I'm trying to resolve this problem: "How many trailing zeros does $(2015!)_6$ have?"
I checked this discussion (how many $0$ does $150!$ have when it transform to base $7$?) and I followed the same reasoning. I counted $969$ trailing zeros assuming that whenever a $2 *3$ factor appears, it adds a trailing zero (the same that happens with $2*5$ in decimal system).But other people counted different values.
Can you please check the result? thanks
$2015!$ is divisible by $2^a3^b$ where $$ a=\left\lfloor\frac{2015}2\right\rfloor+\left\lfloor\frac{2015}4\right\rfloor+\left\lfloor\frac{2015}8\right\rfloor+\ldots =1007+503+251+\ldots=2005 $$ and $$ b=\left\lfloor\frac{2015}3\right\rfloor+\left\lfloor\frac{2015}9\right\rfloor+\left\lfloor\frac{2015}{27}\right\rfloor+\ldots =671+223+74+\ldots=1002 $$ and neither $a$ nor $b$ can be increased. It follows that $2015$ is divisible by $(2\cdot3)^{1003}$ and no more.