I saw a solution in this website but it didn't solve for greater than 2018 coconuts. Please consider this things whne you solve this problem, 1. Use modified version of Chinese remaider theorem where remainder is always same in each case. 2. The number of coconut is greater than 2018. Thanks.

D r . B o k l a n s T h e M o n k e y a n d t h e C o c o n u t s P r o b l e m

Four sailors are shipwrecked on a deserted island - deserted, that is, save a single monkey named Heisenberg who hangs about in the trees near the grove. The sailors gather together all of the coconuts on the island into one big pile and leave them in the grove. There are more than 2018 coconuts! The first sailor awakens at midnight. She doesnt trust the other three and divides the pile of coconuts into quarters and finds that there is one coconut left over. She throws that one to Heisenberg, takes her pile (her fair share) and hides it away, pushes the remaining three piles into one big one and goes back to bed. An hour later, another sailor arises from his slumbers and approaches the pile, divides it into quarters, finds there is one left over which he tosses to the happy monkey, takes his one pile and hides it and pushes the remainder back into one pile. An hour later, the third sailor does the same (Again finding there is one left over which she gives to Heisenberg) and hides away her share... and the last sailor too divides the pile into fourths - with an extra for Heisenberg. Whats the smallest possible total number of initial coconuts on the island (ie, before midnight) if all that is known is that it was larger than 2018? Find, with justification, a formula for all possible initial totals.