AustriaMO2011

Let the prime factor decomposition of $x$ be given by $2^a{3}^b{5}^c{p}_4^{e_4}\dots p_r^{e_r}$ (with $a,b,c\ge 0$). We conclude that $a$ is a multiple of $15$ and is odd, $b$ is a multiple of $10$ and $b+1$ is a multiple of $3$, * $c$ is a multiple of $6$ and $c+1$ is a multiple of $5$.

The smallest positive integers with these properties are $a=15$, $b=20$, $c=24$. All other exponents in the prime factor decomposition of $x$ must be multiples of $30$, thus we obtain the smallest value for $x$ when all other exponents vanish.

Therefore, the smallest solution is given by $x=2^{15}\cdot 3^{20}\cdot 5^{24}$.

qed