4 weighings
and the puzzle b4 this reminded me of that damsel in distress puzzle presented in one of the earliest posts…
dat was the post that hooked me onto this blog … so i am grateful to this puzzle

Sir in the 1st explained problem the notes one, cant we solve by binary logic just convery it into binary and shift lsb to the right and thats the ans?

See sir in the 500 children game 500 is binary 111110100 right? and as i am saying move the left most to right then 111101001 this is 489. Right so simple and fast these type of problems comes under a theory. I forgot about it ! but they are fast once you understand the type!

Even in the other problem such as the children playing the game where the 2nd person leaves the group and the remaining it continues we can always solve this by changing it into binary and then lsb to rsb and then to decimal. I guess its a fast and easy process.

With the notes probles yes it’s work too i have checked because the logic is same 81 binary is 1010001 and shifting 0100011 is 35 and i am correct. I think you should use this in your blog also to explain students. As this is most recent used ones. and also it;s in wiki explain how this works

Oh yes. Not seen that since a long time, and never realised it was applicable here ðŸ™‚ Nice one.

Will have to see if I can understand it in enough detail to write a post explaining it; I generally prefer to avoid “direct formulae” without the reasoning (and if one is asked such a question in an interview, the reasoning is necessary!)

Since that conversation took place 3 years back, I doubt he would check this ðŸ™‚ However, the theory in question is basically that of the Josephus problem: https://en.wikipedia.org/wiki/Josephus_problem

do we need to weigh 23 times?

A lot less, Snehashis!

regards

J

4 weighings

and the puzzle b4 this reminded me of that damsel in distress puzzle presented in one of the earliest posts…

dat was the post that hooked me onto this blog … so i am grateful to this puzzle

Yes 4 weighings it is ðŸ™‚

regards

J

By inducction: #weighings = n that verifies 3^(n-1)< #coins <= 3^n.

3^4=81 and thus n=4.

Sir in the 1st explained problem the notes one, cant we solve by binary logic just convery it into binary and shift lsb to the right and thats the ans?

I’m not sure what you mean, Harsh. Could you explain?

regards

J

See sir in the 500 children game 500 is binary 111110100 right? and as i am saying move the left most to right then 111101001 this is 489. Right so simple and fast these type of problems comes under a theory. I forgot about it ! but they are fast once you understand the type!

Even in the other problem such as the children playing the game where the 2nd person leaves the group and the remaining it continues we can always solve this by changing it into binary and then lsb to rsb and then to decimal. I guess its a fast and easy process.

With the notes probles yes it’s work too i have checked because the logic is same 81 binary is 1010001 and shifting 0100011 is 35 and i am correct. I think you should use this in your blog also to explain students. As this is most recent used ones. and also it;s in wiki explain how this works

Oh yes. Not seen that since a long time, and never realised it was applicable here ðŸ™‚ Nice one.

Will have to see if I can understand it in enough detail to write a post explaining it; I generally prefer to avoid “direct formulae” without the reasoning (and if one is asked such a question in an interview, the reasoning is necessary!)

regards

J

Its explained properly in wiki sir ðŸ™‚ Very well with diffrent cases. Some more complicated case ðŸ™‚

Can you post the wiki link here, Harsh?

Since that conversation took place 3 years back, I doubt he would check this ðŸ™‚ However, the theory in question is basically that of the Josephus problem: https://en.wikipedia.org/wiki/Josephus_problem

regards

J

Thanks a ton J.