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.

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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.

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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

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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 

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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.

The Fearsome Logical Challenge

You and your two friends Pip and Blossom are captured by an evil gang of logicians. In order to gain your freedom, the gang’s chief, Kurt, sets you this fearsome challenge.

The three of you are put in adjacent cells. In each cell is a quantity of apples. Each of you can count the number of apples in your own cell, but not in anyone else’s. You are told that each cell has at least one apple, and at most nine apples, and no two cells have the same number of apples.

The rules of the challenge are as follows: 

The three of you will ask Kurt a single question each, which he will answer truthfully ‘Yes’ or ‘No’. Every one hears the questions and the answers. He will free you only if one of you tells him the total number of apples in all the cells.

    Pip: Is the total an even number?

    Kurt: No.

    Blossom: Is the total a prime number?

    Kurt: No

You have five apples in your cell. What question will you ask?

THIS should be the question that you need to ask!

Logical Response to The Fearsome Challenge

What was the challenge?

Remember, all you have to do is that ask one crucial question to logicians and not necessarily deduce the total count of apples.

Since, each cell has 1 to 9 apples and no two cells have same number of apples, the lowest count of apple is 1 + 2 + 3 = 6 and the highest count would be 7 + 8 + 9 = 24.

That is the total number of apples could be between 6 to 24.

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Now, Pip and Blossom already have gathered some information about the total.

1. The total is not an even number - Hence, only numbers  7,9,11,13,14,15,17,19,21,23 can represent the total count.

2. The Total is not a prime number - Out of the number above, only 9, 15, 21 are non-prime number.

Hence, the total count must be among 9,15 or 21.

Now, your task is easier. All you need to ask the Kert below question -

"Is total is 15?"
 
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CASE 1 : Total is really 15 -

Then Kert would reply with YES to your question and all of you know the total now.

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CASE 2 : Total is 9 -

The Kert's answer to your question would be NO.

If the total is 9 and you have 5 apples then rest of 4 apples must be distributed among Pip and blossom as (1,3) or (3,1) but can't be (2,2) since no 2 cells can have same number of apples.

Now, the friend having 1 apple (or 3 apples) can think that the total can't be 21 as in that case other 2 must have total of 20 (or 18) apples. But the maximum that other two can have is 9 + 8 = 17 apples.

So any of them can deduce that the total is 9.

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CASE 3 : Total is 21 -

Since you have 5 apple other 2 must be having total of 16 apples. One of them must be having 7 apples and other having 9 apples.

The friend having 9 apples can easily deduce the count as 21 since 9 as a total count is impossible in the case as for that the other must have 0 apples.

And the friend with 7 apples know that other can't have 1 + 1 or 2 + 0 (as per given data) apples in order to have total count of 9. Hence, he too can deduce that the total must be 21.

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To conclude, depending on the what Kert answers to your question and the count of apples that each of other 2 friends have one of them (or you too if count is 15) can deduce the total number of apples easily. And eventually, logicians have to set you free as promised.

A Letter Delivery to the Leader

There is a 50m long army platoon marching ahead. The last person in the platoon wants to give a letter to the first person leading the platoon. So while the platoon is marching he runs ahead, reaches the first person and hands over the letter to him and without stopping he runs and comes back to his original position.

In the mean time the whole platoon has moved ahead by 50m.

The question is how much distance did the last person cover in that time.

Assuming that he ran the whole distance with uniform speed. 

Skip to the answer!

Distance Covered by Letter Delivery Person

What was the puzzle?

Let's suppose the first person of the platoon move X meters ahead by the time the last person reaches to him.

To reach at him, the last person has to cover distance of 50 + x.

Assume M be the speed of last person and P be the speed of platoon i.e. of first person.

If T1 is time needed for the last person to get to at the first person,

T1 = (50 + X)/M

T1 = X/P

(50 + X)/M = X/P

M/P = (50 + X)/X  ..........(1)

As per given data, by the time the last person gets back to it's original position (i.e. end of platoon), the platoon moves 50m ahead from it's position that was when the last person started his journey towards first person.

That is end of platoon is now at position at which the start of platoon was initially.

Since, the first person has already moved X meters ahead, he has to move only 50 - X meter to lead the platoon 50m ahead of it's original position.

And, the last person has to move only X meters to get back to  original position i.e. the end of platoon.

If T2 is the time taken by last person to get back to original position (i.e. time taken by first person to move ahead 50 - x) then,

T2 = X/M

T2 = (50 - X)/P

X/M = (50 - X)/P

M/P = X/(50 - X)  ..........(2)

Equating (1) and (2),

(50 + X)/X = X/(50 - X)

X^2 = (50 + X)(50 - X)

X^2 = 2500 - X^2

X^2 = 1250

X = 35.355 meters.

So,

the distance traveled by the last person = (50 + X) + X 
                                                          
                                                          = (50 +35.355) + 35.355 

the distance traveled by the last person = 120.71 meters.

The ratio of their speeds = M/P = (50 + X)/X = (50 + 35.355)/35.355 = 2.41

M = 2.41 P

That is the speed of last man is 2.41 times the speed of first man or the speed of platoon itself.

And that's the speed the last man needed to reach at the first person of the platoon.

"When is the train?"

Lonnie is taking the train to the Library. He tells Rosti the hour of his train’s departure and he tells Ann at which minute it leaves. He also tells them both that the train leaves between 0600 and 1000.

They consult the timetable and find the following services between those two time:

0632 0643 0650 0717 0746 0819 0832 0917 0919 0950
 
Rosti then says “I don’t know when Lonnie’s train leaves but i am sure that neither does Ann”

Ann Replies “I didn’t know his train, but now I do”

Rosti responds “Now I do as well!”

When is Lonnie’s train and how do you know?

THIS is how you should know the correct time!

Similar Kind of Puzzle! 

Scheduled Time of Lonnie's Train

What was the puzzle?

Lonnie tells Rosti the hour of his train’s departure and he tells Ann at which minute it leaves.

Rosti and Ann consult the timetable and find the following services between those two time: 

0632 0643 0650 0717 0746 0819 0832 0917 0919 0950

Rosti then says “I don’t know when Lonnie’s train leaves but i am sure that neither does Ann”

Ann Replies “I didn’t know his train, but now i do”

Rosti responds “Now I do as well!”

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1. If Ann had 43 as a number representing the minutes then she would have an idea of exact time of Lonnie's train as 0643 straightaway as 43 appears only once in the list.

But she says in her statement that she didn't know his train (initially). Hence, she must not had number 43.

2. Similarly, Ann must not had 46 as well as it appear only once in the list in form of 0746.

3. Now, let's think from Rosti's point of view for a moment. How he was sure that Ann too doesn't know the exact time.

Had he got number 6 (or 7) then what he would have thought - 

"Ann might have got 43 (or 46) and hence may know the exact time as 0643 (or 0646). Or she might not have the exact time if she has got any other number. 

I'm not sure whether Ann knows or doesn't know the exact time"

Since, Rosti is sure that Ann too doesn't know the exact time he must not have got number 6 or 7. 

So all timing with 6 and 7 hours in are eliminated leaving behind timing with 8 and 9 hours as - 

0819 0832 0917 0919 0950

4. Ann is smart enough to list out above timings as possible timings after Rosti's first statement.

Now if Ann had number 19 then she would have been confused with 2 timings 0819 and 0919. 

Since, she is sure that she has got correct time it must be among - 0832 0917 0950 where minutes are appearing uniquely.

5. What if Ann had got 17 (or 50) and she deduced correct time as 0917 (or 0950) ? 

That means Rosti must had number 9. With that number, how he would know the correct time whether it is 0917 or 0950?

Since, in his next statement he say that he too know the correct time he must have got number 8 and that's how he know the correct time is 0832.

And since Ann got number 32 with which she is sure that the correct time is 0832.  

 

Story of Man Having 2 Girlfriends

A man who lives in Middletown has two girlfriends, one in Northtown and one in Southtown. 

Trains from the Middletown train station leave for Northtown once every hour. Separate trains from the station also leave for Southtown once every hour. No trains go to both Northtown and Southtown.

Each day he gets to the Middletown train station at a completely random time and gets onto the first train that is going to either Northtown or Southtown, whichever comes first.

After a few months, he realizes that he spends 80% of his days with his girlfriend from Northtown, and only 20% of his days with his girlfriend from Southtown.

How could this be?

THIS could be the reason behind it! 

Behind Unfair The Number of Visits

What's the story behind the title?

The man arrives at Middletown train station at a completely random time of the day. 

Let's take a look at what happens when he arrives at random time of the day.

After arrival on the station, he is likely to get the train in next hour for sure.

After arriving at random time, there are 80% chances that the first train arriving at the station is heading towards Northtown and 20% chances are there the train is heading towards the Southtown. 

That is there has to be 80% minutes of hour (80% of 60 = 48 minutes) where the first train after is heading towards Northtown and 20% minutes of hour where the next train is heading towards Southtown (20% of 60 = 12 minutes).

So, the trains heading towards the Southtown must be scheduled 12 minutes apart from train heading towards the Northtown.

For example, if trains heading to Northtown are scheduled at 9:00 AM, 10:00 AM, 11:00 AM.......etc then the trains to the Southtown must be scheduled at 9:12 AM, 10:12 AM, 11:12 AM.....etc.

With arrival in 48 minutes past 9:12 AM, 10:12 AM etc, he must be getting the Northtown train and if arrived in 12 minutes past 9:00 AM,10:00 AM etc, he would be getting the Southwest train.

Remember, the timing given are for examples only. The Northwest trains may be scheduled at 9:48 AM, 10:48 AM,......etc and Southwest may be scheduled at 10:00 AM, 11:00 AM.

Key is they leave 12 minutes apart, so that 60 minutes of hour are divided into 48 minutes ahead of Northwest train and 12 minutes ahead of Southwest train. 

Challenge of Father to Son

A man told his son that he would give him $1000 if he could accomplish the following task. 

The father gave his son ten envelopes and a thousand dollars, all in one dollar bills. He told his son, "Place the money in the envelopes in such a manner that no matter what number of dollars I ask for, you can give me one or more of the envelopes, containing the exact amount I asked for without having to open any of the envelopes. If you can do this, you will keep the $1000."

When the father asked for a sum of money, the son was able to give him envelopes containing the exact amount of money asked for. 

How did the son distribute the money among the ten envelopes?

THIS is how son accepts the challenge!

Son's Response to the Father's Challenge

What was the challenge?

For a moment, let's suppose father had given $30 to son and provided 5 envelopes and put the same challenge.
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Here, use of the binary number system helps in matter.

The son would distribute 15 dollars into 4 envelops like - 

2^0 = 1, 2^1 = 2, 2^2 = 4, 2^3 = 8.

Now, for any amount asked between 1 to 15, son can produce some of these 4 envelops wherever 1 is there in envelop column for that particular amount.

For example, if father asks for $10 (Binary - 1010), son would give envelop 4 and 2 (8+2=10).

After putting $15 dollar in 4 envelops, he puts remaining $15 in 5th envelop so that he can cover rest of amount between 16 to 30.

If father asks amount greater than 15 then he would take envelop of $15 first and depending on how much the amount asked is greater than this $15 he would pick some of those 4 envelops.

For example, if father asks for $24, then he picks envelop 5 having $15 and envelops for amount 24 - 15 = 9 (Binary - 1009) i.e. envelop 4 and 1 (8+1=9)  i.e. total of 15 + 9 = 24.

So, what we observe from this is that the number of envelops needed for such arrangement is equal to the number of binary bits needed to represent the amount itself or nearest power of 2 greater than the amount.

In above case, to represent 30 in binary we need 5 bits or nearest power of 2 greater than 30 is 32 which needs 5 bits for representation.

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Now, let's turn to the actual challenge where father has asked son to distribute $1000 rightly in 10 envelopes. 

The reason for selecting 10 as a number of envelops is clear now as 1000 needs 10 bits in binary or nearest power of 2 greater than 1000 is 1024 which needs 10 bits for binary representation.

So, the son puts 256, 128, 64, 32, 16, 8, 4, 2, 1 dollars in 9 envelops (envelop numbered as Envelop 9, Envelop 8.........Envelop 1 in order)
 and 1000 - 511 = 489 dollars in 10th envelop.

First 9 envelops will cover amounts from 1 to 511 and for amounts greater than 511 inclusion of 10th envelop having 489 dollars is mandatory.

Again selection of envelops for the amount 511 to 1000 depends on how much the amount exceeds the $489. The binary representation of that difference and selection of envelop accordingly is all that needed.
 
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Let's make sure this distribution with couple of examples.

If father asks for amount of $109 (binary - 1101101) then son picks 

Envelop 1($1) + Envelop 3 ($4) + Envelop 4 ($8) + Envelop 6 ($32) + 
Envelop 7 ($64) i.e. having amount = 1 + 4 + 8 + 32 + 64 = 109 dollars.

If father asks for $525 then son gives $489 via Envelop 10 and rest of amount 
530 - 489 = 40 (Binary - 101000) in form of Envelop - 6 ($32) and Envelop 4 ($8).   

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The Secret Word - Puzzle

A teacher writes six words on a board: “cat dog has max dim tag.” She gives three students, Albert, Bernard and Cheryl each a piece of paper with one letter from one of the words.

Then she asks, “Albert, do you know the word?” Albert immediately replies yes.

She asks, “Bernard, do you know the word?” He thinks for a moment and replies yes. 

Then she asks Cheryl the same question. She thinks and then replies yes. 

What is the word?

THIS must be the given word! 

The Secret Word - Solution

What was the puzzle?

A teacher writes six words on a board: CAT, DOG, HAS, MAX, DIM, TAG 

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1. Albert knows the word right away because he must have received unique letter from above words. Had he received the letter A then he wouldn't have figured out the exact word as A appears in CAT, HAS, MAX and TAG. 

Similarly, he must not have received letters T, D, M and G as those appears multiple times in the list of words.


That is, he must have letter from unique letters C, O, H, S, X, I as they appears only once in the above list of words.

With that the word TAG is eliminated out of the race since any of unique letters doesn't appear in the word TAG.

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2. Now, words left are - CAT, DOG, HAS, MAX, DIM 

In a moment of thinking, Bernard can conclude that the TAG can't be the word and Albert must have got some unique letter from the given words.

After providing letter from C, O, H, S, X, I to Albert, teacher provides letter from rest of letters to Bernard.


Now, if she had given letter -

A - Bernard wouldn't have idea whether the word is CAT, HAS or MAX

D - Bernard wouldn't have idea whether the word is DOG or DIM

M - Bernard wouldn't have idea whether the word is MAX or DIM

That is she must had given letter from unique letters T, O, G, H, S, X.

So, at start, if she provides letter I to the Albert then she can't provide D or M of word DIM to Bernard to give him equal chance to identify the word. 

Hence, the word DIM is also eliminated out of race.

If she had provided letter O to Bernard, then Albert would have been with either D or G with which he wouldn't have been able to figure out the exact word whether it is DOG/DIM or DOG/TAG respectively. Hence, Bernard must not be with letter O.

If Bernard is with letter X then Albert must had either M or A with which he couldn't have figured out the exact word among MAX/DIM or CAT/HAS/TAG/MAX. Hence, Bernard can't have X as well. With that MAX is also eliminated.

Now, if Bernard has letter - 

T - Albert might be with C

G - Albert might be with O

H - Albert may have S

S - Albert may have H. 

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3. The words left are - CAT, DOG, HAS.

Again, in a moment of thinking, Cheryl deduced list of 3 possible words as above.

Now, if she had letter A then she wouldn't have idea of exact word whether it is CAT or HAS. 


If she had unique letter among C/T (or H/S), then either Albert or Bernard with letter A would have been unsuccessful in guess.

If she had letter O/G then either Albert of Bernard would have been with letter D by which they wouldn't know the exact word. 

Hence, Cheryl must had letter D and the word must be DOG.

Albert must have got letter O and the Bernard must have received letter G.

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