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.

How much does each container weigh?

There are five containers of oil of different weight. They are weighed in pairs of two with all possibilities. The weights in kgs are 165, 168, 169.5, 171, 172.5, 174, 175.5, 177, 180, and 181.5 . 

How much does each container weigh?

Weight Calculation of each container is here! 

Weight Calculation of Each Container

What was the given data?

Let's name five containers as C1, C2, C3, C4 and C5 with their increasing order of weights. That is weight of C1 <= C2 <= C3 <= C4 <= C5.


As per given data, weight of these containers when weighed in pairs -

165, 168, 169.5, 171, 172.5, 174, 175.5, 177, 180, and 181.5 . Here, weights too are in the increasing order.

It's quite obvious that the the container having lowest weight and the second lowest weight will weigh lowest in all possible pairs.That is 165 kg must be weight of pair C1-C2.

C1 + C2 = 165  

And the next higher pair weight of 168 kg must be involving lowest weighing container C1 and third lowest weighing container C3.

C1 + C3 = 168

Similarly, C4 and C5 together must weigh highest when weighed in pair.

C4 + C5 = 181.5

And same way, C3 and C5 makes second highest weighing pair.

C3 + C5 = 180.

It's quite clear that each of the container is weighed 4 times as each of them appearing in 4 pairs.

So, if weight all above pairs are added then, 

4 ( C1 + C2 + C3 + C4 + C5 ) = 165+168+169.5+171+172.5+174+175.5+177+180+181.5= 1734 kg


( C1 + C2 + C3 + C4 + C5 ) = 433.5 kg     .......(1)

Putting C1 + C2 = 165 and C4 + C5 = 181.5 in above,

165 + C3 + 181.5 = 433.5 

C3 = 87 kg.

Since, C1 + C3 = 168, 

C1 = 168 - 87

C1 = 81 kg.

Also, we have, C3 + C5 = 180

C5 = 180 - 87

C5 = 93 kg.

Since, C4 + C5 =  181.5 

C4 = 181.5 - 93 = 88.5 kg. 

Again, C1 + C2 =  165

C2 = 165 - 81 = 84 kg.

Hence, the weights of five containers are 81, 84, 87, 88.5 and 93 kg.

Optimize Weighing Balance

You can place weights on both side of weighing balance and you need to measure all weights between 1 and 1000. For example if you have weights 1 and 3,now you can measure 1,3 and 4 like earlier case, and also you can measure 2,by placing 3 on one side and 1 on the side which contain the substance to be weighed.

So question again is how many minimum weights and of what denominations you need to measure all weights from 1kg to 1000kg. 

We require only......Click here to know! 

Source 

Optimisation Of Weighing Balance

What was the task given? 

Just for a moment, let's assume we have to weigh 1 to 30 Kg. Now you can weigh all those with

1,          2,          4,         6,          8,         10..........30

1           2, (2+1), 4, (4+1),6, (6+1), 8,(9+1),10..(29+1),30   .......For middle weights.


Now 6 can be weighed as 2 + 4 & 10 can be weighed as 2 + 8. We can eliminate those. That means we require only

1,        2,         4,          8,         16,

So we need weights of powers of 2.

Now if subtraction is allowed then,we require

1,        3,         6,          9,          12,          15,         18,...............30

For all weights ,

1, (3-1),3, (3+1), (6-1), 6, (6+1), (9-1),9,         12,         15,............30   

But 6 can be weighed as 9-3 , 12 as 9+3, 15 as 27-(9+3). So we can eliminate 6,12,15... This leaves only

1,         3,        9,        27,

In short, we need power of 3 only.      

For the given problem we need to weight 1 to 1000 Kg with subtraction allowed. So the maximum power of 3 that is less than 1000 is 7. To conclude, we require only 7 weights as below.

1,3,9,27,81,243,729