Find all numbers that have 30 factors and have 30 as one of their factors.

Question

Find all numbers that have 30 factors and have 30 as one of their factors. Thank you. Note: please show way if possible.

Answer 1

If $$ N= p_1^{\alpha_1}\cdot\ldots\cdot p_k^{\alpha_k}$$ then the number of its divisors is given by $(\alpha_1+1)\cdot\ldots\cdot(\alpha_k+1)$. So, in order that $30\mid N$ and $ d(N)=30$ we must have $\{2,3,5\}\subseteq\{p_1,\ldots,p_k\}$ and $(\alpha_1+1)\cdot\ldots\cdot(\alpha_k+1)=30.$ Now it is straightforward to check that the choice $k=3,p_1=1,p_2=2,p_3=5$ works as soon as $\{\alpha_1,\alpha_2,\alpha_3\}=\{1,2,4\}$, so: $$ N = 2^4\cdot 3^2\cdot 5 = 720 $$ is a solution. I leave to you to prove that all the solutions are made that way, since there is a unique way to write $30$ as the product of at least three positive integers.

Answer 2

Let $n \in \mathbf N$. Writing the primefactor decomposition as $$ n = \prod_p p^{\alpha_p(n)}, $$ we know that $n$ has $$ \tau(n) = \prod_p \bigl(\alpha_p(n)+ 1\bigr) $$ divisors. (For any $p$ in a divisor we can choose $p$ appearing form $0$ to $\alpha_p(n)$ times), we have as $30 \mid n$, that $\alpha_p(n) \ge 1$, for $p \in \{2,3,5\}$. On the other hand, we must have that $2$, $3$ and $5$ appear in some $\alpha_p(n)+1$, as three of them are greater or equal to two, these must equal $2$, $3$ and $5$. So we are left with six possibilities $$ 2^1 3^2 5^4, 2^1 3^4 5^2, 2^2 3^1 5^4, 2^2 3^4 5^1, 2^4 3^1 5^2, 2^4 3^2 5^1 $$

Answer 3

We write the number as $2^{1+a}3^{1+b}{5^{1+c}}n$ with $(n,30)=1$.

The number of divisors of the number is $30=(2+a)(2+b)(2+c)\tau(n)$. Hence $\tau(n)=1\implies n=1$ ($\tau(n)$ is the number of divisors of $n$).

So $2+a,2+b,2+c$ are $2,3,5$ in some order.

There are $6$ solutions:

$2\cdot3^2\cdot5^4=11250$

$2\cdot3^4\cdot5^2=4050$

$2^2\cdot3\cdot5^4=7500$

$2^2\cdot3^4\cdot5=1620$

$2^4\cdot3\cdot5^2=1200$

$2^4\cdot3^2\cdot5=720$