2021/054)Find all integer solutions of the equation $\lfloor \frac{x}{1!}\rfloor + \lfloor \frac{x}{2!}\rfloor +\cdots + \lfloor \frac{x}{10!}\rfloor = 1001$

 Let us define

$f(x) =  \lfloor \frac{x}{1!}\rfloor + \lfloor \frac{x}{2!}\rfloor +\cdots + \lfloor \frac{x}{10!}\rfloor$

So we have 

$f(x) > \lfloor \frac{x}{1!}\rfloor$ for $x>=2$

as $f(x) = 1001$ so 

$x < 1001$

As $7!>1000$ so we have   $\lfloor \frac{x}{n!}\rfloor =0$ for $x< 1000$ and $n>6$

Further 

$f(kn!+i)= kf(n!) + f(i)\cdots(1)$ for k = 1 to n and i is less than n!

This is so because for m < n 

$\lfloor \frac{kn!+l}{m!} \rfloor =  \lfloor\frac{kn!}{m!}\rfloor  + \lfloor \frac{l}{m!} \rfloor$  

additionally  

$f(kn!+i)= f(kn!) + f(i)\cdots(2)$ for any $k$ and $i \le n!\cdots(2)$

and $f((n+1)!) = (n+1)f(n!) + 1$

to use the facts let us calculate f(k!) for k = 1 to 6.

f(1) = 1, f(2) = 3, f(6) = 10, f(24) = 41, f(120) = 206

and f(720) = 1237

so the value of  x is less than 720 so let us look at next lowest factorial that is 120

f(120) = 206

$\lfloor \frac{1000}{206}\rfloor = 4$ 

using (1) $f(120*4) = 206 * 4 = 824$

Or f(480) =  824

So we have to account for 1001 - 824 = 177

now f(24) = 41  

$\lfloor \frac{177}{41}\rfloor = 4$ 

so we f(96) = 41 * 4 = 164 

so $f(576) = f(480+96) = f(120 * 4 + 96)

= f(480) + f(96) = 824 + 164 = 988$

Now we have to account for remaining 13 and f(3!) = f(6) = 10

so 6 goes one more time and we ahve

$f(582) = f(576 + 6) = f(24 * 24 + 6) = f(576) + f(6) = 988 + 10 = 998$ using (2)

now we need to account for 3 and as f(2!) = f(2) =3 so we get

E$f(584) = f(582+2) = f(97 * 6) + 2 = f(682) + f(2) = 988 + 3 = 1001$      

so x = 584