The greatest integer contained in (3+1)6 is
208
416
415
207
Let R=(3+1)6 and R−[R]=f.
Put F=(3−1)6 , then 0<F<1 and
R+F=(3+1)6+(3−1)6
=233+6C232+6C4(3)+1=2[27+135+45+1]=416
⇒ f+F=416−[R] is an integer
Now, show that f+F=1