@Absinthe
def coctake (n,m):
for i in range (m):
n = n*1.2
n = n+1
return n
def hcoctake (n,m):
for i in range (m):
n = n*0.2
n = n+1
return n
for i in range(1000) :
j = coctake(i*5,5)
j = float(j)
if (j).is_integer() is True :
print int(j)
for m in range (5):
print int(i+hcoctake(i*5,m))
else :
continue
My approach return the original number of coconut to 7771, which is a bit absurd so I don't know if this approach is correct or not. What it does is that it tries to guess the number of the morning pile divided by 5. It then reverse engineer that number to find the possible number of coconut at the start. If the number of coconut that if found is a whole number, it will print that number and find the number of coconut the other 5 people get.