Single Floor

Number Theory Level 2

Solve for natural number $n$ , such that $n+\frac{n}{2!}+\frac{n}{3!}+\frac{n}{4!}+\left\lfloor\frac{n}{5!}\right \rfloor = 2018$

2 solutions

X X
Apr 24, 2018

Let $\frac{n}{4!}=m,24m+12m+4m+1m+\lfloor\frac{m}{5}\rfloor=41m+\lfloor\frac{m}{5}\rfloor =2018$ , $m$ is near $\frac{2018}{41}$ ,so $m=49,n=1176$

Giorgos K.
May 10, 2018

using $Mathematica$

n=1;While[n+n/2!+n/3!+n/4!+Floor[n/5!]!=2018,n++];n

returns 1176