Find the highest natural number which is divisible by $30$ and has exactly $30$ different positive divisors.

Question

Find the highest natural number which is divisible by $30$ and have exactly $30$ different positive divisors.

What I Tried: I am not sure about any specific approach to this problem. Of course as the number is divisible by $30$ , some of the factors of the number would be $1,2,3,5,6,10,20,30$ , but that only makes $8$ divisors and there will be $12$ more which I have no idea of .

Can anyone help me?

Answer 1

(I'm assuming you know the number theoretic way to count the number of divisors of an integer $ n = 2^a \times 3^b \times 5^c \ldots $, which is $ \sigma_0 (n) = (a+1)(b+1)(c+1) \ldots$. If not, go read up on it.)

Let $n$ be such a number.

Hint: Since $ 30 = 2 \times 3 \times 5 $, show that $ n = 2^a \times 3^b \times 5^c$.
In particular, conclude that $n$ cannot have any other prime factors.

Note that every distinct prime factor of $n$ contributes at least one prime factor (not necessarily distinct) of $ \sigma_0 (n)$.
Since $30 \mid n$, so $n$ has at least 3 prime factors (2, 3, 5), which contribute at least 3 (not necessarily distinct) prime factors to $ \sigma_0 (n) = 30$. Since this has 3 prime factors, we conclude that $n$ can have no other prime factors.
Hence, $n = 2^a \times 3^b \times 5^c$.

Hence, the largest natural number is $ 5^4 \times 3^2 \times 2^1$.

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Generlization: If $N = \prod p_i$ is the product of distinct primes, then the highest natural number which is divisible by $N$ and has exactly $N$ different positive divisors is $\prod p_i ^ { p_i - 1 } $.