Puzzle : The Fake Coin Challenge!

There are 101 coins out of which 1 is fake, the fake coin is identical to a genuine coin but differs in weight. Using weight balance only twice how can we determine whether the fake coin is heavier or lighter than a genuine coin?

 

 

This is how to do it! 

Fake Vs Genuine Coin Weigh Comparison

This was the challenge!

Remember we are asked to determine whether the fake coin is lighter or heavier when compared with the genuine coin and not to identify the fake coin itself.

Keep aside any one coin. Divide remaining 100 coins into 2 groups of 50 coins each. Put these 2 groups on 2 pans of the balance.

1. If they weigh equal the the coin that is kept aside is fake. Weigh it against any genuine among 100 coins to know whether fake coin is lighter or heavier than genuine.

2. If they are not equal then that means the fake coin either made one side heavier or the other side lighter.

3. Take the heavier group of 50 coins for the next test. Divide them into 2 groups of 25 coins each. 

4. Put 25 - 25 coins on weighing balance. If they weigh equal then that means no fake coin among them which also means the fake coin was in the other group of 50 coins which was lighter in the first weighing. 

Hence, the fake coin is lighter present in the other group of 50 coins making the group slightly lighter compared to group of 50 genuine coins.

4.1. And if the result of weighing 25 - 25 coins is unequal then it's clear that the fake coin is among these 50 coins. Also, it must be heavier making this group to weigh more than the other group of 50 genuine coins in the first weighing.

This way, we can determine whether the fake coin is heavier or lighter than genuine one using the weighing balance only twice.

The Jungle On The Plot of Land

There is a plot of land that is 16 square acres, arranged in a 4 X 4 grid. There are 16 mammals, two each of the following: beaver, cat, dog, goat, horse, lion, tiger, and walrus. There is one mammal in each acre. The acres are labeled A through P, as shown below.

A B C D
E F G H
I J K L
M N O P

Given the following clues, determine where each mammal is located. The terms "next to" and "connected together" mean vertically, horizontally, or diagonally.

1) A and D do not contain the same mammal, but one of them has a dog.

2) E and O do not contain the same mammal, but one of them has a horse.

3) B and L contain the same mammal, but not a tiger.

4) C and M contain the same mammal, but not a beaver.

5) F, K, N, and P are four different mammals.

6) The tigers are in different columns.

7) The goats are located in two of the following locations: B, C, K, and O.

8) The lions are located in two of the following locations: A, E, F, and M.

9) Cats don't get along with dogs, so neither cat is next to a dog.

10) Tigers have goats on their menu, so each tiger must be next to at least one goat and each goat must be next to at least one tiger.

11) The beavers and walruses live on the same body of water, so all four must be connected together in some fashion.

Click to know how animals are located! 

Oraganised Jungle On The Plot of Land

What was the puzzle?

As per given data - 

There is one mammal in each of 16 acres arranged in 4 x 4 grids. The acres are labeled A through P, as shown below.

A B C D
E F G H
I J K L
M N O P

Given the following clues, determine where each mammal is located. The terms "next to" and "connected together" mean vertically, horizontally, or diagonally.

1) A and D do not contain the same mammal, but one of them has a dog.

2) E and O do not contain the same mammal, but one of them has a horse.

3) B and L contain the same mammal, but not a tiger.

4) C and M contain the same mammal, but not a beaver.

5) F, K, N, and P are four different mammals.

6) The tigers are in different columns.

7) The goats are located in two of the following locations: B, C, K, and O.

8) The lions are located in two of the following locations: A, E, F, and M.

9) Cats don't get along with dogs, so neither cat is next to a dog.

10) Tigers have goats on their menu, so each tiger must be next to at least one goat and each goat must be next to at least one tiger.

11) The beavers and walruses live on the same body of water, so all four must be connected together in some fashion. 

STEPS : 

1] As per Hint 3, blocks B & L contains the same mammal.  As per Hint 4, blocks C & M have the same mammal. So, as per Hint 7 suggests, the goats must be in K & O because if they are in B & C then L & M also supposed to have them but there are total 2 goats only. K - GOAT, O - GOAT

2] If O has a goat then as per Hint 2, E must have a horse.  E - HORSE

3] As per Hint 8, if Lion is located at M then as per Hint 3 the second lion must be in B. But Hint 8 doesn't list B as possible location of lions. Hence, it must not be in M. Also, E is already occupied by horse, hence lions must be at A & F.
A - LION, F - LION

4] So if A has lion, then as per Hint 1, D must have a dog. D - DOG

5] Since C is just near to the D having a dog, as per Hint 9, C can't have cats. As per Hint 10, tigers should be in the 3x3 grids around the goats and goats are located at K and O. But M is out of these grids, hence must not have tigers.

6] And as per Hint 4, C & M having same mammal, but not beaver. As deduced above, it can't be cats or tigers either. With 1 dog already at D, 1 Horse at E, both C & M can't have dogs or horses. Hence, C & M must have walruses (only left mammal out of 8 kind of mammals). C - WALRUS, M - WALRUS

7] As per Hint 11, for walruses (at C & M) to be connected with beavers in some fashion, they must be at G and J. So, G & J have beavers.
G - BEAVER, I - BEAVER

8] Since Hint 3 suggests that B & L are not having tigers, the only animals left in terms of numbers of 2 for to be at B & L are cats. So B & L have cats. B - CAT, L - CAT

9] So far, we have, A - Lion, B - Cat, C - Walrus, E - Horse D - Dog, F - Lion, 
G -  Beaver, J - Beaver, K - Goat, L -  Cat, M - Walrus, O - Goat. Blocks left are  H, I, N, P

10] As per Hint 10 & with goats at K and O, tigers can't be at I. So possible blocks for tigers are H, N and P.

11] But as per Hint 6, tigers has to be in different columns and H & P are in the same column. Also, as per Hint 5, N & P can't have same mammals. And since, H & P falls under same column, the tigers must be occupying N and H. 
 H -  TIGER, N - TIGER

12] Since one cat is at L, P can't be a dog. Hence, I must be having a dog and P must be having a horse.  I - DOG, P - HORSE.

SUMMARY : 

A = lion, B = cat, C = walrus, D = dog
E = horse, F = lion, G = beaver, H = tiger
I = dog, J = beaver, K = goat, L = cat
M = walrus, N = tiger, O = goat, P = horse

 

Angles on Christmas Tree in Puzzle!

Four angels sat on the Christmas tree amidst other ornaments. Two had blue halos and two - yellow. However, none of them could see above his head.

Angel A sat on the top branch and could see the angels B and C, who sat below him. Angel B, could see angel C who sat on the lower branch. And angel D stood at the base of the tree obscured from view by a thicket of branches, so no one could see him and he could not see anyone either.

Which one of them could be the first to guess the color of his halo and speak it out loud for all other angels to hear?

THESE angels have maximum chance of correct guess!


Similar Puzzle 

Logical Angels on Christmas Tree

How their logical skills are challenged?

Case 1 :

The angel A observes that the aureoles of B and C are of the same color. Then, A can be sure that the color of own aureoles must be other than that of B and C.

So, A will be the first to guess the correct color in case he observes 2 same colored aureoles B and C.

Case 2 :

The angle D notices that the color of aureoles of B and C are different. Now, A has to remain silent. The silence of A suggests B that the color of aureole of B must be different than the C.

So, in that case, B will be the first one to guess correct color of own aureole.

However, C has no chance to guess the color of own aureole unless, A or B reveals color of own aureole. And D has absolutely no chance of correct guessing the color of own aureole.

"Spot on the Forehead" Sequel Contest

After losing the "Spot on the Forehead" contest, the two defeated Puzzle Masters complained that the winner had made a slight pause before raising his hand, thus derailing their deductive reasoning train of thought. 

And so the Grand Master vowed to set up a truly fair test to reveal the best logician among them.


He showed the three men 5 hats - two white and three black. 

Then he turned off the lights in the room and put a hat on each Puzzle Master's head. After that the old sage hid the remaining two hats, but before he could turn the lights on, one of the Masters, as chance would have it, the winner of the previous contest, announced the color of his hat. 

And he was right once again.

What color was his hat? What could have been his reasoning? 

The winner is wisest for a reason! 

Master of Logic For a Reason!

How the master was challenged?

Let's assume once again A, B and C are those logicians and C has guessed the color of own hat correctly. Here is what he must have thought - 

============================================================================

"I'm assuming the grand master is conducting this test fairly denying any sort of advantage to any participant.

With that assumption, the grand master can't put 2 white and 1 black hat on heads. In that case, the person having black hat and watching 2 white hats on others' head would know the color of own hat immediately.

For fair play, he can't put 2 blacks and 1 white hat either. That will give unfair advantage to the logicians wearing black hats. Suppose A and B are wearing black and I'm wearing white hat. Now, what A (or B) would be thinking - 

         ---------------------------------------------------------------------------------------------------

       " I'm A (or B) and I can see 1 black and 1 white hat (on head of C). If I have white 
         hat on my head then B (or A) would know color of his hat as black as there are 
         only 2 white hats available and those would be on my head and C's head.

         Moreover, 1 black and 2 white hats already eliminated as it's unfair distribution.

         That means I must be wearing black hat."

         ---------------------------------------------------------------------------------------------------

That's how the combination of 2 black and 1 white hats also eliminated from fair play.

Hence, all of three must be wearing black hats is only fair distribution giving all of us equal chance of winning and hence I must be wearing black hat only. 

============================================================================ 

Note : Here, C is assumed as a winner for only sake of convenience, otherwise either A or B whoever is wisest can be winner.

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.

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.
  ----------------------------------------------------------------------------------------------------

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.
    
     ------------------------------------------------------------------------------------

     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. 

----------------------------------------------------------------------------------------------------

Case 3 : Red + White < Blue + White.

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

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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.

--------------------------------------------------------------------------------------------------

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.

-------------------------------------------------------------------------------------------------- 

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.

Wrong Address By Liar

Mr. House would like to visit his old friend Mr. Street, who is living in the main street of a small village. The main street has 50 houses divided into two blocks and numbered from 1 to 20 and 21 to 50. Since Mr. House has forgotten the number, he asks it from a passer-by, who replies "Just try to guess it." Mr. House likes playing games and asks three questions:

1. In which block is it?

2. Is the number even?

3. Is it a square?

After Mr. House has received the answers, he says: "I'm still doubting, but if you'll tell me whether the digit 4 is in the number, I will know the answer!". Then Mr. House runs to the building in which he thinks his friend is living. He rings, a man opens the door and it turns out that he's wrong. The man starts laughing and tells Mr. House: "Your advisor is the biggest liar of the whole village. He never speaks the truth!". Mr. House thinks for a moment and says "Thanks, now I know the real address of Mr. Street".

 
What is the address of Mr. Street?

This is how Mr.House found correct address! 

Source 

Correct Address Identified !

How pointed towards the wrong one? 

Since Mr. House was able to run at one house after answers of passer-by, he must have got clear clues from that.

3. Is it square ?
  
First thing is sure that, the number must not be a non-square otherwise Mr.House wouldn't be sure as there are plenty of non-square numbers between 1 to 50. So it must be either 4,9,16,25,36,49. (1 is omitted for a reason)

-----------------------------------------------------------------------------------------

1. In which block is it?

Two possible answers here & 2 possible conclusions.

Block 1  :  4,9,16

Block 2  :  25,36,49

------------------------------------------------------------------------------------------

2. Is the number even ?

Now had passer-by  answered Block 1 in 1st question & odd now then Mr. House would have come to know one exact number 9 (that's why 1 omitted here).

Or had he answered Block 2 in 1st question & even now then also Mr. House would have 1 number i.e. 36.

So in both cases, Mr. House would have got 1 fixed number with no point in asking extra question.


That means the passer-by must have told following answers & their possible conclusions are-

Block 1  :  Even  :  Square  :  4,16

Block 2  :  Odd   :  Square  :  25,49

The Honest and Dishonest Guards

You’re in a room with two doors. There’s a guard at each door. One door is the exit, but behind the other door is something that will kill you. You’re told that one guard always tells the truth and the other guard always lies. You don’t know which guard is which. You are allowed to ask one question to either of the guards to determine which door is the exit. 

What question should you ask?

Exit door

Here is that question! 

Source 

Question to Find The Exit Door

What was the situation? 

What if I tell you that you have to ask only one question to anyone & choose the opposite door? Yes, be patient & read further.

Ask anybody this question.

"What would other guard answer if I ask him which one is exit door?"

Now there are two possibilities.

1. If the guard is truth telling then he knows other guard would lie & point the door leading towards the death. So he will point that door.

2. If the guard is lying then he knows other truth telling guard would have pointed right door.But in order to lie, he would point to opposite door! 

Unlock The Lock

Unlock it

It's cracked down here!

Code To Unlock The Lock

Here is what you need to crack!

First let's number the clues as 1,2,3,4 & 5. Now let's start with 4th clue.

4.Nothing is correct.

That puts numbers 7,3,8 out of the picture.

5.One number is correct but wrong placed.

Now with 7 & 3 already out of picture only 0 is the part of the code but current position is wrong.

3.Two numbers are correct but wrong placed placed.

Among 2 correct numbers 0 must be there but again it's position is wrong. So 0 must be at first place. Other correct number which is the part of the code must be either 6 or 2.

From clues 1 & 2 there is 1 number that is correct. For a moment we assume it's 6; both code 6 is sitting at the same place. But as per clue 1 it is correctly placed but as per clue 2 it is wrongly placed. So 6 is also eliminated.We are left with the only number 2.


1.One number is correct & well placed.

We learned so far that 6,8 already eliminated. So this has to be number 2 with it's correct position at 3.


2.One number is correct but wrongly placed.

So far we have got 0 at first place & 2 at third place. Now if 1 is part of the code at second position then it can't be said that it is wrongly placed. So only number left in this clue is 4 with it's position wrong.

Unlocked Lock

 
SO THE CRACKED CODE IS 042