A Brillian Deception

A witch owns a field containing many gold mines. She hires one man at a time to mine this gold for her. She promises 10% of what a man mines in a day, and he gives her the rest. Because she is blind, she has three magic bags who can talk. They report how much gold they held each day, and this is how she finds out if men are cheating her. 

Upon getting the job, each man agrees that if he isn't honest, then he will be turned into stone. So around the witch's mines, many statues lay! 

Now comes an honest man named Garry. He accepts the job gladly. 

The witch, who didn't trust him said, "If I wrongly accuse you of cheating me, then I'll be turned into stone."  

That night, Garry, having honestly done his first day's job, overheard the bags talking to the witch. He then formulated a plan... 

The next night, he submitted his gold, and kept 1.6 pounds of gold. Later, the witch talked with her bags.

The first bag said it held 16 pounds that day. The second one said it held 5 pounds. The third one said it held 2 pounds. 

Beaming, the witch confronted Garry. "You scoundrel, you think you could fool me. Now you shall turn into stone!" the witch cried. One second later, the witch was hard as a rock, and very grey-looking.

How did Garry brilliantly deceive the witch? 

Here is the Garry's Master Plan!
 

Mathematics Behind The Brilliant Deception!

What was the deception?

As per the honest man, he must have mined 16 Pounds of gold since he kept 1.6 Pounds (10% of 16) gold for himself.

And as per magical bags, since Bag no.1 said it had 16 Pounds, Bag no.2 said 5 Pounds and Bag no.3 said 2 Pounds the total gold mined 16 + 5 + 2 = 23 Pounds.

We know, honest man Garry would never do any fraud and neither of magical bag would lie. 

So, it's clear that some of pounds are counted multiple times by magical bags. There are 23 - 16 = 7 Pounds gold extra found by those bags.

Now, Bag No.3 must have 2 Pounds in real.

And to make count of Bag No.2 as 5, Garry must had put 3 Pound + Bag No.3 itself in Bag No.2. Hence, the Bag No.2 informed witch that it had total 5 Pound of gold.

Finally, to force Bag No.1 to tell it's count as 16, Garry must had put 
11 Pounds + Bag No.2 (5 Pounds = 3 + Bag No.1) = 11 + 5 = 16 Pounds.

In short, Garry put 2 pounds of gold in Bag No.3 and put that Bag No.3 in Bag No.2 where he had already added 3 Pounds of gold. After that, he put this Bag No.2 in Bag No.3 where he had already added 11 Pounds of gold (somewhat like below picture).

 

This way, 2 Pounds of gold in Bag No.3 are counted 2 extra times and 3 Pounds gold of Bag No.2 are counted 1 more extra time. That is 2 + 2 + 3 = 7 Pounds of extra gold found by bags.

Sequel : Story of Distribution of Loot

The five pirates mentioned previously are joined by a sixth, then plunder a ship with only one gold coin.

After venting some of their frustration by killing all on board the ship, they now need to divvy up the one coin. They are so angry, they now value in priority order:

1. Their lives
2. Getting money
3. Seeing other pirates die.

How can the captain save his skin?

This is how he should save himself! 

 The Prequel of the story!

Captain's Life Saving Proposal in Sequel

First read the story of sequel!

Let's name all the pirates as Pirate 6,5,4,3,2,1 as per their seniority. Now, the captain should respond with the logic below to save his skin.

Let's consider the cases where there are different number of pirates left on the ship after getting rid of seniors one by one.

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CASE 1 : 2 Pirates

The captain i.e. Pirate 2 can keep coin with him & obviously vote for himself (1/2 = 50% vote) to approve the proposal.

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CASE 2: 3 Pirates.

The captain i.e. Pirate 3 offers coin to Pirate 1 to get his support (2/3 = 66%) on proposal. Since, Pirate 1 knows what is going to happen if Pirate 3 dies as crew reduced to 2 Pirates as in CASE 1.

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CASE 3: 4 Pirates.

The captain i.e. Pirate 4 offers coin to Pirate 2 thereby getting his support (2/4 =50% votes) to get approval on proposal. Again, here Pirate 2 is smart enough to agree on this proposal as he know what will happen if Pirate 4 is eliminated leaving behind 3 pirates on sheep as in CASE 3.

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CASE 4: 5 Pirates.
Now the captain i.e. Pirate 5 always will be in danger as he can give only coin to only 1 of remaining 4. So he can 'earn' only 1 vote in support of his proposal i.e. only 2/5 = 40% votes. Hence, there is no way his proposal get approval & he should be ready to die.

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CASE 5: 6 Pirates.

Here, Pirate 5 has to agree whatever captain i.e. Pirate 6 offers as if Pirate 6 dies the case reduces to CASE 4 where Pirate 5 can't save himself. However, if both Pirate 6 & Pirate 5 die, then CASE 3 comes into reality where Pirate 2 gets coin. So Pirate 2 would never agree on proposal offered by Pirate 6.

However, if Pirate 6 offers coin to Pirate 4 then he would take it happily as he knows that what will happen of both Pirate 6 & 5 are killed as CASE 3 comes into picture where he has to give coin to Pirate 2 to save himself.

If Pirate 6 offers coin to Pirate 3 as well & earn his support as Pirate 3 also knows what is going to be the case of both 6 & hence 5 get eliminated. The case will be reduced to 4 Pirates as in CASE 3 where Pirate 4 will offer coin to Pirate 2.

The Pirate 6 even choose Pirate 1 to offer coin. Again, Pirate 1 smart to think that what will be the case if 6 & hence 5 are killed. It will be the scenario as in CASE 3 where Pirate 4 offers coin to Pirate 2.

In short, the Pirate 6 will always get support from Pirate 5 always in any case & any one from Pirate 4/3/1 if he offers coin to any one of these three. 

That's how he will get support of 50% (3/6) group to get approval for his proposal thereby saving his own life on approval of proposal offered.

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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. 

The Unfair Arrangement!

Andy and Bill are traveling when they meet Carl. Andy has 5 loaves of bread and Bill has 3; Carl has none and asks to share theirs, promising to pay them 8 gold pieces when they reach the next town.

They agree and divide the bread equally among them. When they reach the next town, Carl offers 5 gold pieces to Andy and 3 to Bill.

“Excuse me,” says Andy. “That’s not equitable.” He proposes another arrangement, which, on consideration, Bill and Carl agree is correct and fair.

How do they divide the 8 gold pieces?

This is fair arrangement of gold distribution! 

Source 

Correcting The Unfair Arrangement!

How unfair the arrangement was?

First we need to know how 8 loaves (5 of Andy & 3 of Bill) are equally distributes among 3.

If each of them is cut into 2 parts then total 16 loaves would be there which can't be divided equally among 3.

Suppose, each of loaves is divided into 3 parts making total 24 loaves available.

Now, Andy makes 15 pieces of his 5 loaves. He eats 8 and gives the remaining 7 to Carl.

Bill makes 9 pieces of his 3 loaves. He eats 8 and gives the remaining 1 to Carl.

This way, Carl too gets 8 pieces and 8 breads are distributed equally among 3.

 
Obviously, Carl should pay 7 gold pieces to Andy for his 7 pieces and 1 gold piece to Bill for the only piece offered by Bill. 
 

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.

The Seven Rings

You arrive at a hotel and have 3 sets of golden rings. The first set of rings has 4 rings, the second set has 2 rings and the third only has one ring. You cannot take these sets of rings apart, exchange them for a different form of currency, and the hotel clerk has no change. You want to stay at the hotel for 7 nights, and you have to pay one gold ring for each night that you stay. You cannot pay in advance, or all at once at the end of your stay.

 How do you pay for your 7 nights at the hotel?

This is how should you pay! 

Source 

Paying Rings At Hotel

What was the condition? 

You can pay 7 rings in 7 days in following sequence.

Day 1 : 

Give the only ring that is in first set. Paid 1 ring.

Day 2 : 

Take back ring given on Day 1 & give second set of rings having 2 rings. Paid 2 rings

Day 3 :

Give 1 ring back again. Total rings paid = 2 + 1 = 3

Gold For Every Date

Once Mandar forgot his mother in law's birthday. Obviously, as a universal rule, he got the punishment. His wife ordered, I want you to present gold weighing equal to date of day on which my mother would surprise visit to our home.

Now Mandar went to jeweler & ordered only 5 gold coins of different weights. What are the weights of those gold coins?

This is how he does!