Question

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

Answer 1

Hint: As the prime factorization of 540 is given. Use the concept of prime factorization to find the divisors of 540. After that subtract 1 from the total number of divisors to find the number of divisors of 540 other than 1.

Formula Used: If a number $n$ has the prime factorization as $n=p^a\times q^b\times r^c$. Then the number of divisors of $n$ is $\left(a+1\right)\left(b+1\right)\left(c+1\right)$.

Complete step-by-step answer:
We know that a divisor of a number $n$ is any number that leaves no remainder when $n$ is divided by it (or completely divides $n$).
We will find the total number of divisors of 540. For this, we need the prime factorization of 540:
As it is given that, the prime factorization of 540 is $2\times 2\times 3\times 3\times 3\times 5$.
So, write the factorization in the exponent form,
$\Rightarrow 540=2^2\times 3^3\times 5^1$
We will calculate the total number of divisors of 540. We will substitute 2 for a, 3 for b, and 1 for c in the formula for the total number of divisors:
$\Rightarrow \left(2+1\right)\left(3+1\right)\left(1+1\right)$
Add the terms in the brackets,
$\Rightarrow 3\times 4\times 2$
Multiply the terms,
$\Rightarrow 24$
The total number of divisors of 540 is 24.
Now subtract 1 from the total number of divisors to find the number of divisors of 540 other than 1.
$\Rightarrow 24-1=23$

Hence, the total number of divisors to find the number of divisors of 540 other than 1 is 23.

Note:
Whenever we face such a problem statement the key concept involved is simply to find all the factors of the given number now through those factors we can directly find the total number of divisors using the above-mentioned concept. The noted point here is the number of divisors obtained will always include 1 and the number itself.