Floor function of a factorial

Question

Compute $$\left\lfloor \frac{1000!}{1!+2!+\cdots+999!} \right\rfloor.$$ How can I start with the problem? I thought of dividing by some number, but then I thought that some small numbers when added could also give an integer. Thanks.

Answer 1

$\bf{My\; Solution:}$ Let us estamate the function $\displaystyle \frac{1!+2!+.............+999!}{1000!}$

For Upper bond::

Here $1!+2!+3!+............+998!+999!>998!+999! = 998!\cdot (1+999)=998!\cdot 1000$

So we get $\displaystyle \frac{1!+2!+...........+998!+999!}{1000!}>\frac{998!\cdot 1000}{1000!}=\frac{1}{999}$

So We get $\displaystyle \frac{1000!}{1!+2!+........+998!+999!}<999$

Similarly

For Lower Bond.

Here $\displaystyle 1!+2!+..........+998!+999!=999!\left\{\frac{1}{999!}+\frac{2!}{999!}+.......+\frac{997!}{999!}+\frac{998!}{999!}+1\right\}$

$\displaystyle <999!\left\{\underbrace{\frac{997!}{999!}+\frac{997!}{999!}+.........+\frac{997!}{999!}}_{\bf{997-times}}+\frac{1}{999}+1\right\}=999!\left\{\frac{997\times 997!}{999!}+\frac{1}{999}+1\right\}$

$\displaystyle = 999!\left\{\frac{997}{998\times 999}+\frac{1}{999}+1\right\}<999!\left\{\frac{998}{998\times 999}+\frac{1}{999}+1\right\}=999!\left\{\frac{2}{999}+1\right\}$

$\displaystyle <999!\left\{\frac{2}{998}+1\right\} = \frac{999! \cdot 1000}{998}$

So We get $\displaystyle 1!+2!+..........+998!+999! < \frac{999! \cdot 1000}{998}$

So we get $\displaystyle \frac{1!+2!+..........+998!+999!}{1000!} < \frac{999! \cdot 1000}{998\cdot 1000!}=\frac{1}{998}$

So We get $\displaystyle \frac{1000!}{1!+2!+3!+...........+998!+999!}>998$

So Our final Inequality is $\displaystyle 998< \frac{1000!}{1!+2!+3!+...........+998!+999!}<999$

So We Get $\displaystyle \lfloor \frac{1000!}{1!+2!+3!+...........+998!+999!}\rfloor = 998$