21-ball riddle

Welcome Greko; if this is considered a problem in universities; heavens help us all!


First: 9/9…. (3 left over)

Second (a) If even:
1/1….1


Second (b) If not:
3/3…..3


Third (c) Then:
1/1…1


You have not been charged for this remarkable insight into Quantum mechanics, which allows for the calculation of probabilities of outcomes of concrete experiments, insofar as it allows one to compute the probability of finding an answer in a particular region around the nucleus of probability at a particular time. Contrary to classical mechanics, one can never make simultaneous predictions of conjugate variables, such as position and momentum, with arbitrary accuracy. :wink:

Try: Your method works only if you know that the ball is lighter (or heavier). In that case, you can have up to 27 balls.

I don't think this can be done. It takes 2 weighings for 9 balls IF you know how the odd ball differs from the rest. The "standard" problem of this variety uses 12 balls with 3 weighings.

mark If you are correct then I feel much better because I couldn't figure the answer out.

Mark is absolutely correct; it would have to take four.

Thanks anyway guys!
I also think, and am 99% sure that it can´t be done. I have also tried it and I can get the weight in 4 Weighings, not in three (it´s imppossible), or how markr says, if they tell you previously that the one ball that varies either is heavier or lighter.

Thank you very much anyway and I´ll see you.

heavens help us all!

Howdy doody Miss Chai, are you funnin' me! No; wait, what is that aroma? Woah! That is a turn on... Have you changed your perfume or do you exude charm naturally?

oh try....I'll still respect you in the morning.

Oh Chai, that is what they all say! Now, it may all be down to my overactive thyroid secreting excessive amounts of the hormones: thyroxine and triiodothyronine, but, I have noticed that you have become far more outgoing since you changed your hairstyle.

Call me old fashioned if you will, but after an evening spent scrabbling on the floor, as soon as I lay quartzy on ya (at least 82 double letter on the 'r' and connected to a 3-point word on the u,a or t and upwards of 165 points for playing across one of the four spots for a 7-letter word to collect 2 double word scores).

You would kick me out and console yourself with a younger pool boy! :wink:

Never!

this scheme should work:

3 pilesof 7 marbels. weigh 2 of them.
one is lighter: use that one from now on.
they are the same: use the left over pile.

divide the 7 into 3 piles. one with one and 2 with 3
weigh the 2 with 3
if equal: the one left over is the one
if unequal:

divide the 3, weigh 2 of them
if equal: the one left over is the one is the one
if unequal, the one higher up is the one

Verita