Question

What are the prime factors of the number \[216?\]
(1) $2 \times 2 \times 3 \times 3 \times 3$
(2) $2 \times 2 \times 2 \times 3 \times 3 \times 3$
(3) $2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3$
(4) None of these

Answer 1

Hint: Prime factorization is a process of finding prime numbers, which are multiplied together to get the original number. In easy words, the method of finding prime factors of a number is called prime factorization. For example the prime factor of $4$ is $2 \times 2$ . Similarly the prime factors of the number $16$ is $2 \times 2 \times 2 \times 2$ . Again the prime factors of $15$ are $3 \times 5$ .

Complete step-by-step solution:
First let us break the given number $216$ into smaller factors. For this sake, we first divide $216$ by $2$ and we get $108$ . Hence $216$can be expressed as $216 = 2 \times 108$ . Now we repeat the same process i.e. we break $108$ into smaller factors and we get $108 = 2 \times 54$ . Therefore we have $216 = 2 \times 108$ and then $216 = 2 \times 2 \times 54$ .
Continuing , $216 = 2 \times 2 \times 2 \times 27$
i.e. $216 = 2 \times 2 \times 2 \times 3 \times 9$
i.e. $216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3$.
Here all the factors of $216$ are primes and therefore this is the required solution.
Hence option $(3)$ is correct.

Note: Note that prime factors are factors of a number that are, themselves, prime numbers. In this problem, all the factors of $216$ are relatively prime. We also see that $216$ is a composite number as it has factors other than $1$ and $216$ itself namely as $2$ and $3$ . Another interesting fact is that the given number $216$ is also expressible as a perfect cube because $216 = {6^3}$.