**Factors of 108** are basically the numbers that divide it evenly or exactly without leaving any remainder i.e if a number divides 108 with a remainder of zero, then that number is called a factor.

**All Factors:** 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54 and 108**Factors in pairs: **(1,108),(2,54),(3,36),(4,27),(6,18),(9,12)**Prime Factors:**2, 3

Now, let us see the prime factors and prime factorization of the number.

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## Prime Factorization of 108

**Prime factorization** is a method of “**expressing**” or **finding** the given number as the product of prime numbers. If a number occurs more than once in prime factorization, it is usually expressed in exponential form to make it more compact.

The **Prime factorization** comes out to be 2 x 2 x 3 x 3 x 3

### Prime Factorization of108 by Upside-Down Division Method

**Upside-Down Division** is one of the techniques used in the Prime factorization method to find factors of numbers.

In this method, you will divide a given “composite” number evenly by the several prime numbers(starting from the smallest) till it gets a prime number.

It is called **Upside-Down Division** because the symbol is flipped upside down.

Here, 108 is an even number. So it is undoubtedly divisible by 2 with no remainder.

Thus we do 108÷ 2 = 54. Now find the prime factors of the obtained quotient.

Repeat Step 1 and Step 2 until we get a result of prime number as the quotient. Here, 54 is the quotient. Now find the prime factor of 54

54÷ 2 = 27

Similarly 27÷ 3 = 9

9÷ 3 = 3. Here, 3 is the prime number.

So we can stop theprocess.

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So, now the prime factorization of 108 with the upside-down division method is2× 2 × 3 × 3 × 3.

### Prime Factorization of108 by Factor Tree Method

The **Factor tree method** is another technique for producing the prime factorization and all factors of a given number.

**To use this method for a number x,**

Firstly consider two factors say **a,b** of x such that a*b is equal to x and at least one of them (a, b) is a prime factor say a.

Then consider two factors of b say c, d such that again at least one of them is a prime factor. This process is repeated until both the factors are prime i.e if we get both the factors as prime at any step, we stop the process there.

Following is the factor tree of the given number.

Here we can get the prime factorisation of 108 as 2 * 2 * 3 * 3 * 3 and the two prime factors are 2, 3

## FAQs

**What is the Sum of the Factors of 108?**

Sum of all factors = (22 + 1 – 1)/(2 – 1) × (33 + 1 – 1)/(3 – 1) = 280

**What are the Prime Factors of 108?**

The prime factors of 108 are 2, 3.

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**What are the Factors of 108?**

The factors are 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 108.