1[x/1!]+[x/2!]+...[x/2007!]=1005
=[([x]+{x})/1!] +[([x]+{x})/2!]+...[([x]+{x})/2007!]
=[[x]/1!]+[[x]/2!]...[[x]/2007!]=1005 (1)
if [x]=k*n! ,k<n+1,
we see [x] should clearly lie between 5! and 6!
so k<6
n=5.
we have [x]=k*5!
substitute it in (1)
find the nearest rounded integer k to get [x].
cheers!
