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Sixty-one marbles or fewer

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A sealed, opaque and weightless box contains marbles of uniform weight and no more than 61 in number. I give you a precision two-pan balance and offer you $50 to tell me how many marbles the box contains. You may purchase any number of calibrated weights at $10 each and in units of your choice to assist you. Postage and handling for the calibrated weights is a flat $2 charge.

Will you accept the offer, and what is your reasoning?

Edited by bonanova
Change "bag" to "box."
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Posted · Report post

I do not accept.

If I find out the full weight is i.e. 40, there can be 10 marbles weighting 4 each, or 1 marble weighting 40 and so on

.

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Posted · Report post

I do not accept.

If I find out the full weight is i.e. 40, there can be 10 marbles weighting 4 each, or 1 marble weighting 40 and so on

.

choose to order calibrated weights in marble units. Units are your choice. Then a calibrated weight of seven marble units would balance seven marbles. Would you accept the offer then?
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Posted (edited) · Report post

I do not accept.

If I find out the full weight is i.e. 40, there can be 10 marbles weighting 4 each, or 1 marble weighting 40 and so on

.

choose to order calibrated weights in marble units. Units are your choice. Then a calibrated weight of seven marble units would balance seven marbles. Would you accept the offer then?

I cannot purchase more than 4 weigths.

I do not need to weight 61 (nor any other odd weight) if I can weight 60.

a+b+c+d=60

a+b+c=58, => d=2

This should be solvable by brute force or even by the Monte Carlo method, but I hope to find something better.

Edited by harey
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Posted (edited) · Report post

With d=2 and c=6, I can weight:


2=d
4=c-d=6-2
6=c=6
8=c+d=6+2=8

(It determinates odd numbers are between 1 and 9.)

To weight 10, I need a 3rd weight, I can go up to: b=10+c+d=18

10=18-6-2
12=18-6
14=18-6+2
16=18-2
18=18
20=18+2
22=18+6-2
24=18+6
26=18+6+2

To weight 28, I need a 4th weight, I can go up to: a=28+b+c+d=28+18+6+2=54

That's quite strange because with c,b,d, I can make any weight up to 26, so we can go with the marbles up to 81 (and usually problems are designed this way). So I continue.

28=54-18-6-2
30=54-18-6
32=54-18-6+2
34=54-18-2
36=54-18
38=54-18+2
40=54-18+6-2
42=54-18+6
44=54-18+6+2
46=54-6-2
48=54-6
50=54-6+2
52=54-2
54=54
56=54+2
58=54+6-2
60=54+6

4 weights, 3 postages, I did not earn much today ;)

I checked and rechecked... A typo? Or a try to misslead us?
Edited by harey
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Posted · Report post

It appears that I would need 5 weights to discriminate the number of marbles, which would put me at a $2 loss due to postage.



I need a weight equal to 2 marbles in order to distinguish the cases of 1 marble from 2 marbles from more than two marbles.

To distinguish 3 from 4 from more, however, I'd need a 2nd weight equal to 4 marbles.

Then to distinguish 5 from 6 from more I can combine previous weights, but to discriminate 7 from 8 from more I need a 3rd weight equal to 8 marbles.

The binary pattern continues: I need a 4th weight equal to 16 marbles, and a 5th equal to 32.
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Posted · Report post

It appears that I would need 5 weights to discriminate the number of marbles, which would put me at a $2 loss due to postage.

I need a weight equal to 2 marbles in order to distinguish the cases of 1 marble from 2 marbles from more than two marbles.

To distinguish 3 from 4 from more, however, I'd need a 2nd weight equal to 4 marbles.

Then to distinguish 5 from 6 from more I can combine previous weights, but to discriminate 7 from 8 from more I need a 3rd weight equal to 8 marbles.

The binary pattern continues: I need a 4th weight equal to 16 marbles, and a 5th equal to 32.

I do. I can do it with 4 weights, of values 2, 6, 18, and 54.

I can use the 2w to discriminate cases of 1 marble, 2 marbles, or more. If I then add the 2w to the same side of the scale as the box, I can use the 6w to discriminate 3 marbles from 4 from more. Removing the 2w, I can use the 6w to discriminate 5 from 6 from more. Then, adding the 2w and 6w together I can tell if there are 7 marbles, 8 or more.

But now I need a third weight that can give me a net weight of 10 marbles, to distinguish cases of 9 from 10 from more. So I choose 18w, since I can counterbalance it with the 2w and 6w to yield a net of 10.

By similar combinations, these three weights allow me to discriminate up to 2+6+18 = 26 marbles. If there are still more than this, I need a 4th weight that can yield an equivalent of 28 marbles. So I choose 54, since I can counterbalance it with the 2w, 6w, and 18w to yield the desired weight.

By combining these weights I can measure precisely how many marbles are in the bag, at an $8 profit.

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