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Math Help - Three sailors...

  1. #1
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    Question Three sailors...

    Three sailors were shipwrecked on an island. To make sure that they would
    have enough to eat, they spent the day gathering bananas. They put the
    bananas in a large pile and decided to divide the bananas equally among
    themselves the next morning. However, each sailor distrusted the other
    two, so after they were asleep, one of the sailors awoke and divided the
    pile of bananas into three equal shares. When he did so, he found that he
    had one banana left over; he fed it to a monkey. He then hid his share and
    went back to sleep. A second sailor awoke and divided the remaining bananas
    into three equal shares. He too found that he had one banana left over and
    fed it to the monkey. He then hid his share and went back to sleep. The
    third sailor awoke and divided the remaining bananas into three equal shares.
    Again, one banana was left over, so he fed it to the monkey. He then hid his
    share and went back to sleep. When the sailors got up the next morning, the
    pile was noticeably smaller, but since they all felt guilty, none of them
    said anything and they all agreed to divide the bananas. When they did so,
    one banana was left over and they again gave it to the monkey.
    What was the minimal possible number of bananas at the beginning?
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  2. #2
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    Hello, mathcalculushelp!

    This is a classic problem, usually called "The Monkey and the Coconuts";
    . . often there are five or more sailors.

    Let N = the original number of bananas.

    The first sailor divided them into 3 equal groups and had 1 banana left over.
    . . . N \;=\;3A + 1 .[1]
    He hid A bananas and put the remaining 2A bananas back in a pile.

    The second sailor divided the 2A bananas into 3 groups and had one left over.
    . . . 2A \:=\:3B + 1 .[2]
    He hid B bananas and put the remaining 2B bananas back in a pile.

    The third sailor divided the 2B bananas into 3 groups and one left over.
    . . . 2B \:=\:3C + 1 .[3]
    He hid C bananas and put the remaining 2C bananas back in a pile.

    In the morning, they divided the 2C bananas into 3 groups and had one left over.
    . . . 2C \;=\;3D + 1 .[4]


    From [4], we have: . C \:=\:\frac{3D+1}{2}

    Substitute into [3]: . 2B \;=\;3\left(\frac{3D+1}{2}\right)+1 \quad\Rightarrow\quad B \;=\;\frac{9D+5}{4}

    Substitute into [2]: . 2A \;=\;3\left(\frac{9D+5}{4}\right) + 1 \quad\Rightarrow\quad A \;=\;\frac{27D + 19}{8}

    Substitute into [1]: . N \;=\;3\left(\frac{27D+19}{8}\right) + 1 \quad\Rightarrow\quad N \;=\;\frac{81D+65}{8}


    So we have: . N \;=\;10D + 8 + \frac{D+1}{8}

    Since N is an integer, D+1 must be divisible by 8.

    The first time this happens is when D = 7.


    Therefore: . N \;=\;10(7) + 8 + \frac{7+1}{8} \quad\Rightarrow\quad \boxed{N \:=\:79}

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  3. #3
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    Question D+1?

    I understood till N=(81D+65)/8

    But where is the next step of D+1 coming from?
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  4. #4
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    81D + 65 = 80D + 64 + (D + 1). It's just an expansion making most of the terms divisible by 8. Dividing the expanded terms gives 10D + 8 + (D + 1)/8
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