The last one is [x/10!], right?

eliza2003 wrote: The last one is [x/10!], right? Yes. I edited the statement. Sorry.

x=6!a+5!b+4!c+3!d+2!e+f , then easyily find the expression is 1237a+206b+41c+10d+3e+f=2019. a=1,b=3,c=4 and x=1176

If $x>1200$ you get that $LHS>1200+600+200+50=2050$, contradiction. It's now just bashing and as said above the answer seems to be $x=1176$.

$$2019= \left \lfloor \frac{x}{1!} \right \rfloor + \left \lfloor \frac{x}{2!} \right \rfloor + \dots + \left \lfloor \frac{x}{10!} \right \rfloor \le \frac{x}{1!}+ \frac{x}{2!}+...+ \frac{x}{10!}<x(e-1)$$So $x> \frac{2019}{e-1}> \frac{2019}{1.719}>1174.$ $1175$ doesn't work, but $1176$ does. For $x>1176,$ we clearly have $\left \lfloor \frac{x}{1!} \right \rfloor + \left \lfloor \frac{x}{2!} \right \rfloor + + \dots + \left \lfloor \frac{x}{10!} \right \rfloor>\left \lfloor \frac{1176}{1!} \right \rfloor + \left \lfloor \frac{1176}{2!} \right \rfloor + \dots + \left \lfloor \frac{1176}{10!} \right \rfloor =2019$ (the inequality being strict because of the first term) and so $x=1176$ is the unique solution.

这都配的上叫第三题？ 目测x+x/2+x/6+x/24<~2019 -> x 略小于 1182。以列举法可得x=1176 Sent from Montgomery Blair HS