Find a $\tau(6n)$, such that $\tau (2n)=28$ and $\tau (3n)=30$.
Since $n\in \mathbb N$ $n=p_1^{\alpha_1} \cdot p_2^{\alpha_2} \cdot p_3^{\alpha_3}\cdots p_n^{\alpha_n}$, then $2n=p_1^{\alpha_1+1} \cdot p_2^{\alpha_2} \cdot p_3^{\alpha_3}\cdots p_n^{\alpha_n}, p_1=2$, $3n=p_1^{\alpha_1} \cdot p_2^{\alpha_2+1} \cdot p_3^{\alpha_3}\cdots p_n^{\alpha_n}, p_2=3$,
If $gcd(n,3)=1$ and $gcd(n,2)=1$ then $\tau (2n)=\tau (2) \tau (n)=28$ and $\tau (3n)=\tau (3)\tau (n)=30$ then $\tau(n)=14\not =15= \tau (n)$
then $\tau (2n)=(2+\alpha_1)(1+\alpha_2)\ldots(1+\alpha_n)=28=2\cdot 2 \cdot 7=2\cdot 14$ and $\tau (3n)=(1+\alpha_1)(2+\alpha_2)\ldots(1+\alpha_n)=30=2\cdot 2 \cdot 3=5\cdot 6$, since $2+\alpha_1 >2$ and $2+\alpha_2>2$ then, from first equation $2+\alpha_1=7$ or $2+\alpha_1=14$, so $\alpha_1=5$ or $\alpha_1=12$ because of other equation we need $\alpha_1=5$ and then $6(2+\alpha_2)=30$, so $\alpha_2=3$.
Then $\tau(6n)=(2+\alpha_1)(2+\alpha_2)=(2+5)(2+3)=35$
Is this ok?