Question

The prime factorisation of $540$ is $2\times 2\times 3\times 3\times 3\times 5$. How do you find the number of the divisor of $540$ other than $1$?

Answer 1

Hint: We are given the prime factorisation of $540$ and are asked to find the number of the divisor of $540$. To answer this, we have to learn about prime factorisation and the divisor of any number. Then we will consider an example to learn it better. And then lastly, we will use it in our problem.

Complete step by step solution:
We are given a number along with its prime factorisation. We are asked to find the number of the divisor. To answer this, let us first understand about prime factorisation and the divisor. Prime factorisation means to break the given number into a multiple for the prime multiples, for example, if we have $10$, we can write it as $10=2\times 5$, where $2$ and $5$ are prime. So, prime factorisation of $10=2\times 5$. Now, as far as divisor is concerned, divisors are those numbers which divide the given number. For example, if we look at the divisor of $4$. So, $4$ is divisible by $2,1$ and $4$. So, the divisors of $4$ are $1,2$ and $4$.
Now we will work on our problem. We are given the prime factorisation of $540$ is $2\times 2\times 3\times 3\times 3\times 5$. Here the multiple of $2$ is $2$, multiple of $3$ is $3$ and multiple of $5$ is $1$. So,
$540={{2}^{2}}\times {{3}^{3}}\times {{5}^{1}}$
Now to find the number of divisor, we will use the formula which says that for any number $x$ which is written as product of prime raised to some power as, $x={{p}_{1}}^{{{e}_{1}}}{{p}_{2}}^{{{e}_{2}}}{{p}_{3}}^{{{e}_{{}}}}.....{{p}_{n}}^{{{e}_{n}}}$ , then the number of divisor is given as,
$d\left( x \right)=\left( {{e}_{1}}+1 \right)\left( {{e}_{2}}+2 \right)\left( {{e}_{3}}+3 \right).....\left( {{e}_{n}}+n \right)$
 Now in our problem, we have,
$540={{2}^{2}}\times {{3}^{3}}\times {{5}^{1}}$
So, here,
${{e}_{1}}=2,{{e}_{2}}=3,{{e}_{3}}=1$
So,
$\begin{align}
  & d\left( 540 \right)=\left( 2+1 \right)\left( 3+1 \right)\left( 1+1 \right) \\
 & \Rightarrow d\left( 540 \right)=3\times 4\times 2 \\
 & \Rightarrow d\left( 540 \right)=24 \\
\end{align}$
So, the total number of divisors of $540$ is $24$. If we leave out 1, then we have the number of divisors of $540=24-1=23$.

Note: When the number is larger, we cannot find the number of divisors very easily as it will take time to find it one by one. So, we use the formula. Also, we must remember that 5 is actually 5 raised to the power of 1. It has 1 in the exponential and we must not miss this fact.