Marbleous 2
You have a collection of bags of marbles. Some of the bags contain marbles of weight 10 g, the other bags contain marbles of weight 11 g. Your mission is to determine which bags are which in one weighing. You have a suitably accurate weighing machine. I won't state the other assumption, in case I give too much away.
This is probably another familiar golden oldie.
This is probably another familiar golden oldie.
Labels: logic, mathemagic
posted by Chris at
4:15 PM
19 Comments:
Take one marble from bag 1, 4 from bag 2, and so on up to n^2 from the last bag n.
Weigh this collection. If all marbles weigh 10 grams the total weight would be 10(n^3-1)/(n-1).
If some marbles weigh 11 grams the extra weight N will be the sum of the squares of the bag numbers of 11 gram marbles.
Off the top of my head I can't think of a quick way to solve this, and I have to go out now.
I guess your assumption is that the bags can be arranged in such a way that there are enough marbles in each to perform the necessary selections 1 to n^2.
Put all bags onto scale at once, record weight. As each bag is removed, take note of the difference in weight left on scale and place bags into piles accordingly.
And before you say that would be more than one weighing, the bags have to come off one by one to sort. Once on, once off. One weighing ;->
I believe the bags with the most marbles out weigh the bags that are almost empty.
Oh, wait, did you say the bags had the same number of marbles?
I posted on the floating hourglass.
Maybe the bag with the floating marbles was the lightest.
Hi WoO (would you I abbreviate you to Wiz?). You got the right idea about the secret assumption.
You've caught me off guard with that answer. I'm sad to say it doesn't quite work. If bags 3 and 4 contained 11 g marbles, then the excess weight would be 9+16=25 g. So it would be the same as if only bag 5 contained 11 g marbles. The same problem will occur whenever the three possibilities form a Pythagarean triple.
Anononymous. Nice try ;)
LOL. I love your humour.
Chris, let's make this easy, shall we?
Collection of bags = 10 (could be less or more but this is making it easy )
Each bag contains 10 marbles (again, could be less or more but, easy does it)
Let's leave the ratio of 11g marble bags to 10g marble bags to chance
Place all bags onto scale, note weight. As each bag is removed from the scale, note the weight of remaining bags. Subtract from weight of total, if divisible by 11, place in 11g slot. If not, place in 10g.
See, this is where the 'Trick Of Mind' comes in. I only used 'one weighing' ;->
LOL. You're very insistent. Thanks for drawing it to my attention. LOL.
OOOps. Crossed posts. my 6:24 PM post was intended for Ragknot. But the same goes for you Anonymous, we must go on meeting like this. LOL.
Hi Chris,
I realised after I went out that I'd overlooked Pythagoras in my rush to be the first to post.
Maybe this approach works for higher powers? Are sums of cubes unique? What about fourth or higher powers?
Feel free to call me Wiz. No-one out here knows who I am.
Hi Wiz. I knew there was a good chance you'd realise the problem with your post. The cubes and higher will work courtesy of Fermat's last theorem. I hadn't expected Fermat to crop up when I posted the problem - so that's a result, thank you for that. So that means: there's yet another way. The other way uses less marbles.
I'm knocking off very soon (it's almost 5 am here). Greetz.
Hi Wiz. I remain sure that your solution is good, especially as Fermat's last theorem is true. It has been known for a considerable time that cubes will do the trick.
It is possible to make a table of all possible results (for a finite collection of bags of marbles). The table will look strange though.
But I'd still like to see someone post the solution I was expecting.
I think this is the solution that you were looking for - I first saw this puzzle in a maths textbook (without an answer).
Take 1 marble from bag one, 2 from bag two, 3 from bag three and so on until you have marbles from all bags (the marbles from each bag must stay together and not mix).
Weigh. The bag that has the lighter marbles is indicated by the weight in grams less than what the weight would have been if the marbles had all weighed 11 grams.
e.g you have 8 bags
8+7+6+5+4+3+2+1 marbles = 36 marbles
36*11=396
if bag 6 is lighter then the weighing will say 390.
The essayist
Hi Anonymous/the essayist. You can post using a name.
Unfortunately your method doesn't work. If bags 1+2 were heavy, you'd get the same reading as if only bag 3 were heavy. Thanks for trying though.
Perhaps this puzzle wan't as well known as I had anticipated :)
Slightly enigmatic clue: it's all 0s and 1s.
Ah, of course, thanks for that last clue, Chris.
Take 1 from the first bag, 2 from the second, 4 from the third and so on up to 2^(n-1) from the last of n bags. Total number of marbles taken from these bags is 2^n - 1.
Weigh the ones you've taken out. The excess of this weight over 10 x (2^n - 1) in grams will, when converted to binary, give a string of 0's and 1's.
A zero in any position, starting from the right, means that the corresponding bag has 10 gram marbles, and a 1 in any position mrans that that bag has 11 gram marbles.
The assumption, as I said for my earlier attempt, is that the bags can be arranged so that there are enough marbles in each to make these selections.
Yes, I had seen puzzles like this before. I just went for the wrong power series!
Incidentally, the "Fermat" approach mentioned earlier doesn't work any better for powers of 3 than for 1 or 2. Because 3^3 + 4^3 + 5^3 = 216 = 6^3 you couldn't distinguish bags 3, 4 and 5 being heavier from just bag 6 being heavier. Maybe higher powers might work, but you'd need an awful lot of marbles and scales that could weigh many kilograms to 1 gram accuracy.
Hi Wiz. Did you remember to kick yourself when you got it. I liked your home goal. I've obviously mis-remembered something, and believed that Fermat's LT immediately generalized to u³+v³+w³=z³ etc., also wasn't possible. Thanks for spotting that one. You've sorted me out.
You look like you are about to stumble on a new conjecture yourself. I suspect that the higher powers are no better either, but I really don't know.
I was really stunned at your first try, in part because I thought you had a valid answer, but also in part because I thought you would have got the binary one first. Greetz.
I just spotted your Wizardean quadruple: 3,4,5,6. The mind boggles at where that could lead.
Perhaps Ragknot might like to play with that. (hint, hint).
whern the wizard said at 5.53 'The excess of this weight over 10 x (2^n - 1) in grams'
what do you mean by the 'excess' of the weight ?
is it the mass, or just the part after the decimal point?
or summat else?
Hi Anonymous. The Wizard meant the excess 1 gram for each of the the heavy marbles.
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