Math Problem Statement

Three sailors were marooned on a deserted island that was also inhabited by a band of monkeys. The sailors worked all day to collect coconuts but were too tired that night to count them. they agreed to divide them equally the next morning. During the night, one sailor woke up and decided to take his share. He found that he could make three equal piles, with one coconut left over, which he threw to the monkeys. Thereupon, he put his own share in a pile down the beach, and left the remainder in a single pile near where they all slept. Later that night, the second sailor awoke and, likewise, decided to take his share of coconuts. He also was able to make three equal piles, with one coconut left over, which he threw to the monkeys. Somewhat later, the third sailor awoke and did exactly the same thing with the remaining coconuts. In the morning, all three sailors noticed that the pile was considerably smaller, but each thought that he knew why and said nothing. When they then divided what was left of the original pile of coconuts equally, each sailor received seven and one was left over, which they threw to the monkeys. How many coconuts were in the original pile?

Solution

Let's solve the problem step-by-step, analyzing how each sailor took his share and how the remainder behaved in terms of divisibility.

1. Setting up the scenario:

Let NN be the initial number of coconuts collected by the three sailors. The scenario unfolds as follows:

  1. First Sailor:

    • When the first sailor woke up, he found that the total number of coconuts could be divided into 3 equal piles with one left over.
    • He throws the one leftover coconut to the monkeys and takes his share, which is N13\frac{N - 1}{3}.
    • Thus, the remaining pile has N1N13=2(N1)3N - 1 - \frac{N - 1}{3} = \frac{2(N - 1)}{3} coconuts.
  2. Second Sailor:

    • The second sailor wakes up and sees the new pile size 2(N1)3\frac{2(N - 1)}{3}.
    • He also finds that it can be divided into 3 equal piles with one left over.
    • He throws the leftover to the monkeys and takes his share, which is 2(N1)313=2(N1)39\frac{ \frac{2(N - 1)}{3} - 1}{3} = \frac{2(N - 1) - 3}{9}.
    • The remaining pile now is ( \frac{2(N - 1) - 3 - \frac{2(N - 1)ـ

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Math Problem Analysis

Mathematical Concepts

Number Theory
Modular Arithmetic
Algebra

Formulas

Division with Remainder
Algebraic Equations

Theorems

Properties of Modulo Operation
Integer Division

Suitable Grade Level

Grades 10-12