Factors of 39 | Prime Factorization of 39 - Explained Simply

To find the factors of 39, we look for integers that divide 39 exactly, leaving no remainder. We can test numbers starting from 1 up to the square root of 39, which is approximately $$\sqrt{39} \approx 6.24$$.

1. Testing 1: $$39 \div 1 = 39$$. Factors: \(1\) and \(39\).

2. Testing 2: 39 is an odd number, so it is not divisible by 2.

3. Testing 3: The sum of the digits of 39 is $$3 + 9 = 12$$. Since 12 is divisible by 3, 39 is divisible by 3. $$39 \div 3 = 13$$. Factors: \(3\) and \(13\).

4. Testing 4: $$39 \div 4 = 9$$ with a remainder of 3.

5. Testing 5: 39 does not end in 0 or 5, so it is not divisible by 5.

6. Testing 6: $$39 \div 6 = 6$$ with a remainder of 3.

Since we have reached 6, and the next integer is 7 (which is greater than \(\sqrt{39}\)), and we have already found the pair corresponding to 3 (which is 13), we have found all the factors. We can also perform the prime factorization of 39:

$$39 = 3 \times 13$$

The factors are 1, 3, 13, and 39.

Thus, the factors of 39 are: 1, 3, 13, 39.

Prime Factorization of 39

To find the prime factorization of 39, we must express it as a product of prime numbers.

Step 1: Check for divisibility by the smallest prime, 2.

Since 39 is an odd number, it is not divisible by 2.

Step 2: Check for divisibility by the next smallest prime, 3.

We sum the digits of 39: $$3 + 9 = 12$$. Since 12 is divisible by 3, 39 is divisible by 3.

$$39 \div 3 = 13$$

Step 3: Analyze the quotient.

The quotient is 13. We need to determine if 13 is a prime number. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. 

13 is not divisible by 2, 3, or 5. Since $$5^2 = 25$$ is greater than 13, we only need to check primes up to 5. Therefore, 13 is a prime number.

Step 4: Write the prime factorization.

Since both 3 and 13 are prime numbers, the prime factorization of 39 is:

$$39 = 3 \times 13$$