Thursday, April 28, 2011

Weighting Puzzles One Side Weighing & two side Weighing



Puzzle 1:

The puzzle is if the shopkeeper can only place the weights in one side of the common balance. For example if shopkeeper has weights 1 and 3 then he can measure 1, 3 and 4 only. Now the question is how many minimum weights and names the weights you will need to measure all weights from 1 to 1000. This is a fairly simple problem and very easy to prove also. Answer for this puzzle is given below.

Solution :


This is simply the numbers 2^0,2^1,2^2 ... that is 1,2,4,8,16... So for making 1000 kg we need up to 1, 2, 4, 8, 16, 32, 64, 128, and 512. Comments your suggestions or other answers.

Puzzle 2:


This is same as the above puzzle with the condition of placing weights on only side of the common balance being removed. You can place weights on both side and you need to measure all weights between 1 and 1000. For example if you have weights 1 and 3,now you can measure 1,3 and 4 like earlier case, and also you can measure 2,by placing 3 on one side and 1 on the side which contain the substance to be weighed. So question again is how many minimum weights and of what denominations you need to measure all weights from 1kg to 1000kg. Answer for this puzzle is given below.

Solution:


For this answer is 3^0, 3^1, 3^2... That is 1,3,9,27,81,243 and 729.

5 comments:

Anonymous said...

why did u use 2 as base in the 1st problem and 3 as base in 2nd problem??

Zyler said...

dude i agree with the solution but bro why 2 and 3 can u explain....why only 2 and 3 in problem 1 and 2 respectively?

Shashank Mani said...

@Zyler..I Don't Think Anything New , Its Game of Number Theory & How You Can Apply Maths to Solve Puzzles :)

Thanks
Shashank :)

Zyler said...

@shashank mani.....agreed its fun to solve puzzle but one should also know how they solved the puzzle then only we can benefit from it.
so thats why iw as asking u tok 2^0,2^1....so on in 1st problem and 3^0,3^1 and so on in 2nd problem why 2 & 3??...i understand the answer except 2 and 3

Anonymous said...

In the first case we had only two options for each of the weight,
1)Place the weight on the left side
2)Don't place the weight
So the amount of weights required were in the terms of 2^0,2^1,...
But In the second case we had three options for each of the weight,
1)Place the weight on the left side
2)Place the weight on the right side
3)Don't place the weight anywhere
So the amount of weights required were in the terms of 3^0,3^1,...