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# Weighing II.

17 replies to this topic

### #11 spoxjox

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Posted 19 December 2007 - 09:52 PM

I don't think so ... what about 15
8+4+2+1=15
8+4+3+0=15

You're right, of course, rookie. The original answer was indeed the optimal solution. Sorry.
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### #12 rookie1ja

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Posted 20 December 2007 - 11:10 AM

I don't think so ... what about 15
8+4+2+1=15
8+4+3+0=15

You're right, of course, rookie. The original answer was indeed the optimal solution. Sorry.

no worries ... if you know some other good brain teasers, you can post it in the new puzzles section
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### #13 miya

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Posted 22 February 2008 - 01:39 AM

Another solution:
spike stands on the scale, holding all the boxes. He drops on box at a time, and sees the difference in weight change for each individual box, and can infer which box contains the real gummy bears, and which contain fake ones.

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### #14 lexam

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Posted 04 April 2008 - 02:33 AM

correct me if im wrong but, do you not need to be told how many each carton contains, what if a carton only contains 20 gummy drop bears? your solution would not be correct - but if you were to asume that it has in excess of this, as you have then you could be correct. But yeh, would'nt you need to be a bit more specific?
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### #15 rookie1ja

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Posted 04 April 2008 - 10:33 AM

correct me if im wrong but, do you not need to be told how many each carton contains, what if a carton only contains 20 gummy drop bears? your solution would not be correct - but if you were to asume that it has in excess of this, as you have then you could be correct. But yeh, would'nt you need to be a bit more specific?

good point

Edit: each carton contains 50 bears
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### #16 Jarod997

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Posted 30 April 2008 - 09:30 PM

Weighing II. - Back to the Water and Weighing Puzzles
Real gummy drop bears have a mass of 10 grams, while imitation gummy drop bears have a mass of 9 grams. Spike has 7 cartons of gummy drop bears, 4 of which contain real gummy drop bears, the others imitation. Using a scale only once and the minimum number of gummy drop bears, how can Spike determine which cartons contain real gummy drop bears?

Edit: scales (for this case) = weighing machine (like in science labs, so no pair of scales) and the plates are big enough for all possible weighings
Edit: each carton contains 50 bears

Spoiler for Solution

Excellent puzzle. =8o) Although in my google gadget where I get a feed of these things, it does not pick up your edits, and I was puzzling away for a good while before "giving up" and going to the BrainDen page -- only to find the addendums. hehe
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### #17 mayiko

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Posted 04 July 2008 - 05:18 AM

Is this a possible solution?

1 2 3 4 7 12 21 <-- 50 gummies

so you take the number of grams below 50 x 10 = 500

if the difference is
6 -> carton 3 and 2 of the cartons to carton 3's left are immitations
7-9 -> carton 4 and 2 of the cartons to carton 3's left are immitations
10-14 -> carton 5 and 2 of the cartons to carton 3's left are immitations
15-23 -> carton 6 and 2 of the cartons to carton 3's left are immitations
24-40 -> carton 7 and 2 of the cartons to carton 3's left are immitations

and since the difference would be a unique sum of 3 of the numbers in the sequence, it should be easy to find the remaining 2 immitations, that is, assuming my sequence is correct.
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### #18 PtGrill

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Posted 16 September 2008 - 09:39 PM

Is this a possible solution?

1 2 3 4 7 12 21 <-- 50 gummies

so you take the number of grams below 50 x 10 = 500

if the difference is
6 -> carton 3 and 2 of the cartons to carton 3's left are immitations
7-9 -> carton 4 and 2 of the cartons to carton 4's left are immitations
10-14 -> carton 5 and 2 of the cartons to carton 5's left are immitations
15-23 -> carton 6 and 2 of the cartons to carton 6's left are immitations
24-40 -> carton 7 and 2 of the cartons to carton 7's left are immitations

and since the difference would be a unique sum of 3 of the numbers in the sequence, it should be easy to find the remaining 2 immitations, that is, assuming my sequence is correct.

I don't think you meant to put "carton 3's left" over and over so I changed it above but anyway:
No it is not a solution because 2 + 3 = 1 + 4.
For example if you find the difference to be twelve grams that means carton 5 and two to its left. The two to the left must add up to 12 - 7 = 5 and that could mean carton 1 and carton 4 or carton 2 and carton 3.
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