Question

How many positive divisors does 120 have?

Answer 1

Hint: We first try to find the prime factorisation of 120. This gives the factors to form the divisors of 120 which is created by those factors. We use the concept of combinations for those factors to either take a prime factor or not take it. Finally, we find the number of divisors in the form of $(x+1)(y+1)(z+1)$ for the prime factorisation $a^x\times b^y\times c^z$.

Complete step by step solution:
We have to find the number of positive divisors of 120.
We first find the prime factorisation of 120 and then use the combinational approach to find the number of divisors 120 can have.
So, the prime factorisation of 120 is
$\begin{array}{rl}& 2|\underset{―}{120}\\ & 2|\underset{―}{60}\\ & 2|\underset{―}{30}\\ & 3|\underset{―}{15}\\ & 5|\underset{―}{5}\\ & 1|\underset{―}{1}\end{array}$
So, $120=2^3\times 3\times 5$.
Now to form a divisor we have to take the factors of the divisor from the slot of $2^3,3,5$.
The choices for every prime number are that it can be taken and not taken. Particularly for prime 2, the choices will be 4 types.
So, if the prime factorisation of any number is in the form of $a^x\times b^y\times c^z$, then the number of combinations for divisors will be $(x+1)(y+1)(z+1)$.
For finding the divisors of 120, we have $(3+1)(1+1)(1+1)=4\times 2\times 2=16$ number of divisors.
Therefore, 120 have 16 numbers of divisors.

Note: We need to remember that the divisors 1 and 120 is included in the calculation of $(3+1)(1+1)(1+1)=4\times 2\times 2=16$. We have 1 when we are not taking any primes and we have 120 when we are taking all the primes with their highest indices number. In special cases we can omit it if it’s been asked to find non-trivial divisors.