Saturday, 6 August 2016

mathematics - How many chocolates?


One Christmas night, 3 kids were sleeping in their room. Their father came and kept a bag full of chocolates in the center of room with a card that says: "Sweets for my three sweet kids."





  1. At midnight, the first kid wakes up, and sees the bag of chocolates. He takes exactly one third of the chocolates in the bag, hides them under his pillow, and leaves the remaining chocolates in the bag.




  2. After some time the second kid wakes up. He does the same with the chocolates that were left by the first kid.




  3. Then, the third kid wakes up after some time and does the same as well.





Finally, at morning they all wake up, looked at each other and said "Look, we got chocolates!" They divided the chocolates equally again.


So, what is the least number of chocolates their father would have kept in the bag?


Hint:



Each time at night a kid divided the chocolates, there was no remainder. The same happened in morning, it all divided equally.




Answer



Let's say X is the number of chocolates.
This means.




Xmod3=0.
how the first kid divided them

23Xmod3=0
the second kid had only 23 of the total chocolates to divide.
He takes 13 of 23X so the last kid counts
49X chocolates.
This means 49Xmod3=0.

The third kid takes 13 of 49 of X...so 427 of chocolates.
S0 on the last morning there are left (49427)X=827X.
And this 827Xmod3=0.
This means 827X=3k.

in order for x to be integer, 3K must be divisible by 8.
The smallest possible k is 8, so 827X=24.



Solving the equation we get X =



81.



How it all happened.
The fist kid wakes up:




Divides 81 chocolates into 3 piles of 27.
He hides his 27 and puts 54 back.



Second kid wakes up.



Divides 54 chocolates into 3 piles of 18.
puts his 18 under the pillow and puts back 36.



The last kid wakes up




Divides 36 chocolates into 3 piles of 12.
puts his 12 under the pillow and puts back 24.



They all wake up and



divide 24 chocolates. 8 each.



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