let me try again on other page xD------------------------Nop, I give up, or what I posted is the answer, or I cant find it xD 37/3 does leave 12+1, but if next sailor would divide again, he would get 12 so.......
Now 3rd sailor has 28+9+13=50, that shares as 16+16+16+1 and 1 is left.
Best I could think of xD
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Ahahaha, Np, I calculate fast, but "next morning was crucial" lol, let me try again on other page xD
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Nop, I give up, or what I posted is the answer, or I can't find it xD 37/3 does leave 12+1, but if next sailor would divide again, he would get 12 so.... good luck xD
Hint: Check question again, they must be either stealing or giving some, if its same stockpile, only once there would be +1 banana.
Your typos make it hard to read.
Basically you want a number, n0, such that dividing it by 3 gives remainder of 1
Then subtracting n0 number by 1 gives a new number n1 which has the property such that
dividing n1 by 3 gives a remainder of 1 also, and subtracting n1 by 1 gives n2 such that
n2 divided by 3 has a remainder of 1, and subtracting n2 by 1 gives n3 and finally,
n3 divided by 3 has remainder of 1
So in short:
n0/3 has remainder 1 (First sailor divides it)
n1 = n0 - 1 (First sailor gives leftover to monkey)
n1/3 has remainder 1 (Second sailor divides remaining pile)
n2 = n1 - 1 (Gives to monkey)
n2/3 has remainder 1 (3rd sailor divides pile)
n3 = n2 - 1 (Gives to monkey)
n3/3 has remainder 1 (All sailors divide the pile)
Also n0, n1, n2, and n3 must be integers.
37 doesn't seem to work. 37/3 = 12 remainder 1
Subtract 1, you get 36.
36/3 = 12 (no remainder)
These types of questions are the kind you can expect in math competitions. Doesn't require highly advanced mathematics, but they are very tricky and not like any routine math problems you may typically see in school.
An even more elegant way of interpreting this is to find 3 consecutive numbers such that each of these numbers, when divided by 3, have a remainder of 1.
I'm not sure if it's even possible to get a number according to this rule. Once you get a number with remainder 1 from division by 3, subtracting 1 from it automatically makes it remainder 0, so I must be interpreting the question wrong.
