Answer:
Sure, let's go through finding the Least Common Multiple (LCM) of 330, 495, and 297 step by step:
**Step 1: Prime Factorization**
To find the LCM, we start by finding the prime factorization of each number:
- \( 330 \):
\[
330 = 2 \times 3 \times 5 \times 11
\]
- \( 495 \):
\[
495 = 3^2 \times 5 \times 11
\]
- \( 297 \):
\[
297 = 3^3 \times 11
\]
**Step 2: Identify highest powers of all prime factors**
Now, we look at the highest powers of each prime factor present in any of the numbers:
- For \( 2 \): Highest power is \( 2^0 \) (since it appears in 330 but not in the others).
- For \( 3 \): Highest power is \( 3^3 \) (from 297).
- For \( 5 \): Highest power is \( 5^1 \) (common in 330 and 495).
- For \( 11 \): Highest power is \( 11^1 \) (present in all three numbers).
**Step 3: Calculate the LCM**
To find the LCM, multiply together these highest powers:
\[
\text{LCM} = 2^0 \times 3^3 \times 5^1 \times 11^1
\]
Calculate each part:
- \( 2^0 = 1 \) (since any number to the power of 0 is 1).
- \( 3^3 = 27 \)
- \( 5^1 = 5 \)
- \( 11^1 = 11 \)
Now, multiply these together:
\[
\text{LCM} = 1 \times 27 \times 5 \times 11
\]
Perform the multiplications step by step:
- \( 27 \times 5 = 135 \)
- \( 135 \times 11 = 1485 \)
Therefore, the Least Common Multiple (LCM) of 330, 495, and 297 is \( \boxed{1485} \). This means 1485 is the smallest number that is divisible by each of 330, 495, and 297 without leaving a remainder.
