of 12, 15, and 18 is **180**.

### Step 2: Find the smallest five-digit number
The smallest five-digit number is **10000**. We need to find the smallest number greater than or equal to 10000 that is divisible by 180.

### Step 3: Divide 10000 by 180
We perform the division:

\[ 10000 \div 180 \approx 55.55 \]

### Step 4: Round up to the nearest whole number
Since we need a whole number, we round up \(55.55\) to \(56\).

### Step 5: Multiply by 180
Now we multiply \(56\) by \(180\) to find the smallest five-digit number that is divisible by 180:

\[ 56 \times 180 = 10080 \]

### Conclusion
The smallest five-digit number that is completely divisible by 12, 15, and 18 is **10080**.

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