Spot On The Forehead!

Three Masters of Logic wanted to find out who was the wisest among them. So they turned to their Grand Master, asking to resolve their dispute.

"Easy," the old sage said. "I will blindfold you and paint either red, or blue dot on each man's forehead. When I take your blindfolds off, if you see at least one red dot, raise your hand. The one, who guesses the color of the dot on his forehead first, wins."


And so it was said, and so it was done. 

The Grand Master blindfolded the three contestants and painted red dots on every one. 

When he took their blindfolds off, all three men raised their hands as the rules required, and sat in silence pondering.

Finally, one of them said: "I have a red dot on my forehead."

How did he guess? 

And this is how his logical brain responded! 


Similar kind of puzzles are - 

The Greek Philosophers  

Real Test Of Genius  

Logical Response By Master of Logic!

What was the test?

Let A, B and C be the names of three logicians and C be the logician who correctly guessed the color of dot on forehead. 

Now, this could be the C's logic behind his correct guess - 

"If I had blue dot on my forehead then A and B must had raised hands after looking red dots on the foreheads of other. In case, what A (or B) would have thought? His logic would be -

    "If C is with the blue dot then B (or A) must have raised hand after noticing 
    red dot on my forehead, hence I must have red dot."

So A (or B) would have successfully guessed color of dot on own forehead easily.

But neither A or B not responding that means I must have red dot on my forehead!"  

The logician who guess it correctly could be either A or B not necessarily be C; here it is assumed C is wisest for the sake of convenience.

Solve The Picure Equation!

Can you find correct number for '?' ?

Here is the solution!

Solution of Picture Equation!

What was the equation?

From picture, it is clear that,

Pair of shoes = 20, single shoe = 10

Man = 5

Goggles = 2

Single Glove = 20

Now, equation in question has man with 2 shoes, 2 gloves and 1 goggles hence his value in equation = 20 + 40 + 5 + 2 = 67

Hence, final equation appears as

10 + 67 x 2 = 144.

 
The answer is 144.

Heavier Vs Lighter Balls

We have two white, two red and two blue balls. For each color, one ball is heavy and the other is light. All heavy balls weigh the same. All light balls weigh the same. How many weighing on a beam balance are necessary to identify the three heavy balls? 

You need only 2 weighings! Click here to know how!

Identifying The Heavier/Lighter Balls

What was the task given?

Actually, we need only 2 weighing. 

Weigh 1 red & 1 white ball against 1 blue & 1 white ball. Weighing result between these balls helps us to deduce the relative weights of other balls.
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Case 1 : 

If this weighing is equal then the weight of red ball and blue ball are different.
That's because the weighing is equal only when there is combination of 
H (Heavy) + L (Light) against L (Light) + H (Heavy). 

Since, one white ball must be heavier than the other white ball placed in other pan, the red & blue balls place along with them must be of 'opposite' weights with respect to the white balls place along with them.

So weigh this red against blue will give us the heavier red (or blue) leaving other red (or blue) as lighter. 

If red (or blue) weighs more in this weighing then the white ball placed with this red (or blue) in first weighing must be lighter than the other white ball placed with blue (or red) ball.  

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Case 2 : Red + White > Blue + White.

The white ball in Red + White must be heavier than the white ball in Blue + White.

The reason is if this white was lighter then even heavier red in the Red + White can't weigh more than Blue + White having heavier white. That is H + L can't beat H + H or obviously would equal with H + L!

So we have got the heavier and lighter white balls for sure.

Now, weigh red from Red + White and blue from Blue + White against other red and blue balls left.

     Case 2.1/2.2 : Red + Blue > or < Red + Blue 

     Obviously, red and blue balls in left pan are heavier (or lighter) than those in other.
    
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     Case 2.3 :  Red + Blue = Red + Blue 

     Obviously, that's because of H + L = L + H.

      But the red is taken from Red + White which was heavier than the Blue + White. 
      Since, L + H (white is heavier as concluded) can't weigh more than H + L
      (other white is lighter as concluded) or L + L, the red in Red + White 
      must be heavier making H + H combination in that pan in first weighing (case 2). 

      So, we got heavier red and lighter blue obviously leaving lighter red 
      and heavier blue in other pan. 

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Case 3 : Red + White < Blue + White.

Just replace Red with blue & vice versa in the deduction made in case 2. 

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Larger, Smaller or Similar?

Which of the yellow areas is larger?

Here is comparison of areas! 

Both Sharing Equal Area!

Which areas are into comparison?

Actually, areas of both are equal. The diagonal divide the rectangle into 2 halves. So triangle A and A' or B and B' have equal areas.

When diagonal divides the area of main rectangle into 2 halves, area of triangles A (or A') and area of triangle B (or B') are further subtracted from each half to get the areas of the shaded region.

Since equal areas are subtracted from triangles formed by diagonal to get the shaded area, the area of shaded parts are equal. 

 That is from each half area subtracted = A + B = A' + B'.

The Numbered Hats Test!

One teacher decided to test three of his students, Frank, Gary and Henry. The teacher took three hats, wrote on each hat an integer number greater than 0, and put the hats on the heads of the students. Each student could see the numbers written on the hats of the other two students but not the number written on his own hat.

The teacher said that one of the numbers is sum of the other two and started asking the students:

— Frank, do you know the number on your hat?

— No, I don’t.

— Gary, do you know the number on your hat?

— No, I don’t.

— Henry, do you know the number on your hat?

— No, I don’t.

Then the teacher started another round of questioning:

— Frank, do you know the number on your hat?

— No, I don’t.

— Gary, do you know the number on your hat?

— No, I don’t.

— Henry, do you know the number on your hat?

— Yes, it is 144.

What were the numbers which the teacher wrote on the hats?

Here are the other numbers!

Source 

Cracking Down The Numbered Hats Test

What was the test?

Even before the teacher starts asking, the student must have realized 2 facts.

1. In order to identify numbers in this case, the numbers on the hats has to be in proportion i.e. multiples of other(s). Like if one has x then other must have 2x,3x etc.

2. Two hats can't have the same number say x as in that case third student can easily guess the own number as 2x since x-x = 0 is not allowed.

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Now, if numbers on the hats were distributed as x, 2x, 3x then the student wearing hat of number 3x would have quickly responded with correct guess. That's because he can see 2 number as x and 2x on others hats and he can conclude his number as x + 2x = 3x since 
2x - x = x is invalid combination (x, x, 2x) where 2 numbers are equal.

Other way, he can think that the student with hat 2x would have guessed own number correctly if I had x on my own hat. Hence, he may conclude that the number on his hat must be 3x.

But in the case, all responded negatively in the first round of questioning. So x, 2x, 3x combination is eliminated after first round.

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That means it could be x, 3x, 4x combination of numbers on the hats.

In second round of questioning, Henry guessed his number correctly.

If he had seen 3x and 4x on other 2 hats then he wouldn't have been sure with his number whether it is x or 7x.

Similarly, he must not have seen x and 4x as in that case as well he couldn't have concluded whether his number is either 5x or 3x.

But when he sees x and 3x on other hats he can tell that his number must be 4x as 2x (x,2x,3x combination) is eliminated in previous round!

So Henry can conclude that his number must be 4x.

Since, he said his number is 144,

4x = 144

x = 36

3x = 108.

Hence, the numbers are 36, 108, 144.

Divide The Cake Into Equal Parts!

I have just baked a rectangular cake when my wife comes home and barbarically cuts out a piece for herself. The piece she cuts is rectangular, but it’s not in any convenient proportion to the rest of the cake, and its sides aren’t even parallel to the cake’s sides. 

 
I want to divide the remaining cake into two equal-sized halves with a single straight cut. How can I do it?



This is how it can be cut! 

Cutting The Cake Into Equal Parts!

What was the problem? 

Generally, a line drawn through the center of rectangle divides it into 2 equal parts.
Hence, a line drawn through the centers of both rectangles would divide each of them into 2 equal parts as shown below. (To get the center of each rectangle, all we need to do is draw diagonals of both).

Different Kind of Dice Game!

Timothy and Urban are playing a game with two six-sided dices. The dice are unusual: Rather than bearing a number, each face is painted either red  or blue.

The two take turns throwing the dice. Timothy wins if the two top faces are the same color, and Urban wins if they’re different. Their chances of winning are equal.

The first die has 5 red faces and 1 blue face. What are the colors on the second die?

The second die must have 'these' colors!

Die Needed For Different Kind of Dice Game!

What was the different in dice game?

Throwing two six-sided dice produces 36 possible outcomes. Since both Timothy and Urban have equal chances of winning, there are 18 outcomes where top faces are of same color.

Let's assume there are 'x' red faces and (6-x) blue faces on the other dice.

Remember, the first die has 5 red faces and 1 blue face. Then there are 5x ways by which top faces are red & 1(6-x) ways by which they both are blue. But as deduced above, there are 18 such outcomes in total where faces of dice matches color.

Therefore,

5x + 1(6-x) = 18

4x = 12

x = 3

Hence, other die have 3 red colored & 3 blue colored faces.

"Go The Distance"

There are 50 bikes with a tank that has the capacity to go 100 km. Using these 50 bikes, what is the maximum distance that you can go? 

Here is the maximum distance calculation!

Maximizing The Distance!

What was the challenge?
 
Remember, there are 50 bikes, each with a tank that has the capacity to go 100 kms. 

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SOLUTION 1 : 

Any body can think that these 50 bikes together can travel 50 x 100 = 5000 km. But this is not true in the case as all bikes will be starting from the same point. And we need to find how far we can we go from that point. 

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SOLUTION 2 : 

Just launch all 50 bikes altogether from some starting point and go the distance of only 100 km with tanks of all bikes empty in the end.

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SOLUTION 3 : 

1. Take all 50 bikes to 50 km so that tank of each is at half.

2. Pour fuels of 25 bikes (half filled) into other 25 bikes so that their tanks are full.

3. Now, move these 25 bikes to another 50 km so that again their tanks are at half.

4. Pour fuel of 12 bikes into other 12 so that we have 12 bikes with full fuel tank. Leave 1 bike with half filled fuel tank and repeat above.

So for every 50 km distance, half of bikes are eliminated as - 

50 ---> 25 ---> 12 ---> 6 ---> 3 ---> 1

The last bike left with it's tank full can go 100 km. So. the total distance that can be traveled in the case is 5 x 50 + 100 = 350 km. 

However, we have wasted 1/2 fuel each whenever odd number of bikes are left i.e. at 25 and at 3. 

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SOLUTION 4 :

Let's optimize little further so that the 1/2 fuel is not wasted whenever odd bikes are left.


1. Take all 50 bikes to 50 km so that tank of each is at half.

2. Pour fuels of 25 bikes (half filled) into other 25 bikes so that their tanks are full.

3. Now take these 25 bikes to another 20 km using 1/5th (20/100) fuel of each. 

4. Make 5 groups of 5 bikes each. From each group, use 4/5th fuel of 1 bike to fill tank 1/5th emptied tanks of other 4 bikes.

5. Leave bike with empty tank and take 20 bikes to next 50 km. And again after 50 km, pour fuel of 10 bikes into other 10 to eliminate 10.

6. After moving 10 bike for another 50 km, again pour fuel of 5 bikes into another 5.

7. Now take these 5 bikes to another 20 km using 1/5th (20/100) fuel of each.

8. Use 4/5th fuel of 1 bike to fill tank 1/5th emptied tanks of other 4 bikes. 

9. Now these 4 bikes again taken to another 50 km where 2 more are eliminated by taking half of their fuel to fill tanks of other 2.

10. After taking those 2 bikes for another 50 km distance, 1 can be eliminated by taking away it's half fuel to fill up the tank of other bike.

11. The last bike can now go another 100 km distance as it's tanks is full.

To summarize,

50 ---50km---> 25 ---20km---> 20 ---50km---> 10 ---50km--- > 5 ---20km--- > 4 ---50km ---> 

--->2 ---50km---> 1 ---100km ---||

Total distance that can be traveled = 5 x 50 + 2 x 20 + 100 = 390 km.  

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SOLUTION 5 : 

Now we have got the idea from SOLUTION 4 how to maximize the distance further.

Instead of waiting for tanks to be at half or 4/5th we should empty the tank of 1 bike into others at the point where that bike has sufficient fuel for this process.

For example, to have 49/50th fuel in tank of 1 bike at some point, all bikes need to be taken so that 1/50th of each is used up. Since the bike goes 100 km with full tank, with 1/50th fuel it can go 100 x 1/50 = 2km distance.

In short, after 2km distance 49/50th fuel of 1 bike can be used to fill 1/50th empty tanks of other 49 bikes. Now, that 1 bike with empty tank can be left there.

For next phase, we have, 49 bikes. Now, after using up another 1/49th fuel for another distance of (1/49) x 100 = 100/49 km, the 48/49th fuel left in any one bike can fill up the tanks of other 48 bikes (each with 1/49th part is empty). Then, these 48 bikes can be taken for the next phase.

Now, again after consuming 1/48 fuel for another distance of 100/48km, 47/48th of fuel from 1 bike can be used to fill tanks of other 47 bikes (each bike with 1/48th tank empty after traveling 100/48km). So, now 47 bikes can be taken for the next phase.

This way, we are making sure that at each phase 1 bike uses it's all fuel to make tanks of other full.

Repeating this process, till 1 bike left which can go further 100km with full tank.

So the total distance that can be covered is - 

100/50 + 100/49 + 100/48 +.................100/1 = 449.92 km.

And this is the maximum distance that we can go with 50 bikes.


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So – who stole the apple?

During lunch, 5 of Mr. Bryant’s students visit the supermarket.

One of the 5, stole an apple.

When questioned…

  Jim said: it was Hank or Tom.
  Hank said: neither Eddie or I did it.
  Tom Said: you’re both lying
  Don said: no one of them is lying, the other is speaking the truth.
  Eddie said: no Don, that’s not true.

When the shop owner asked Mr. Bryant, he said that three of the boys are always truthful, but two lie all the time.

So – who stole the apple?

And the name of the person who stole the apple is......! 

Tom is a Apple Thief!

What's the story behind the title? 

Let's recall who said what.

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  Jim said: it was Hank or Tom.
  Hank said: neither Eddie or I did it.
  Tom Said: you’re both lying
  Don said: no one of them is lying, the other is speaking the truth.
  Eddie said: no Don, that’s not true.

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Clearly, Don and Eddie are making contradicting statements. Hence, one of them must be liar.

So there must be 1 more liar among Jim, Hank and Tom (since there are 2 liars & 3 truth tellers).

Tom's statement - you're both lying points Jim and Hank as liars. But there are total 2 liars with one being from either or Eddie as deduced above.

Hence, Tom himself must be that other liar.

Now we are sure that Jim and Hank must be telling the truth & as told by Hank, Eddie or Hank himself is not thief.

Since, Hank is not the one who stole the apple, the statement made by Jim suggests that Tom is the person who stole that apple.

With that, since 1 liar found among Jim, Hank and Tom, the immediate statement made by Don must be lie and the statement made by Eddie is truth.

Divide 1 Cube into 20 Cubes!

From a 1987 Hungarian math contest for 11-year-olds:

How can a 3 × 3 × 3 cube be divided into 20 cubes (not necessarily the same size)?

Cut this way to get 20 cubes....

Division of 1 Cube into 20 cubes

What was the challenge?

Mark cube for cutting 3 x 3 x 3 = 27 cubes. Cut any section of 2x2x2 = 8 cubes & cut rest of 27-8 = 19 cubes. So these 19 cubes plus 1 cube of 2x2x2 give us total number of 20 cubes.