Determine the number of positive integers such that $[\frac{a}{2}] + [\frac{a}{3}] + [\frac{a}{5}] = a$

$$ \frac{a}{2} + \frac{a}{3} + \frac{a}{5}-3<\Big[\frac{a}{2}\Big] + \Big[\frac{a}{3}\Big] + \Big[\frac{a}{5}\Big]=a\leq \frac{a}{2} + \frac{a}{3} + \frac{a}{5}$$

Hint: use $$x-1<[x]\leq x$$ then you get $$ \frac{a}{2} + \frac{a}{3} + \frac{a}{5}-3\leq a\leq \frac{a}{2} + \frac{a}{3} + \frac{a}{5}$$ I get 29 solutions...