The smallest positive integer $n$ with $24$ divisors (where $1$ and $n$ are also considered as divisors of $n$) is?
As far as I know it can be solved like this:
prime factors of $24$ are : $(2,2,2,3)$
So, the smallest number will be $a^1 \times b^1 \times c^1 \times d^2$.
where $a=7$, $b=5$, $c=3$ and $d=2$.
I got the exponents of $a$, $b$, $c$, $d$ by subtracting $1$ from the prime factors of $24$.
This way I am getting the answer as $420$ but the answer is $360$.
Now I need help as to if I am not doing the question correctly or I am skipping some bit of information (I think $1$ and $n$ being inclusive in divisors is something that is giving me the wrong answer).
Please help me…..!
Best Answer
There are a lot of numbers having exactly $24$ factors because if $n=p_1^{\alpha_1}p_2^{\alpha2}....p_k^{\alpha_k}$ then the total number of factors of $n$ is equal to $(\alpha_1+1)(\alpha_2+1)....(\alpha_k+1)$
You try with $n=2^a3^b5^c$ and $$(a+1)(b+1)(c+1)=24$$
The number $24=2^3\cdot3$ has $(3+1)(1+1)=8$ factors and some easy calculation gives $$a+1=4\\b+1=3\\c+1=2$$ Thus $$\color{red}{n=2^33^25=360}$$