Puzzle : An Evil Troll on A Bridge

A bridge was guarded by an evil troll. The troll was very intelligent, but he was also a coward. He was afraid of anyone smarter than him. So every time anyone tried to cross the bridge, the troll would set up a test. If the traveler passed the test, he would be allowed to cross. Otherwise, the troll would eat him.

A traveler came across the bridge. 

The troll said, "You may only cross my bridge if you know the password." 

He then wrote thirteen pairs of letters on a rock:

A-V
B-W
C-Q
D-M
E-K
F-U
G-N
H-P
I-O
J-R
L-X
S-T
Y-Z

"These thirteen pairs consist of all 26 letters of the alphabet," said the troll. 

"The password contains thirteen letters, no two of which are the same. Each pair consists of one letter that is in the password and one other letter. If you wrote out the "other" letters in alphabetical order and then wrote each "password" letter under each one's corresponding "other" letter, you would have the correct spelling of the password."

Then the troll wrote five short words on the rock: FACE, QUEST, QUICK, SWITCH, and WORLD. 

"Each short word contains exactly the same number of letters with the password," he said.

So, what is the password? 

HERE is that PASSWORD!

Source

Solution : An Evil Troll on A Bridge Puzzle : Solution

What was the Puzzle?

The thirteen pairs of letters given by an evil troll are -

A-V
B-W
C-Q
D-M
E-K
F-U
G-N
H-P
I-O
J-R
L-X
S-T
Y-Z

And 5 short words given by troll are -  FACE, QUEST, QUICK, SWITCH, and WORLD.  

As described in the given details, we'll refer letter from password as PASSWORD letter & other as OTHER letter.

As per troll, those short words are having same number of PASSWORD letters.

STEPS :

1] Both S & T are appearing in the pair with each other. Hence, either S or T must be a PASSWORD letter but not both. Since, both letters are appearing in short word QUEST, that is QUEST having at least 1 PASSWORD letter for sure hence, all 5 must have at least 1 PASSWORD letter.

2] Suppose every short word has 1 PASSWORD letter. With S or T as 1 PASSWORD letter from QUEST, other letters Q, U, E can't be PASSWORD letters. 

If Q, U, E are not PASSWORD letters then C (from C-Q pair), F (from F-U pair) and K (from E-K pair) must be PASSWORD letters. 

In that case, FACE will have 2 PASSWORD letters viz. C & E which goes against our assumption of having exactly 1 PASSWORD letter in each short word. 

3] Let's assume along with S or T the second PASSWORD letter is E i.e each short word has 2 PASSWORD letters. Again, Q, U can't be PASSWORD letters but C (from C-Q pair) & F (from F-U pair) must be. Still then FACE will have 3 PASSWORD letters which goes against our assumption of exactly 2 PASSWORD letter in each short word. 

4] Now, let's assume along with S or T the second PASSWORD letter is U. Again, Q, E can't be PASSWORD letters but C (from C-Q pair) & K (from E-K pair) must be. Still then QUICK will have 3 PASSWORD letters which goes against our assumption of exactly 2 PASSWORD letter in each short word. 

5] Let's assume there are 4 PASSWORD letters in each short word. So apart from S or T, the letters Q, U, E of short word QUEST must be PASSWORD letters. 

If Q, U, E are PASSWORD letters then C (from C-Q pair), F (from F-U pair) and K (from E-K pair) must NOT be the PASSWORD letters. 

In the case, the short word FACE will have maximum only 2 PASSWORD letters (not sure about A from A-V pair) which again goes against our assumption of exactly 4 PASSWORD letter in each short word. 

6] Hence, each short word must be having 3 PASSWORD letters. 

If Q, E are the PASSWORD letters with S or T in QUEST, then C & K can't be PASSWORD letters. With that, Q, U, I will be 3 PASSWORD letters in QUICK. And if U too is the PASSWORD letter then QUEST will have 4 PASSWORD letters. 

If Q, U are the PASSWORD letters with S or T in QUEST, then C & F can't be PASSWORD letters. With that, FACE can have maximum of only 1 PASSWORD letter. 

7] Hence, U & E must be other 2 PASSWORD letters apart from S or T in short word QUEST. So Q must not be the PASSWORD letter but C must be. Also, F and K can't be the PASSWORD letters.  Hence, FACE will have E, C and A as PASSWORD letters. 

If A is PASSWORD letter then V (from A-V pair) can't be the PASSWORD letter.

8] Next, from QUICK we will have, C, U and obviously I as 3 PASSWORD letters after Q, K are ruled out. If I is PASSWORD letter then O (from I-O pair) can't be the PASSWORD letter.

9] Just like QUEST, SWITCH too have either S or T as PASSWORD letter. Moreover, it has I & C as PASSWORD letters. Hence, H & W must not be the PASSWORD letters.

10] So, if W & O are not the PASSWORD letters then other 3 letters of WORLD i.e. R, L, D must be PASSWORD letters. With that M (from D-M pair), J (from J-R pair) and X (from L-X pair) are ruled out.

11] So far we have got - 

PASSWORD letters - U, E, C, A, I, R, L, D, Either S or T.

OTHER letters - Q, F, K, V, O, H, W, M, J, X  

12] Arranging every OTHER letter in alphabetical order & writing down corresponding PASSWORD letter below it -

OTHER :  F   H   J   K   M   O   Q   V   W   X
PASS.  :  U   P   R   E   D   I    C   A    B   L 

13] Now, S-T, G-N, Y-Z are the only 3 pairs left. And correct placement for these pairs must be like.

OTHER :  F   G   H   J   K   M   O   Q   S   V   W   X   Z
PASS.  :  U   N   P   R   E   D   I    C   T   A    B    L   Y

CONCLUSION : 

The PASSWORD that an evil troll has set must be UNPREDICTABLY. 

 
 

Puzzle : The Case of Missing Servant

A king has 100 identical servants, each with a different rank between 1 and 100. At the end of each day, each servant comes into the king's quarters, one-by-one, in a random order, and announces his rank to let the king know that he is done working for the day. 

For example, servant 14 comes in and says "Servant 14, reporting in." 

One day, the king's aide comes in and tells the king that one of the servants is missing, though he isn't sure which one.

Before the other servants begin reporting in for the night, the king asks for a piece of paper to write on to help him figure out which servant is missing. 

Unfortunately, all that's available is a very small piece that can only hold one number at a time. The king is free to erase what he writes and write something new as many times as he likes, but he can only have one number written down at a time. 

The king's memory is bad and he won't be able to remember all the exact numbers as the servants report in, so he must use the paper to help him.

How can he use the paper such that once the final servant has reported in, he'll know exactly which servant is missing?

Mathematical Trick to know the missing servant! 

Solution : The Missing Servant in the Case

What was the case?

When the first servant comes in, the king should write his number on the small piece of paper. For every next servant that reports in, the king should add that servant's number to the current number written on the paper, and then write this new number on the paper while erasing old one.

Addition of numbers from 1 to 100 = 5050.

Hence, 

Missing Servant Number = 5050 -  Addition of ranks of 99 Servants.

So, depending on how far the addition of 99 servants' rank goes to near 5050, the king can easily deduce the rank of missing servant.

For example, if the addition that king has after 99 servants report in is 5000 then the servant having rank = 5050 - 5000 = 50 must be missing. 

 

Story of Farmer's 3 Sons

A farmer’s wife made some chapatis…The farmer had 3 sons…. 

The first son came, gave one chapati to the dog,and made three equal parts of remaining chapatis, ate one part of it and left the other two parts for his brothers…. Other two sons came one after the other and did the same thinking that they came first…

Then at night all three came to the house, one of them gave one chapati to dog and made three equal parts and the three brothers ate one-one part of it… 

If no chapati was broken in pieces then how many minimum number chapatis did the mom made?

Here are MATHEMATICAL steps for solution! 

Mathematics in the Story of Farmer's 3 Sons

What was the story?

Let X be the number of chapatis that farmer's wife made.

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

First son gave 1 chapati to dog. Chapatis left = (X - 1)

Made 3 equal parts of remaining. Chapatis in each part = (X - 1)/3

He Ate one part of it. Chapatis left = 2(X - 1)/3.

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

Second son gave 1 chapati to dog. Chapatis left = 2(X - 1)/3 - 1

Made 3 equal parts of remaining. Chapatis in each part = [2(X - 1)/3 - 1]/3

He Ate one part of it. Chapatis left = 2[2(X - 1)/3 - 1]/3 

                                                  = [4(X-1) – 6]/9

                                                  = (4X - 10)/9

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

Third son gave 1 chapati to dog. Chapatis left = (4x - 10)/9 - 1

Made 3 equal parts of remaining. Chapatis in each part = [(4X - 10)/9 - 1]/3

He Ate one part of it. Chapatis left = 2[(4X - 10)/9 - 1]/3

                                                  = (8X – 20 – 18)/27 
       
                                                  = (8X – 38)/27 

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

At that night, one of them gave 1 chapati to dog.

Chapatis left = (8X – 38)/27 - 1

After that they made 3 equal parts of remaining parts and each son ate one part.

If we assume that each son ate Y chapatis then,

Y = [(8X – 38)/27 - 1]/3

Y = (8X - 38 – 27)/81

Y =  (8X – 65)/81

81Y= 8X - 65

X = (81Y + 65)/8 

X = 81Y/8 + 65/8

X = 10Y + Y/8 + 8 + 1/8

X = 10y + 8 + (y+1)/8

For X to be integer, Y has to be 7,15,23.....

So possible values of X are - 79, 160, 241.

Since, question asks minimum number, the farmer's wife must have made 79 chapatis in total.

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.

Whose Number is Bigger?

Ali and Zoe reach into a bag that they know contains nine lottery balls numbered 1-9. They each take one ball out to keep and they look at it secretly. Then, they make the following statements, in order:

Ali: "I don't know whose number is bigger." 

Zoe: "I don't know whose number is bigger either." 

Ali: "I still don't know whose number is bigger." 

Zoe: "Now I know that my number is bigger!"

Assuming Ali and Zoe are perfectly logical, what is Zoe's smallest possible number?

'This' is that smallest possible number!

My Number is the Bigger One!

First you can read what happened?

Recalling what Ali and Zoe said - 

Ali: "I don't know whose number is bigger." 

Zoe: "I don't know whose number is bigger either." 

Ali: "I still don't know whose number is bigger." 

Zoe: "Now I know that my number is bigger!" 

First statement of Ali indicates that she doesn't have either 1 or 9. If she had 1 (or 9) then she would have an idea that Zoe must have bigger (or smaller) number.

Now Zoe is smart enough to know that Ali doesn't have 1 or 9 which is clear from Ali's first statement. Zoe's first statement indicates that she doesn't have 2 or 8 (& neither 1 or 9). If she had 1/2 (or 8/9) then she could have concluded that Ali has bigger (or smaller) number.

Till now Ali has an idea that the Zoe doesn't have 1,2,8,9. So still Ali can't have 2/8 as in that case too she could have made a different statement. Further if Ali had 3 or 7 (and knowing the fact that Zoe doesn't have 1,2,8,9); Ali could have an idea whose number is bigger as 3 is smallest while 7 is biggest among remaining numbers. That means she doesn't have 3 or 7 ( and 1,2,8,9).

From all the statement Zoe can conclude that Ali doesn't have 1,2,3,7,8,9. In short, Ali must have either 4,5,6. 

Now when Zoe says she has bigger number then it must be either 6 or 7 and Ali having 4 or 5. It can't be 5 as in that case Zoe wouldn't be confident as Ali could have 6.

So the Zoe's smallest possible number is 6.

     

"Hear & Identify What the Time is!"

Your grandma’s wall clock chimes the appropriate number of times at every whole hour, and also once every 15 minutes. If you hear the wall clock chime once, how much more time do you need to figure out what the time is, without looking at it?

This is the way to figure out the exact time!

Source 

The Count Will Tell The Exact Time!

What was the challenge?

Since the clock chimes appropriate number of times at every whole hour, it's not difficult to predict the exact time after counting sounds but the exception of 1:00 AM/PM.

For example, if the first sound you heard at 4:30 AM then you will hear another sound at 4:45 AM and clock will strike 5 times at 5 AM by which you will easily know the exact time. So, even if you heard it at 15 minutes past any hour, you will need only 45 minutes to figure out the exact time.

But at 1:00 AM/PM the clock will strike only once & there the problem starts.

In worst condition, if you have heard first sound at 12:15 then clock will strike once for next 6 times at 12:30, 12:45, 1:00, 1:15, 1:30, 1:45. That is you will hear clock strike once for 7 consecutive times. So in worst condition, you need 1 hour and 30 minutes to figure out the exact time.


And if you hear the single strike for less than 7 consecutive times, then you can easily figure out the exact time.

 

Develop an Unbeatable Strategy!

Consider a two player coin game where each player gets turn one by one. There is a row of even number of coins, and a player on his/her turn can pick a coin from any of the two corners of the row. The player that collects coins with more value wins the game. 

Develop a strategy for the player making the first turn, such he/she never looses the game.

Note : The strategy to pick maximum of two corners may not work. In the following example, first player looses the game when he/she uses strategy to pick maximum of two corners.

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

Example :

  18 20 15 30 10 14

First Player picks 18, now row of coins is
  20 15 30 10 14

Second player picks 20, now row of coins is
  15 30 10 14

First Player picks 15, now row of coins is
  30 10 14

Second player picks 30, now row of coins is
  10 14

First Player picks 14, now row of coins is
  10 

Second player picks 10, game over.

The total value collected by second player is more (20 +
30 + 10) compared to first player (18 + 15 + 14).
So the second player wins.
===================================================================
 
This is the unbeatable strategy! 

The Unbeatable Strategy for a Coin Game!

Why strategy needed in the case?

Let's recall the example given in the question.
-------------------------------------------------------------------

Example

  18 20 15 30 10 14

First Player picks 18, now row of coins is
  20 15 30 10 14

Second player picks 20, now row of coins is
  15 30 10 14

First Player picks 15, now row of coins is
  30 10 14

Second player picks 30, now row of coins is
  10 14

First Player picks 14, now row of coins is
  10 

Second player picks 10, game over.

The total value collected by second player is more (20 +
30 + 10) compared to first player (18 + 15 + 14).
So the second player wins.
---------------------------------------------------------------------------------

Here, it's very clear that the player who chooses coins numbered at even positions wins & the one who chooses odd position loses.

So first player who is going to choose coin first need to be smart. All that he/she need to do is make sum of values of all coin at even position, sum of values of all coins at odd positions and compare them. 

If he finds the sum of values of coins at odd position greater then he should choose 1st coin (odd position) followed by 3rd,5th.....(odd positions) & force other to choose even coins

And if he finds the sum of values of coins at even positions greater then he should choose
last coin (which is at even position as number of coins are even).

For example, in above case, 

  18 20 15 30 10 14

first player calculates 

Sum of odd coins = 18 + 15 + 10 = 43
Sum of even coins = 20 + 30 + 14 = 64.  

Since sum of even coins is greater, he should choose 6th coin (which will be followed by 4th and 2nd) and force other player to choose 5th,3rd and 1st coin.

Even in case second player selects 1st coin after 14 at other corner is selected by first, still he can be forced to choose the coin at 3rd position (odd position) if first player selects 2nd coin.

In short, first player need to make sure that he collects all the coins that are at even positions or odd position whichever has greater sum!   

Note that the total number of coins in the case can't be odd as then distribution of coins among 2 will be unequal.

The Dice Date Indicator!

How can you represent days of month using two 6 sided dice? You can write one number on each face of the dice from 0 to 9 and you have to represent days from 1 to 31, for example for 1, one dice should show 0 and another should show 1, similarly for 29 one dice should show 2 and another should show 9.

This is how it can be indicated!

Making of The Dice Date Indicator

What was the challenge?

Dice 1: 0 1 2 3 5 7

Dice 2: 0 1 2 4 6 8

The number 0 has to be present on both the dice. The '0' on first die needed for dates from 1 to 9 to show them as 01,02,03......and the '0' on second die will be used for the dates 10,20,30.

The number 1 and 2 are repeated on the dates 11 and 22 so those 2 numbers has to be there on both dice.

Now, we are left with total 6 positions but 7 numbers - 3 to 9.

However, 6 and 9 can be represented by single die if it is written on one of the side of the die. In normal position it will represent 6 & in inverted position it will show 9 or say vice versa.

For example, the dates 6 and 9 can be indicated as -