**Given that n>1, what is the smallest positive integer n whose positive divisors have a product of n^6?**

\(\text{If $n$ is not a perfect square, the divisors of n are divided into couples.} \\ \text{For example, the divisors of $12$ are $1, 2, 3, 4, 6,$ and $12$.} \\ \text{$d(12)$ is $6$, and $1·2·3·4·6·12=1728=123=1262=12^{\frac{d(n)}{2}}$} \\ \text{In our example, $1$ and $12$ are partners, as are $2$ and $6$, as are $3$ and $4$.} \\ \text{Note that the product of any $2$ partnered numbers is $n$.} \\ \text{Then there are \frac{d(n)}{2} couples. } \\ \text{The product of the elements in any couple is $n$, } \\ \text{so the product of all the divisors of $n$ is $ n^{ \frac{d(n)}{2}}$.}\)

\(\text{The divisors have a product of $n^6$, so $n^6=n^{ \frac{d(n)}{2}}$, or $6 = \frac{d(n)}{2}$ } \\ \text{So we see $\mathbf{d(n) = 12}$. Our number $n$ must have $\mathbf{12}$ divisors! }\)

\(\text{The factorisation of $12$ is ${\color{red}3}\cdot {\color{red}2}\cdot {\color{red}2}$} \\ \text{$d(n) = (\underbrace{{\color{blue}2}+1}_{={\color{red}3}}) (\underbrace{{\color{blue}1}+1}_{={\color{red}2}}) (\underbrace{{\color{blue}1}+1}_{={\color{red}2}}) =12, $} \\ \text{so the factorisation of $ n = p_1^{\color{blue}2}p_2^{ \color{blue}1 }p_3^{ \color{blue}1 } $ } \\ \text{The smallest prime numbers are $p_1=2,~p_2=3$ and $p_3=5$ } \\ \text{So the number $n = 2^{\color{blue}2}3^{ \color{blue}1 }5^{ \color{blue}1 } = \mathbf{60} $ }\)