Hi , I have one expression as follows:-
(b*y) — (r)*floor(b*y/r)
b and r are known. We can set any value of y of our choice >=0 .
Now what are the possible value that this expression can take?
By running for some values of r and b, I observed two things:-
if (b%r==0) then the answer is zero (which seems obvious to me) if(b%r!=0) then answer is the following :- g*K where K>=0 and K<= (r/g)-1.
Can someone provide the proof for second case? Also will the answer for following expression be similer? r*ceil(b*y/r)-b*y
(b*y) — (r)*floor(b*y/r)is the same as(b*y)%r = (b%r) * (y%r) % rIf you are given b, then you know b%r.
So the values will be
((b%r) * 0) % r,((b%r) * 1) % r, ...,((b%r) * (r-1)) % r(some of these may be the same).Got it Thanks!!