Find the smallest integer n > 1 such that the product of the factors of n is $n^{15}$

Question

Find the smallest integer n > 1 such that the product of the factors of n is $n^{15} $

Answer 1

Suppose there is a set of divisors of $n$ say $D_n = {d_1, d_2, d_3, ..., d_k}$.

Note that for each divisor $d \in D_n$, $\frac{n}{d} \in D_n$.

Hence $$\Pi_{d_i\in D_n}d_i= n^{k/2}$$

Thus you need to find $n$ which has $k=30$ factors. (Note that $k=31$ could also do the job but as @OscarLanzi says, we are looking for the smallest $n$, so we use the smaller value)

If $n = p_1^{k_1}p_2^{k_2}...$ where $p_i$ is a prime, the number of divisors is $k = \Pi(k_1+1)(k_2+1)\dots$

Hence, $30 = 30 \cdot 1 = 15 \cdot 2 = 10 \cdot 3 = 6 \cdot 5 = 5 \cdot 3 \cdot 2$ can be represented as a set of powers of primes. Since we need the least, we can consider them to be powers of the lowest primes $2, 3, 5$

Among the following, the least is: Some possible values of $n$ will hence be:

• $2^{29}$ corresponding to $k = 30 \cdot 1$
• $2^{14} \cdot 3^1$ corresponding to $k = 15 \cdot 2$
• $2^5 \cdot 3^4$ corresponding to $k = 6 \cdot 5$

The least of these seems to be $n = 720 = 2^4 \cdot 3^2 \cdot 5^1$ corresponding to $30 = 5 * 3 * 2$