Story of Distiribution of 100 Coins Loot

Five ship’s pirates have obtained 100 gold coins and have to divide up the loot. The pirates are all extremely intelligent, treacherous and selfish (especially the captain).

The captain always proposes a distribution of the loot. All pirates vote on the proposal, and if half the crew or more go “Aye”, the loot is divided as proposed, as no pirate would be willing to take on the captain without superior force on their side.

If the captain fails to obtain support of at least half his crew (which includes himself), he faces a mutiny, and all pirates will turn against him and make him walk the plank. The pirates start over again with the next senior pirate as captain.

What is the maximum number of coins the captain can keep without risking his life?

He can take away 98 coins! How? Read here! 

The Captain's Undeniable Proposal

What was the situation?  

He can keep 98 coins! Surprised? Read how.

Let's number 5 pirates as Pirate 5, Pirate 4.....Pirate 1 as per their descending order of seniority.

Pirate 5 keeps 98 coins with him and gives 1 coin each to Pirate 3 and Pirate 1.

Now Pirate 5 i.e Captain explains his decision -

CASE 1:

 If there were only 2 pirates then Pirate 2 would have taken all 100 coins after obtaining his own vote which accounts to 50% votes (50 % of 2 = 1).

CASE 2: 

 In case of 3 pirates, the Pirate 3 would have offered 1 coin to Pirate 1 & would have kept 99 coins with him. Now Pirate 1 wouldn't have any option other than agreeing on deal with Pirate 3. That's because if he doesn't agree then Pirate 3 would be eliminated & all coins would be with Pirate 2 as explained in above (case 1) of only 2 pirates. So votes of Pirate 1 and Pirate 3 which account to 66% (2 out of 3) votes of group locks this deal and Pirate 2 would be left without any coin.

CASE 3: 

 Now in case of 4 pirates, Pirate 4 would offer coin to Pirate 2 & would keep 99 coins with him. Now, Pirate 2 know what happens if Pirate 4 gets eliminated. Pirate 3 would offer 1 coin to Pirate 1 & will take away 99 coins. So Pirate 2 would definitely accept this deal. That's how votes of Pirate 4 and Pirate 2 makes 50% (2 out of 4) votes of group to pass the proposal. Pirate 3 and Pirate 1 can't do anything in this case.

By now, Pirate 3 and Pirate 1 realizes what happens of Pirate 5 gets eliminated. They won't be getting any coin if Pirate 4 becomes captain as explained above (case 3). So they have no option other than to vote for the proposal of Pirate 5. 

This way, Pirate 5, Pirate 3 and Pirate 1 (3/5 = 60% of crew) agree on proposal of Pirate 5 where he takes away 98 coins with 1 coin each to Pirate 3 and Pirate 1. 

A Mathematical Clue From The Merchant

A rich merchant had collected many gold coins. He did not want anybody to know about them. 

One day, his wife asked, “How many gold coins do we have?”

After pausing a moment, he replied, “Well! If I divide the coins into two unequal numbers, then 32 times the difference between the two numbers equals the difference between the squares of the two numbers.”

The wife looked puzzled. Can you help the merchant’s wife by finding out how many gold coins they have?

Here are mathematical steps to find those!

Using The Mathematical Clue

What was that clue?

Since when divided into 2 unequal numbers difference won't be 0. Let x and y be the 2 unequal numbers.

As per merchant,

32 (x - y) = x^2 - y^2

32 (x - y) = (x - y) (x + y)

Dividing both sides by (x - y) which is non zero as x is not equal to y,

32 = x + y

x + y = 32.

Let's verify with x = 30 and y = 2. So 32 (x - y) = 32 ( 30 - 2) = 896. And x^2 - y^2 = 30^2 - 2^2 = 900 - 4 = 896.

Hence, Merchant had 32 coins in total.

Corrupt Servant Stealing The Gold

You have 10 bags of gold coins.You have appointed 1 servant to carry each bag. One of your servants who were responsible for transport of the money wanted to trick you. He took one of the bags and filed away one gram of gold from each coin. One coin normally weighs 10 grams.

Can you figure out in one scaling which bag contains lighter coins? Which servant should be fired? – using digital scales (shows the exact weight of an item)?

Trick to get down the culprit! 

Source

Master Plan for Gold Stealing Servant

Where story begins? 

We need to take coins from each of 10 bags to test.

Number all the servants & their respective bags from 1-10. Now take 1 coin from first bag, 2 coins from second bag, 3 coins from third bag & so on to 10 coins from tenth bag. Put all those on digital scale. In normal case, with no corruption it would weigh in the case as 55 x 10 = 550 gm. But coins from one of the bag weighs 1 gm less. So if it weighs 1 gm less then 1st coin that came from first bag is filed one & hence first servant is corrupt. If it weighs 548 gm then coins from second bag are reduced ones & hence second servant is corrupt.

Depending on how many gm less than 550 gm it weighs the bag with manipulated coins can be identified & respective servant can be fired.