OliForum Matematico

$n=2^{a_2}\cdot3^{a_3}\cdot5^{a_5}\cdots$ (scompongo $n$ in fattori primi)
$f\left( n\right) :=\left( a_2+1\right)\cdot\left( a_3+1\right)\cdot\left( a_5+1\right)\cdots$ (numero di divisori positivi di $n$)
$k_n:=\dfrac{f\left( n\right)}{\left( a_2+1\right)\cdot\left( a_3+1\right)}$ (per comodità mia)
$f\left( 2n\right) =\left( a_2+2\right)\cdot\left( a_3+1\right)\cdot k_n =28$
$f\left( 3n\right) =\left( a_2+1\right)\cdot\left( a_3+2\right)\cdot k_n =30$
$f\left( 6n\right) =\left( a_2+2\right)\cdot\left( a_3+2\right)\cdot k_n$
$f\left( 3n\right) -f\left( 2n\right) = 2 = \left(\left( a_2+1\right)\cdot\left( a_3+2\right) - \left( a_2+2\right)\cdot\left( a_3+1\right)\right)\cdot k_n$
$\left( a_2-a_3\right)\cdot k_n = 2$

Pongo $k_n =1$ e quindi $a_2 = a_3 +2$:
$f\left( 2n\right) =\left( a_3+4\right)\cdot\left( a_3+1\right) =28$
$a_3=3;a_2=5$
$f\left( 3n\right) =\left( a_3+3\right)\cdot\left( a_3+2\right) =6\cdot 5=30$
$f\left( 6n\right) =\left( a_2+2\right)\cdot\left( a_3+2\right) =7\cdot 5 =35$

Pongo $k_n =2$ e quindi $a_2 = a_3 +1$:
$f\left( 2n\right) =2 \left( a_3+3\right)\cdot\left( a_3+1\right) =28$
Non ha soluzioni intere quindi $k_n$ non può essere $2$.

La risposta è quindi $35$.