Skip to main content
Posts
You are given a unlimited number of coins and 10 pouches. Now, you have to
divide these coins in the given pouches in a manner that if someone asks
you for any number of coins between 1 to 1000, you should be able to give
the amount by just giving the pouches. You are not allowed to open
pouches for that.
How will you do it?
Know here the only efficient way to do that!
Source
What was the challenge?
Once again here binary number system comes in handy. Similar kind of use of binary system in day to day life is here! Another intelligent use is here! We are already provided 10 pouches which is exactly equal to the number of bits required to represent any number from 1 to 1000. Let's number the pouch as Pouch 0 to Pouch 9. So we need to group coins in 10 pouches like below.
Pouch 0 : 1
Pouch 1 : 2
Pouch 2 : 4
Pouch 3 : 8
Pouch 4 : 16
Pouch 5 : 32
Pouch 6 : 64
Pouch 7 : 128
Pouch 8 : 256
Pouch 9 : 512
Now if somebody asks us for 30 coins then we should give Pouch 4, Pouch 3, Pouch 2, Pouch 1. (11110) That's the binary representation of 30 if we assume Pouches as a bits. If another asks for 828 (binary - 1100111100) then we should give Pouch 9, Pouch 8, Pouch 5, Pouch 4, Pouch 3,
Pouch 2.
From a pack of 52 cards ,I placed 4 cards on the table.
I will give you 4 clues about the cards:
Clue 1: Card on left cannot be greater than card on the right.
Clue 2: Difference between 1st card and 3rd card is 8.
Clue 3: There is no card of ace.
Clue 4: There is no face cards (queen,king,jacks).
Clue 5: Difference between 2nd card and 4th card is 7.
Identify four cards ?
Cards identified here!
Source
What was the task given?
Let's list the clues once again here for our convenience.
Clue 1: Card on left cannot be greater than card on the right.
Clue 2: Difference between 1st card and 3rd card is 8.
Clue 3: There is no card of ace.
Clue 4: There is no face cards (queen,king,jacks).
Clue 5: Difference between 2nd card and 4th card is 7.
From Clue 4, it's very clear that there is no King, Queen or Jack card.
From Clue 2 & Clue 3, we have combinations of either 1,9 or 2,10 at first & third place. But Clue 3 eliminates the combination of 1,9.So at first place we have 2 & at third we have 10.
Again from Clue 5 & Clue 3, possible combinations at second & fourth place are 2,9 & 3,10. If it was 2,9 then 4 cards would have been like 2,2,10,9. But according to Clue 1 the card on left can't be greater than that at right. Here, the card at third place (10) is greater than that at fourth place (9) placed at right. Hence, this would be invalid combination.
Hence the correct combination for the second & fourth place is 3,10.
So we have 4 cards as 2,3,10,10.
