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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Started by rookie1ja, Mar 30 2007 05:57 PM

17 replies to this topic

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.

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

rookie1ja (site admin)

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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.

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.

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?

Posted 04 April 2008 - 10:33 AM

good pointcorrect 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?

Edit: each carton contains 50 bears

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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 bearsSpoiler 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

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.

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.

Posted 16 September 2008 - 09:39 PM

I don't think you meant to put "carton 3's left" over and over so I changed it above but anyway: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.

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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