17 replies to this topic

[spoxjox]

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.

[rookie1ja]

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

[miya]

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.

[lexam]

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?

[rookie1ja]

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

[Jarod997]

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

Weighing II. - solution
Spike uses 51 gummy drop bears: from the 7 boxes he takes respectively 0, 1, 2, 4, 7, 13, and 24 bears.
The notion is that each box of imitation bears will subtract its number of bears from the total "ideal" weight of 510 grams (1 gram of missing weight per bear), so Spike weighs the bears, subtracts the result from 510 to obtain a number N, and finds the unique combination of 3 numbers from the above list (since there are 3 "imitation" boxes) that sum to N.
The trick is for the sums of all triples selected from the set S of numbers of bears to be unique. To accomplish this, I put numbers into S one at a time in ascending order, starting with the obvious choice, 0. (Why is this obvious? If I'd started with k > 0, then I could have improved on the resulting solution by subtracting k from each number) Each new number obviously had to be greater than any previous, because otherwise sums are not unique, but also the sums it made when paired with any previous number had to be distinct from all previous pairs (otherwise when this pair is combined with a third number you can't distinguish it from the other pair)--except for the last box, where we can ignore this point. And most obviously all the new triples had to be distinct from any old triples; it was easy to find what the new triples were by adding the newest number to each old sum of pairs.
Now, in case you're curious, the possible weight deficits and their unique decompositions are:
3 = 0 + 1 + 2
5 = 0 + 1 + 4
6 = 0 + 2 + 4
7 = 1 + 2 + 4
8 = 0 + 1 + 7
9 = 0 + 2 + 7
10 = 1 + 2 + 7
11 = 0 + 4 + 7
12 = 1 + 4 + 7
13 = 2 + 4 + 7
14 = 0 + 1 + 13
15 = 0 + 2 + 13
16 = 1 + 2 + 13
17 = 0 + 4 + 13
18 = 1 + 4 + 13
19 = 2 + 4 + 13
20 = 0 + 7 + 13
21 = 1 + 7 + 13
22 = 2 + 7 + 13
24 = 4 + 7 + 13
25 = 0 + 1 + 24
26 = 0 + 2 + 24
27 = 1 + 2 + 24
28 = 0 + 4 + 24
29 = 1 + 4 + 24
30 = 2 + 4 + 24
31 = 0 + 7 + 24
32 = 1 + 7 + 24
33 = 2 + 7 + 24
35 = 4 + 7 + 24
37 = 0 + 13 + 24
38 = 1 + 13 + 24
39 = 2 + 13 + 24
41 = 4 + 13 + 24
44 = 7 + 13 + 24
Note that there had to be (7 choose 3) distinct values; they end up ranging from 3 to 44 inclusive with 7 numbers missing: 4, 23, 34, 36, 40, 42, and 43.

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

[mayiko]

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.

[PtGrill]

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.