Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

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mathsyperson
2006-02-13 23:27:08

Maybe 4x is an abbreviation for forks.

Tigeree
2006-02-13 17:06:49

hu 4x ? there  easy

justlookingforthemoment
2006-02-13 16:54:07

'Tis deleted. I think krrr might have been the same person as berthold, so I wasn't sure whether they meant delete the '4x' post or what.

Is it just me, or does there seem to be quite a few teabag question posts floating around... ?

irspow
2006-02-13 03:38:57

krrr, I didn't remember if this was the case or not so I posted the "delete this message above".  It appears that "Guests" do not have the ability to edit their posts.  Sorry, but it seems that only a moderator could do that for you.  Once you are a member, however, you can edit or delete any posts that you have offered.

bethold
2006-02-13 02:33:05

4x

ryos
2006-02-12 18:49:33

"What do you mean, I can only weigh one time? Forget that. I'll find a job at a factory that isn't managed by lunatics."

Anyway, I think Rod's got it. If you select as your sample sizes successive powers of two, then weigh the lot. Subtract the weight from what it would be if all were running smoothly (10230), convert the result to binary, and you'll be able to to tell based on which bits are set which machines are working properly or not.

Tigeree
2006-02-12 18:25:03

u say tea bags ?

MathsIsFun
2006-02-12 17:14:55

Powers? Select 1,2,4,8,16, etc?

Bethold
2006-02-12 11:01:43

"10 machines filling tea bags. each machine fills 10g tea bags they goes to separate boxes. suddenly one machine start to fill 9 g tea bags. how can find out which machine fill that bag . you can weight only one time."

Now the variation here is... Not just one machine is defected. Some machines are defected. You dont know if it is 1,2,3 or all. You have to find which machines are defected. You can weigh only one time. (Assume that you have an unlimited number of sample bags from each machine)

Since you know the answer for the original riddle this one must be easy