seaglass jewelry wholesale Olympic Question: On a pirate ship, 100 people are divided into 100 gold coins ...

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5 thoughts on “seaglass jewelry wholesale Olympic Question: On a pirate ship, 100 people are divided into 100 gold coins ...”

  1. Mathew

    modern wholesale jewelry The No. 1 robbers were scored to No. 3 gold coins, 2 robbers No. 4 or 5, and 97 unique. The distribution scheme can be written (97, 0, 1, 2, 0) or (97, 0,1,0,2).

    The reasoning process is: pushing forward from the back, if the robber No. 1-3 is fed sharks, only No. 4 and 5 are left. Shark, to swallow all gold coins alone. Therefore, No. 4 can only support the 3rd. If you know this on the 3rd, you will mention the distribution scheme (100, 0, 0), and the existing gold coins on the 4th and 5th hairs of No. 4 and 5, because he knows that there is nothing to get on the 4th, but he will still vote for votes, and then he will vote. Add his own ticket to his plan. However, the proposal from No. 3 to No. 3 will be proposed (98, 0, 1,1), that is, abandon No. 3, and give one gold coin No. 4 and 5. Because the scheme is more favorable for the 4th and 5th than the 3rd distribution, they will support him and do not want him to leave the situation and be allocated by No. 3. In this way, 98 gold coins will be taken on the 2nd. However, the No. 2 scheme will be learned by No. 1, and No. 1 will be proposed (97, 0, 1, 2, 0) or (97, 0, 1, 0, 2), that is, abandon No. 2, and while Give a gold coin No. 3 and give 2 gold coins No. 4 (or 5) at the same time. Because this plan on No. 1 is better for No. 3 and No. 5 (or No. 5), it is better than the distribution of No. 2, they will vote for No. 1 in favor, plus No. 1's own ticket, No. 1 No. 1, No. 1 The scheme can be passed, and 97 gold coins can be easily fell into the bag. This is undoubtedly a solution that can get the maximum income!

    The supplement:
    . The bear_82 friend and Owenannie friend criticized the answer clearly.
    But the "should be for the third, the fifth ... the old ninety -nine -nine -nine -nine -nine -ranking pirates were proposed by Owenannie's friend, and the remaining 51 of the boss" made different opinions. " I think that this idea can't stand the scrutiny. If other pirates don't say, they say that the pirate No. 3 will be so easily. If he seal the No. 1 pirates, will the most ideal result of the most ideal result of less than one gold coin? The answer is no, so he will never support the Pirate No. 1 pirate "Pirates with a number of pirates, the remaining 51, the remaining 51 to the boss"
    !

    The logical reasoning of the so -called "100 pirates and 100 gold coins" should be beyond the imagination of ordinary people and belongs to the difficulty of "mathematical experts".

  2. Suzanne

    sterling silver skull jewelry wholesale "It should be for the third child, the fifth child .... the nine ninety ninety -nine rows of pirates, one person, one, the remaining 51 people who belong to the boss themselves" made different opinions. I think that this idea can't stand the scrutiny. If other pirates don't say, they say that the pirate No. 3 will be so easily. If he seal the No. 1 pirates, will the most ideal result of the most ideal result of less than one gold coin? The answer is no, so he will never support the Pirate No. 1 pirate "Pirates with a number of pirates, the remaining 51, the remaining 51 to the boss"
    !

  4. Marcia

    wholesale americana jewelry The second floor is just an excerpt of the answer to similar questions. There are 100 people in this question, so it is more complicated. The answer is wrong

    Pirates of gold coins

    , the idea on the second floor is still right

    should be for the third child, the fifth child .... the old ninety nineteen -nine rows of pirates One, the remaining 51 returns to the boss

    The application of this question:

    I assume that the first 98 people were thrown into the sea, two left, 99 must be, No. 99 must be there, No. 99 must be there, No. 99 must be there, No. 99 must be there, No. 99 must be there. All 100 will be dominated by all, even if the No. 100 does not agree, there is no way, so when there is a gold coin No. 100 in the 98th allocation scheme, No. 100 will support the scheme on No. 98

    Push forward according to this, everyone will know when everyone is allocated by themselves as long as the votes of pirates with their own strange rankings can be passed by more than half, so as long as they give these people a gold coin r r

    So the first allocated boss will attract pirates who have been separated by their own wonderful individuals, and at the same time retain their own interests

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