An Island Of Puzzles

There is an Island of puzzles where numbers 1 - 9 want to cross the river.

There is a single boat that can take numbers from one side to the other.

However, maximum 3 numbers can go at a time and of course, the boat cannot sail on its own so one number must come back after reaching to another side.

Also, the sum of numbers crossing at a time must be a square number.

You need to plan trips such that minimum trips are needed.

This should be that minimum number! 

Numbers On An Island Of Puzzles

What was the challenge?

We need only 7 trips to send all digits across the river.

1. Send 2, 5, 9 (sum is 16).

2. Bring back the 9.

3. Send 3,4, 9 (sum is 16).

4. Bring back the 9.

5. Now send 1,7,8 (sum is 16).

6. Bring back the 1.

7. And finally send 1,6,9 (sum is 16).

A Door Of Fate Or Logics?

A prisoner is faced with a decision where he must open one of two doors. Behind each door is either a lady or a tiger. They may be both tigers, both ladies or one of each.

If the prisoner opens a door to find a lady he will marry her and if he opens a door to find a tiger he will be eaten alive. Of course, the prisoner would prefer to be married than eaten alive :).

Each of the doors has a statement written on it. The statement on door one says, “In this room there is a lady, and in the other room there is a tiger.”

The statement on door two says, “In one of these rooms there is a lady, and in one of these rooms there is a tiger.”
 
The prisoner is informed that one of the statements is true and one is false.

Which door should the Prisoner open?

This should be his choice!

Logical Choice Of Door

What were the choices?

For a moment, let's assume that first statement is true. The lady is behind the Door 1 and tiger is behind the Door 2. But this makes statement 2 also true where it says there is tiger behind one of these door & lady behind one of these doors. Hence, the statement 1 can't be true.

Hence, statement 2 must be true.

Only possibilities left are -

Door 1 - Tiger
Door 2 - Tiger

Door 1 - Lady
Door 2 - Lady

Door 1 - Tiger
Door 2 - Lady.

Since, the true second statement is suggesting there is lady behind 1 & tiger behind the other door, the possibilities of both tigers or ladies are eliminated.

That's why behind Door 1 is tiger & behind Door 2 is lady.

 

One More CryptArithmatic Problem

What three digits are represented by X, Y, and Z in this addition problem?

  XZY

+XYZ
______
  YZX

Here is the solution!

Solution Of One More CryptArithmatic Problem

What was the problem?

Here is the equation rewritten.

  XZY
+XYZ
______
  YZX
 

Let's start with the tens place. Z + Y = Z is there. That means either Y = 0 or 9 if 1 is carry from ones place.

Since Y is at hundred's place it can't be 0. Hence, Y = 9.

At hundred's column, now we have, X + X = 9. That's only possible if X = 4 and carry 1 forwarded from tens place. So X = 4.

Now, finally, at ones place, we have, 9 + Z = 4. Hence, Z must be 5 with carry 1 being forwarded to the next place.

To sum up, X = 4, Y = 9 and Z = 5.

 

Who Is The Engineer?

On a train, Smith, Robinson, and Jones are the fireman, the brakeman, and the engineer (not necessarily respectively). Also aboard the train are three passengers with the same names, Mr. Smith, Mr. Robinson, and Mr. Jones.

(1) Mr. Robinson is a passenger. He lives in Detroit.
(2) The brakeman lives exactly halfway between Chicago and Detroit.
(3) Mr. Jones is a passenger. He earns exactly $20,000 per year.
(4) The brakeman’s nearest neighbor, one of the passengers, earns exactly three times as much as the brakeman.
(5) Smith is not a passenger. He beats the fireman in billiards.
(6) The passenger whose name is the same as the brakeman’s lives in Chicago.

Who is the engineer?

Want to know who? Click Here! 

That's Why Smith Is An Engineer!

Would you like to read question first? 

Let's list all the clues once again here.

(1) Mr. Robinson is a passenger. He lives in Detroit.
(2) The brakeman lives exactly halfway between Chicago and Detroit.
(3) Mr. Jones is a passenger. He earns exactly $20,000 per year.
(4) The brakeman’s nearest neighbor, one of the passengers, earns exactly three times as much as the brakeman.
(5) Smith is not a passenger. He beats the fireman in billiards.
(6) The passenger whose name is the same as the brakeman’s lives in Chicago.

Since as per (2), the brakeman lives exactly halfway between Chicago and Detroit, locations Chicago or Detroit can't be nearest to him. Hence, the passenger that (4) is suggesting must be nearer to brakeman than Chicago and Detroit.

Now as per (1), Mr. Robinson lives in Detroit, means he is not the nearest to brakeman. Mr.Jones earning is $20,000/year as per (3), which is not evenly divisible by 3. Hence, the passenger (4) is pointing is not Mr.Jones.

So neither Mr. Robinson not Mr.Jones but Mr.Smith is the nearest neighbor. 

Now Mr. Robinson lives in Detroit and Mr.Smith is living in between Chicago and Detroit but nearer to brakeman. Hence, Mr. Jones must be living in Chicago.

According to (6), Jones must be name of the brakeman as he is sharing his name with the man living in Chicago.

And if Smith is not fireman as per (5), he must be an engineer!   

  

Unique Audiance For A Fairy Tale

A sultan has 14 daughters. He decides to tell every night four of his daughters a fairy tale, but in such a way that every night, there will be another combination of four daughters. How many nights will keep the sultan busy telling fairy tales? 

Find number of nights here!

Nights For Fairy Tales!

Story behind the title?

For a moment, let's name all the daughters as A, B, C, D, E.....N. 

There are 14 x 13 x 12 x 11 = 24024 combinations of daughters. But in these combinations, lots of combinations are repeated. For example, ABCD combination is same as ABDC or DABC etc. There can be 4 x 3 x 2 x 1 = 24 combinations while considering group of 4 daughters here for example it is A, B, C, D. For these 24 combination we should count only 1 as a unique combination.

Hence for 24024 combinations, we have 24024/24 = 1001 unique combination. 

In short, Sultan would be busy for 1001 nights in telling fairy tales to his 14 daughters in unique combinations.

Numbers For The Letters?

A - B = B

B * C = A

D : B = E

C * C = E

C + E = A

For which numbers stand A, B, C, D and E? 

Skip To The Answer!  

Correct Numbers For Letters

What was the question? 

Let's first rewrite all equations and number them

1. A - B = B

2. B * C = A

3. D : B = E

4. C * C = E

5. C + E = A

For which numbers stand A, B, C, D and E? 

From 1, A = 2B. Putting this in 2 gives,

C = 2.

Putting C = 2 in 4 gives, E = 4.

So from 5, A = C + E = 2 + 4 = 6.

Equation 2 gives, B = A/C = 6/2 = 3.

And equation 3 gives D = B * E = 3 * 4 = 12.

Conclusion : A = 6, B = 3, C = 2, D = 12 and E = 4. 

Who Lives Where?

There are 4 big houses in my home town.
They are made from these materials: red marbles, green marbles, white marbles and blue marbles.
* Mrs Jennifer's house is somewhere to the left of the green marbles one and the third one along is white marbles.

* Mrs Sharon owns a red marbles house and Mr Cruz does not live at either end, but lives somewhere to the right of the blue marbles house.

* Mr Danny lives in the fourth house, while the first house is not made from red marbles.
Who lives where, and what is their house made from ?

Know about location and home about each!

Owners Of Homes in Home Town

What were the clues? 

To simplify the process, let's number the houses as 1,2,3,4 and separate clues given.

a. Mrs Jennifer's house is somewhere to the left of the green marbles one.
b. The third one along is white marbles.
c. Mrs Sharon owns a red marbles house
d. Mr Cruz does not live at either end.
e. Mr Cruz lives somewhere to the right of the blue marbles house.
f. Mr Danny lives in the fourth house
g. The first house is not made from red marbles.

As per clues (b) and (g), 1 and 3 aren't made of red marbles. Since Danny is living in 4 as per (f), the red house owned by Mrs. Sharon must be positioned at 2.

Since 2 is already occupied, as per (d), Mr. Cruz must be living at number 3 which is of white marbles according to (b).

According to (a), 4 must be green marbles since otherwise Jennifer wouldn't be somewhere at left. As per (f) this 4 th house is owned by Mr.Danny.

The only house left for Jennifer is blue positioned at 1.

To conclude,

Mrs Jennifer - blue marbles at Number 1
Mrs Sharon - red marbles at Number 2
Mr Cruz - white marbles at Number 3
Mr Danny - green marbles at Number 4

Mathematical Talk Between Horse And Camel

A horse and a camel were carrying boxes on their backs. The horse started complaining to the camel that his load is too heavy.

The camel replied 'Why are you complaining? If you gave me one of your boxes I would have double what you have and if I give you one of my boxes we two would have an even load.'

How many boxes do each of the animal (horse & camel) is carrying ?

Find here the load on each of them! 
 

Suggestion From Mathematical Talk!

What was the talk?

Let's assume C be the number of boxes that camel is carrying and H be that being carried by horse.

As per first part of camel's statement i.e. if you gave me one of your boxes I would have double what you have  

C + 1 = 2 (H - 1) 

C + 1 = 2H -2

C = 2H - 3 ........(1)

Now in second part (i.e.if I give you one of my boxes we two would have an even load) of camel's statement suggests,

C - 1 = H + 1 


C = H + 2 

Putting (1) in above,

2H - 3 = H + 2

H = 5

Again putting this value in (1) gives,

C = 2*5 - 3 = 10 - 3 = 7 

C = 7. 

Horse is carrying 5 boxes and the camel is carrying 7 boxes.  

 

Fair Distribution Of Water

In Sahara desert , 3 men found a big 24L Jar is full of water. Since there is shortage of water so they decided to distribute the water among themselves such that they all have equal amounts of it. But they only have a 13L, a 5L and an 11 liter Jar.

How do they do it? 

Here is how to do it!

Source 

Equal Distribution Of Water

What was the challenge?

1. First pour 24L into 13L and 11L jar.There will be no water in 24L jar.

2. Now pour 13L jar into 5L jar till 5L is filled. So 8L of water will be left in 13L jar.

3. This 8L of water from 13L jar is emptied in empty 24L jar.This will leave 13L jar empty.

4. Now pour 11L water from 11L jar into 13L of jar. There is still space for 2L of water in 13L jar.

5. Pour 5L jar into 13L of jar which had space for 2L only. So 3L of water will be left in 5L of jar.

6. This 3L of water is emptied in 11L jar.

7. A 13L jar full of water is again poured into 5L jar leaving behind 8L of water in it.

8. A 5L jar full of water is finally poured in 11L jar already having 3L of water. 

9. This way, 24L, 11L and 13L of jar would have 8L of water each.

Another CryptArithmatic Problem

Replace letters with numbers assuming numbers can't be repeated. 

     SEND
  + MORE
  ----------
 = MONEY

Process of decryption is here! 

Decrypting The CryptArithmatic Problem

What was the problem? 

Let's take a look at the equation once again.

       S  E N D
  +   M O R E
  -----------------
 = M O N E Y

Now letter M must be representing the carry generated & it must be 1. 

And if M = 1 then S must be 9 or 8 if carry is generated from hundreds place. 

In any case, O can be either 0 or 1. But can't be 1 as M = 1 hence O = 0.

If O = 0 then E + 0 = N i.e. E = N if there is no carry from tens place. 

Hence, N = E + 1.

Let's turn towards tens place now. With no carry from units place N + R = 10 + E .

Putting N =  E + 1, we get, R = 9. And with carry from units place, we have,

1 + N + R = 10 + E, gives us R = 8. Hence either R = 9 and S = 8 or R = 8 or S = 9.

For a moment, let's assume R = 9 and S = 8 with no carry from units place, then, 

   8 E N D
+ 1 0 9  E
=========
1 0 N E Y 

For this to be correct, we need carry at thousands place generated from hundreds 

place. That's only possible if E = 9 and carry is generated from tens place

forwarded to hundreds place. Since 9 is already being used for R this combination

is just impossible.

Hence, R = 8 and S = 9 with carry 1 from units place.

Now it looks like,

   9 E N D
+ 1 0 8 E
========
1 0 N E Y 

This means D + E >= 10 i.e. D + E = 10 + Y.  And numbers left are 2,3,4,5,6,7.

If E = 2 then D must be 8 or 9 for D + E >= 10.

Since 8 and 9 already taken, this is not possible.

If E = 3 then D can be 7 but Y would be 0 in the case. 

Since O = 0 already taken this value of E is also not valid. 

For any other value of D, D + E < 10 .

If E = 4 then D = 7 or 6 and Y = 0 or 1.

Both are taken hence this value of E in invalid.

Also,D <= 5 in the case gives  D + E < 10 .

If E = 6, N = 7 then, D <= 5. 

With D = 5 or 4, Y = 0 or 1, both are used for O and M already.

And for D = 2 or 3,D + E < 10 .

In short, this value of E is also not valid.   

So only value of E left is 5. Hence, N = 6 and D = 7. That gives, Y = 2.

To conclude,

  9 5 6 7
+ 1 0 8 5
=========
1 0 6 5 2