Make 7 Using 5 !

In the given picture, you can see that there are two matchsticks that have been used to create five squares. You are allowed to move just two matchsticks and must form seven squares. You can't overlap the matches and you are not allowed to break them. Also, like you can see in the picture, all squares must be closed. 

That's how it can be done!

Steps to Make 7 from 5

What was the challenge?

All we need to do is to re place these 2 match sticks.

So we get,

Mathematical Puzzle On The Chess Board

A white rook and a black bishop of a standard chess set are randomly placed on a chessboard

What is the probability that one is attacking the other?


You can skip to the answer! 
  

Resolving Mathematical Puzzle On Chess Board

But what was the puzzle?

Let's recall the definition of the probability. It's ratio of number of desired combinations to the number of total possible combinations.

A Rook and Bishop can have 64 Permutations 2 = 64 X 63 = 4032 possible combinations on a standard chess board if placed randomly.

Now there are 2 possible cases - The Rook attacking Bishop and the Bishop attacking Rook. Both can't attack each other simultaneously.

CASE 1 : The Rook attacking the Bishop 

The Rook can have 64 possible positions on the chess board and for every position it attacks 14 other position in it's attacking lines. That means 64 X 14 = 896 possible combinations where Rook is attacking Bishop. 

 CASE 2 : The Bishop attacking the Rook

Now imagine 4 hallow co-centrist squares around the center of the chess board with outermost have side 8 units & innermost having side 2 units.(See below)

Here each square has side with thickness of 1 unit.

If the Bishop is anywhere on outermost square which has 28 possible positions then it attack 7 other positions.( See below pics).

So there are 28 x 7 = 196 possible combinations where Bishop attacking the Rook.

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Now if the bishop is anywhere on 20 squares of inner square then 9 other positions will be in it's attacking lines. (See below).

  

In this way, there will be 20 X 9 = 180 such combinations where the bishop will attack the rook. 

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Now if the Bishop is placed anywhere on the 12 squares of more inner square then 11 other positions will be in it's lines of attack.( See below).

In short, there are 12 X 11 = 132 combination where Bishop will be attacking the rook.

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And finally, if the bishop is placed at any of 4 positions of the innermost square then it will attack 13 other positions like below.

That is, there will be 4 X 13 = 52  such combinations where the Bishop will be attacking the Rook.

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Adding all possible combinations of CASE 1 and CASE 2 as - 896 + 196 + 180 + 132 + 52 =  1456. It means there are 1456 possible combinations where either the Rook attacking Bishop or Bishop attacking the Rook.

The Required Probability  = Number of Required Combinations / Number of Total Combinations = 1456 / 4032 = 0.3611

To conclude, 0.3611 is the probability that the Rook or Bishop attacking each other if place randomly on standard chess board.

NOTE : Don't get confused with black Bishop on black square used in illustrations of attacking lines in CASE 2. Even if it was black Bishop on white square then also it would attack same other positions mentioned in that particular consideration. And random placement means it could be anything - on black or on white.

What's the answer?

Just try to find it!

Here is the answer!

'This' Is The Answer!

What was the question? 

Just count the intersecting points.

First one has 9, second one has only 1 & third has 4. 

Hence, the answer is 4.

Lighting Up The Candles

In a group of 200 people, everybody has a non burning candle. On person has a match at lights at some moment his candle. With this candle he walks to somebody else and lights a new candle. Then everybody with a burning candle will look for somebody without a burning candle, and if found they will light it. This will continue until all candles are lit. Suppose that from the moment a candle is lit it takes exactly 30 seconds to find a person with a non burning candle and light that candle.

From the moment the first candle is lit, how long does it take before all candles are lit?

ESCAPE to answer! 

Time Calculation For Lighting Up The Candles

What is the exact situation?

From a moment from first candle is lit, 30 seconds later there would be total 2 candles lit. In next 30 seconds, each of these 2 candle holders will find 1 candle to lit. So there are now 4 candles lit after 60 seconds. In next 30 seconds, these 4 candles would lit another 4 candles making total 8 candles lit. 

In short, for every 30 seconds, the number of candles lit are doubled. So after 7 sets of 30 seconds, 2^7 = 128 candles would be lit. At 8th set of 30 seconds, 256 candles can be lit. But we have only 200 candles. Still 72 of 128 candles would lit another 72 in 8th set of 30 seconds. 

To conclude, 8 X 30 = 240 seconds = 4 minutes required to lit all 200 candles. 

Check Mate in 1 Move

You are playing with white and its your turn. Check mate the opponent in 1 move.

'Here' is that move!

'That' Move to Check Mate The Opponent

First know the current situation on the board!

Just kill the Black Rook by pawn with the move C7 = > B8. Revive KNIGHT at that position to check mate the opponent straightway. Here, opponent can't use bishop at D6 to kill our knight as in that case his king will be in line of attack of our rook at D1.

 

Formations of Special 6-Digit Numbers

How many six digit numbers can be formed using the digits 1 to 6, without repetition such that the number is divisible by the digit at its unit place?

Skip to the count!

Counting The Formations of 6-Digit Special Numbers

You may need to read the question first! 

Let's remind that the numbers can't be repeated. So mathematically there are total 6! (720) unique numbers can be formed.

The number XXXXX1 will be always divisible by 1; so there we have 5! = 120 numbers.

The number XXXXX2 will be always divisible by 2; so there we have 5! = 120 numbers. 

Since sum of all digits is 21 which is divisible by 3; the number XXXXX3 will be always divisible by 3. So we have 5! = 120 more such numbers.

The number XXXXY4 is divisible only when Y = 2 or 6. So in the case we have 2 x 4! = 48 numbers.

The number XXXXX5 will be always divisible by 5; so there we have 5! = 120 numbers.

The number XXXXX6 will be always divisible by 6 (since it is divisible by 2 & 3); so there we have 5! = 120 numbers.

Adding all the above counts - 120 + 120 + 120 + 48 + 120 + 120 = 648.  

So there are 648 six digit numbers can be formed using the digits 1 to 6, without repetition such that the number is divisible by the digit at its unit place.
  

Trip Around The Earth

Professor Fukano plans to circumnavigate the world in his new airplane. But the plane's fuel tank doesn't hold enough for the trip—in fact, it holds only enough for half the trip. But with the help of two identical support planes (which can refuel him in mid-air) piloted by his assistants Fugari and Orokana, the professor thinks he can make it in one trip. But since all three planes have the same problem of limited fuel, how can they work together to achieve the professor's goal without anyone running out of fuel?

1. The professor's plane must make a single continuous trip around the world without landing or turning around.

2. Each plane can travel exactly 1 degree of longitude in 1 minute for every kiloliter of fuel. Each can hold a maximum of 180 kiloliters of fuel.

3. Any plane can refuel any of the others in mid-air by meeting at the same point and instantly transferring any amount of fuel.

4. Fugari and Orokana's planes can turn around instantaneously without burning fuel.

5. Only one airport is available for any of the planes to land, take off, or refuel.

6. All three planes must survive the experiment, and none may run of fuel in mid-air.

'This' is how mission is completed!
  

For The Journey Around The Earth

First read T&Cs of the journey! 

Let's assume that the only airport mentioned is located at the top of the earth. 

Recollect all the data given.
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1. The professor's plane must make a single continuous trip around the world without landing or turning around.

2. Each plane can travel exactly 1 degree of longitude in 1 minute for every kiloliter of fuel. Each can hold a maximum of 180 kiloliters of fuel.

3. Any plane can refuel any of the others in mid-air by meeting at the same point and instantly transferring any amount of fuel.

4. Fugari and Orokana's planes can turn around instantaneously without burning fuel.

5. Only one airport is available for any of the planes to land, take off, or refuel.

6. All three planes must survive the experiment, and none may run of fuel in mid-air.

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As per (2), since each plane travel 1 degree of longitude in 1 minute for every kiloliter of fuel, means that plane need 360 minutes (6 hours) and 360 kiloliters of fuel. But remember plane can hold only 180 kiloliters of fuel.

Let's suppose that, all three planes takes off from airport exactly at 12:00 PM towards the WEST.

We will break this 6 hours journey into 8 parts where each plane travels 45 degree of longitude east or west using 45 kiloliters of fuel.  

START :

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PART 1 :

 Exactly at 12:45 PM, all will be at 45 degree angle with reference to the center of the earth.
At this point, each of them will use 45 kiloliters of fuel. Hence, each will have 135 kiloliters of fuel. Here, Orokana gives away 45 kiloliters to each of Fukano & Fugari. So she is left with the 45 kiloliters which she uses to go back to the starting point.

 
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PART 2 : 

In next 45 minutes, both Fukano and Fugari moves to 90 degree of longitude spending 45 kiloliters of fuel. Now,here both will have 135 kiloliters each in their fuel tank. Here, Fugari refuels Fukano's fuel tank with 45 kiloliters of fuel; leaving 90 kiloliters in own tank for the backward journey towards the starting point.

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PART 3 : 

Fukano travels further, while Fugari is in midway of the backward journey. Again, both spend 45 kiloliters of fuel.

 
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PART 4 :

Exactly at 3:00 PM, Fukano reaches at 180 degree while Fugari reaches back to the starting point. Till then, Orokana refuels her plane & takes off towards EAST. She has to take off as Fukano is left with only 90 kiloliters of fuel by which he could travel half of the rest of journey.

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PART 5 :

In next 45 minutes, Fukano's plane uses 45 kiloliters further while Orokana travels 1/3rd of Fukano's remaining journey in reverse direction so as to meet Fukano in midway. In process, her plane again uses 45 kiloliters of fuel with 135 kiloliters left.

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PART 6 :

At 04:30 PM, Orokana meets Fukano whose plane had no fuel left at the point & refuels his plane with 45 kiloliters of fuel. Remember Orokana's plane consumed 45 kiloliters more till she meets Fukano. Now, since both of them have left only 45 kiloliters in fuel tank, Fugari whose plane standing at airport is refueled at full 180 kiloliters takes of in the direction of EAST.

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PART 7 : 

At 05:15 PM, when fuel tank indicators of both Fukano & Orokana are pointing to 0, Fugari meets them & gives 45 (to Fukano) + 45 (to Orokana)  = 90 out of 135 (180 - 45 used since take off) . Now all are left with 45 kiloliters of fuel & 45 degrees of journey is left.

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PART 8 : 

And this is how, exactly at 06:00 PM all of them reaches back to the starting point safely.

But is this the most efficient way to make trip around the Earth? Certainly not! 

If plane was built with fuel tank of 360 + then the mission wouldn't have required any assistance. Just because of limited fuel tank, 45 + 45 + 90 + 90 + 90 + 90 + 45 + 45 = 270 Kiloliters of fuel burnt to assist Fukano's plane.

Test if You Are a Genius!

People’s Daily,China tweeted out this math puzzle in which each picture represents a number.

The question reads,

"Are you a math genius? Test your brain with this Chinese math puzzle. Did you get it right? (HINT: The correct answer is 16)"

Solve if you can!

Did you figure out how correct answer is 16? 

Did You Prove Yourself Genius?

Look at the question first! 

Let's take a look at the puzzle again.

First thing, we should observe in such puzzles that how objects in the question equation differs from those in given equations.

Here, pair of the sneakers appears as it is in the question while pair of whistles reduced to 1 whistle in question.

And if we look closely the cat, the whistle around the neck is missing in the question.

Now from equation (1), pair of sneakers = 10.

From equation (2), Cat = 5. 

From equation (3), 2 whistles = 4; hence 1 whistle = 2.

Here, we can deduce that cat without whistle in question is CAT - WHISTLE = 5 - 2 = 3.

So the equation in question comes out as, 10 + 3 X 2  = 10 + 6 = 16.

The Green-Eyed Logic Puzzle

In the green-eyed logic puzzle, there is an island of 100 perfectly logical prisoners who have green eyes—but they don't know that. They have been trapped on the island since birth, have never seen a mirror, and have never discussed their eye color.

On the island, green-eyed people are allowed to leave, but only if they go alone, at night, to a guard booth, where the guard will examine eye color and either let the person go (green eyes) or throw them in the volcano (non-green eyes). The people don't know their own eye color; they can never discuss or learn their own eye color; they can only leave at night; and they are given only a single hint when someone from the outside visits the island. That's a tough life!

One day, a visitor comes to the island. The visitor tells the prisoners: "At least one of you has green eyes." 

On the 100th morning after, all the prisoners are gone, all having asked to leave on the night before. 

How did they figure it out?


Here is the solution! 

The Green-Eyed Puzzle Solution

Here is that Puzzle! 

Nobody is going to dare to go the guard unless he is absolutely sure that he is green eyed; otherwise it would be suicidal move.

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For a moment, let's assume there are only 2 prisoners named A & B trapped on island.

On first day, A is watching green eyed B & B can see green-eyed A. But both are not sure what is color of their own eyes. Instead, A(or B) waited B(or A) to escape from island since he is green-eyed. Rather both are sure that other too doesn't know anything about color of own eyes.

On next morning both see each other still on island. Here is what A thinks.

If I was non green-eyed then B would have realized that the person pointed by visitor in his statement ('at least one of you have green eyes') is himself. Hence, B would have realized that he is green-eyed & could have escaped easily. Since B didn't try to escape that means I too must have green eyes.

So A can conclude that he too have green eyes. Exactly same way, B concludes that he too has green eyes. Hence, on next day both can escape from the island.
 
Note that, if the night of the day on which visitor made statement is counted then next day would have 1st morning & 2nd night since visitor's visit. Now since A & B left on 2nd night, we can't see anybody on 2nd morning next day.

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Now let's assume there are 3 prisoners named A,B and C trapped on island.

Let's think from A's point of view as an example.What he thinks.

Let me assume I don't have green eyes.Now each B and C could see 1 green-eyed & other non green-eyed person. But still they don't know color of own eyes.So on that night nobody tries to escape.

On first morning I see both B and C still present there.

Now B can think that if he has no green eyes then C could have concluded that the person pointed by visitor's statement ('at least one of you have green eyes') is C himself (as both A & B are non green-eyed. This way, C would have realized that he is green-eyed.

In a very similar way, B would have realized that he too is green-eyed.

Now both of them could have escaped on that night as they are sure that they are green-eyed.

But on the second morning, I see again both of them are still there. So now I can conclude that I too have green eyes.

If A can conclude then why can't B and C?  So after seeing each other on 3rd day, each of 3 can conclude color of eyes is green. Now on 3rd night they all can escape safely.

This is called as inductive logic.

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If observed carefully, 2 prisoners need 2 nights and 3 prisoners need 3 nights to logically deduce that the each of them is green-eyed.

Hence, 100 prisoners would require 100 nights to absolutely make sure that each of them is green eyed.

That's why on the 100th morning day, there is no prisoner present on island. 

Crack 'This' Logic!

IF

1   1   1   1    =  R

2  2   2   2   =  T

3  3   3   3   = E
 
4  4   4   4   = N

Then

 5  5   5   5   =  ?

Skip To The Answer!

Cracked 'THAT' Logic!

Or look at the question first!


 Let's have a look at the given equations once again.

1   1   1   1    =  R
 
2  2   2   2   =  T

3  3   3   3   = E

4  4   4   4   = N

If rewritten after adding numbers & addition is spelled then,

1   1   1   1    =  FOUR
    
2  2   2   2   =  EIGHT

3  3   3   3   = THREE

 4  4   4   4   = SIXTEEN

 Clearly, last letter of spelling is taken.

Hence,


5  5  5  5 = TWENTY

5  5  5  5 = Y