To find the prime factorization of the number 540, we can follow these steps:
### Step 1: Start with the number 540
We begin with the number we want to factor, which is 540.
### Step 2: Divide by the smallest prime number
The smallest prime number is 2. We check if 540 is divisible by 2.
- **Calculation**: 540 ÷ 2 = 270
- Since 540 is even, it is divisible by 2.
### Step 3: Continue dividing by 2
Now we take the quotient, which is 270, and divide it by 2 again.
- **Calculation**: 270 ÷ 2 = 135
- 270 is also even, so we can divide it by 2 once more.
### Step 4: Move to the next prime number
Now we have 135. Since 135 is odd, we cannot divide it by 2. The next prime number is 3.
- **Calculation**: 135 ÷ 3 = 45
- 135 is divisible by 3.
### Step 5: Continue dividing by 3
Next, we take 45 and divide it by 3 again.
- **Calculation**: 45 ÷ 3 = 15
- 45 is also divisible by 3.
### Step 6: Divide by 3 once more
Now we take 15 and divide it by 3 again.
- **Calculation**: 15 ÷ 3 = 5
- 15 is divisible by 3.
### Step 7: Finish with the last prime number
Now we have 5, which is a prime number itself. We can stop here as we cannot divide 5 by any prime number other than itself.
- **Calculation**: 5 ÷ 5 = 1
### Step 8: Write down the prime factors
Now we can write down all the prime factors we used:
- From the divisions, we have: 2, 2, 3, 3, 3, and 5.
### Step 9: Express in exponential form
We can express the prime factorization in exponential form:
- 2 appears twice, so we write it as \(2^2\).
- 3 appears three times, so we write it as \(3^3\).
- 5 appears once, so we write it as \(5^1\).
### Final Result
Thus, the prime factorization of 540 is:
\[ 540 = 2^2 \times 3^3 \times 5^1 \]
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