RIDDLES -- Dimag ki batti jalao
- Thread starter dkumars
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sanjeev_balyan
New member
ya this is the ans, I was expecting. Touch all the lights in one go. Each light will receive three transactions, one by its own physical touch and two by its two neighbours physical touch. Than it will go like this OFF--ON--OFF. U will get all lights off. Anyway u have cracked the problem, the way i wanted. Congratulation and keep it up:rock:rock
Well, its not so easy also but let me take an attempt in a very general way.
How do we define 0? 0 means nothing which does not exist that means a name given to absence of something.
Now, if something does not exist then how could we apply an operation like divison on that?
Just for an example, I have 1 mango and I have to divide it among some(say 1 here) guys then I have something to distribute and divide. But if I say I have no mango and I have to divide it among no guys then it is just an imaginary situation. We can not define such a situation because the things which do not exist, we can not comment on that.
I guess u have a better answer and I am waiting for that.
sanjeev_balyan
New member
Try this one-
A man has 11 centimeters of a gold bar (of uniform rectangular cross section). He is staying 11 nights in a hotel, and the owner demands 1 centimeter of gold per night as payment. The owner will not accept back payment or advanced payment – so every night the man must give the owner exactly 1 additional centimeter of gold: no more, no less. He has only two rusty saws with which to saw his gold bar, and each saw can only be used once.
How does the man satisfy owner's payment requirement by making only two cuts on his gold bar.
A man has 11 centimeters of a gold bar (of uniform rectangular cross section). He is staying 11 nights in a hotel, and the owner demands 1 centimeter of gold per night as payment. The owner will not accept back payment or advanced payment – so every night the man must give the owner exactly 1 additional centimeter of gold: no more, no less. He has only two rusty saws with which to saw his gold bar, and each saw can only be used once.
How does the man satisfy owner's payment requirement by making only two cuts on his gold bar.
devendra ji...u wrote a lot..But all mathematical operations are valid for 0... When we multiply something by 0 then result is 0....similarly when divided by 0 then result is infinite...
In 0/0, the 0 in numerator wants to make expression 0 while the 0 in denominator wants to make infinite...so a fight goes on between the two 0....because both have the same capacity so result is undefined....
I hope, it is not tough to unterstaned this.....
In 0/0, the 0 in numerator wants to make expression 0 while the 0 in denominator wants to make infinite...so a fight goes on between the two 0....because both have the same capacity so result is undefined....
I hope, it is not tough to unterstaned this.....
devendra ji...u wrote a lot..But all mathematical operations are valid for 0... When we multiply something by 0 then result is 0....similarly when divided by 0 then result is infinite...
In 0/0, the 0 in numerator wants to make expression 0 while the 0 in denominator wants to make infinite...so a fight goes on between the two 0....because both have the same capacity so result is undefined....
I hope, it is not tough to unterstaned this..... But it was not easy to think also
Hmmm dats the best solution ... thanks
loginrahul
New member
Bhai question is quite tough so to agar wo Jaat hi to dusre hotel me chal jayega........:rock:rock:rock:rock
Jat hai toh kabhi doosre hotel mein nahi jaayega ... wahi rahega aur bina paisa/sona diye
sanjeev_balyan
New member
Bhai Rahul, Jat itni jaldi haar thode hi na maan lega.
First hint- first saw will be used to make two pieces and than second saw will be used to cut both the pieces placed together, means finally u will get 4 pieces
now try, i believe, u can crack it this time
loginrahul
New member
lage tum log dimaak lagva k hi maanoge....
to ans hai.... bhai 4 pieces hoge... 1,2,3,5 cm ke (Fibonacci series)
1st cut at 5 cm ( 5,6 )
2nd cut by placing 5 cm bar over 6cm and cut them in 2,3(5) and 1,5(6) cm.
The bar is cut into 2 parts, one having 5 pieces and another 6 pieces. Then those two parts are placed in such a way that we get 4 pieces of lengths 1 cm, 2 cms, and rest two are 4 cms.
But this is not a solution
neetu ji ur answer is partialy right...i am improving it..
first cut into two pieces,one have 3 cm and other have 8 cm...
now put one upon other and cut in a way that we will get four pieces of 1 cm and 2 cm(from 3 cm part) , 4 cm and 4 cm(from 8 cm part)...now transaction will be as following...
1st night---- give 1 cm piece
2nd night---- give 2 cm piece and takes back 1 cm piece.
3rd night----- give 1 cm piece
4th night----- give 4 cm piece and takes back 1 and 2 cm pieces
5th night----- give 1 cm piece
6th night------give 2cm piece and takes back 1 cm piece
7th night------give 1cm piece
8th night------give 4 cm piece and takes back 1 and 2 cm piece
9th night------give 1 cm piece
10th night-----give 2 cm piece and takes back 1 cm piece
11th night-----give 1 cm piece
first cut into two pieces,one have 3 cm and other have 8 cm...
now put one upon other and cut in a way that we will get four pieces of 1 cm and 2 cm(from 3 cm part) , 4 cm and 4 cm(from 8 cm part)...now transaction will be as following...
1st night---- give 1 cm piece
2nd night---- give 2 cm piece and takes back 1 cm piece.
3rd night----- give 1 cm piece
4th night----- give 4 cm piece and takes back 1 and 2 cm pieces
5th night----- give 1 cm piece
6th night------give 2cm piece and takes back 1 cm piece
7th night------give 1cm piece
8th night------give 4 cm piece and takes back 1 and 2 cm piece
9th night------give 1 cm piece
10th night-----give 2 cm piece and takes back 1 cm piece
11th night-----give 1 cm piece
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Ohh $hit ... I even could not understand the problem completely and was hitting somewhere else for 11 single pieces... I was out of competetion this time ... bad luck
Yes the answer was correct, though I was a bit lazy in elaborating it 
!! Thanks for explaining.
!! Thanks for explaining.loginrahul
New member
wow.... ek queston 2 answer ye to bahut badi na insafi hai......
sanjeev_balyan
New member
Rahul bhai, 2 nahi aur bhi bhaut sare answer hai eske.
cut first at 4 and 7 cm. than put together these pieces and cut the second chance at 01 cm from any flushed side. than u will get, 1 cm & 3 cm, 1 cm & 6 cm . now u can do all payments up to 11.
well done all, Rahul to Neetu,
now very simple and based on above problem, what should be the domination of currency notes to give up to 100 Rs with out doubling the notes.
Sanjeev ji, as per my understanding of the question, my take on the puzzle is:-
To have a currency upto Rs. 100, we can use the denominations of powers of 2, i.e., 2^0 =1
2^1 =2
2^2 =4
2^3 =8
2^4 =16
2^5 =32
2^6 =64
A little elaboration on how these denominations could be used:-
3 -> give 1 + 3
5--> give 4 + 1
6 -> give 4 + 2
7 -> 4 +2 + 1
9 -> 8 + 1
10 -> 8 + 2
11 -> 8 + 2 + 1
12 -> 8 + 4
13 -> 8 + 4 + 1
14 -> 8 + 4 + 2
15 -> 8 + 4 + 2 +1
17 -> 16 + 1
18 -> 16 + 2
19 -> 16 + 2 +1
20 -> 16 + 4
21 -> 16 + 4 + 1
22 -> 16 + 4 + 2
99 -> 64 + 32 + 2 + 1