WakeUp wrote: Find the greatest positive integer $x$ such that $23^{6+x}$ divides $2000!$ $v_{23}(2000!)=\sum_{k=1}^{+\infty}\left\lfloor\frac{2000}{23^k}\right\rfloor$ $=\left\lfloor\frac{2000}{23}\right\rfloor$ $+\left\lfloor\frac{2000}{23^2}\right\rfloor$ $=86+3=89$ So $\boxed{x=83}$

I don't understand what the point of $23^{6+x}$. Why not just $23^x$ or something? Is there a typo?

Hmmm why exactly is this in an Olympiad Shortlist

judgefan99 wrote: OlympusHero wrote: Hmmm why exactly is this in an Olympiad Shortlist

Hmmm why exactly are you flexing Are you kidding me? I made a comment. Anyone experienced with Olympiad would agree this is too easy to be in an Olympiad Shortlist. Just look at even the A1, G1, N1, C1 of the Shortlist from recent Olympiad and compare.

it's the junior balkian mathmatical olympiad so it's not supposed to be terribly difficult

Yes, I know, but if you compare this to the recent problems from the Junior Balkan Math Olympiad, the difference is large!

Probably why it got shortlisted.

The amount of multiples of 23 that are in the product $2000!$ are $$\left\lfloor\frac{2000}{23}\right\rfloor+\left\lfloor\frac{2000}{23^2}\right\rfloor=86+3=89$$Therefore, $6+x=89$, so $x=\boxed{83}$.

$e_{23}(2000!)=\left\lfloor\frac{2000}{23}\right\rfloor$ $+\left\lfloor\frac{2000}{23^2}\right\rfloor$ $=86+3=89$ $x=83$ seemed way too easy for a shortlist... even relative to oldest jbmo p1s

We use Legendra's so we have 2000/23 (floor)+2000/529 (floor)=$86+3=89$, so $x=83$

WakeUp wrote: Find the greatest positive integer $x$ such that $23^{6+x}$ divides $2000!$ use legandra done.