[Guest]

There is a new kind of balance scale that has three weighing pans. When you put things in the pans, the pans will indicate the relative heaviness of each thing or group of things being weighed. For example, if something in pan one weighed 3, and the thing in pan two weighed 4 and the thing in pan three weighed 5, pan three would hang lower than pan two which would hang lower than pan one. Likewise, if you weighed things that were 4, 4 and 2, the pan with the 2 in it would hang higher than the other two pans, which would hang at the same height.

There are a group of marbles that all look exactly alike. Genuine marbles weigh 10. But, you know there is a counterfeit marble that weighs 9 and another counterfeit marble that weighs 11. If you can only use the scale twice, what is the maximum number of marbles you can have, where you would be guaranteed to identify the light and heavy marble?

[Guest]

Apologies, bad proofreading job. You can use the scale three times.

[Guest]

If you do loads, and have the 9 and 11 in the same pile, wont it weigh the same as all of the others?

[Guest]

Yea. That's what makes this one tricky.

[Guest]

Then:

6 because then you would be able to work past any weighing issues like the one I mentioned above

[Guest]

Then:

6 because then you would be able to work past any weighing issues like the one I mentioned above

You would be able to do 6 marbles in 2 weighings easily like this. However, if you read my first reply, you are allowed 3 weighings, which will allow for many more:

Put two marbles in each pan. If they are all equal, then remove one marble from each pan. Whichever pan is different will tell you everything you need to know. If it's light, then that's the light marble and the one you took off that pan is the heavy marble.

If the pans are unequal, then you know that one of the marbles is heavy in the heavy pan and one of the marbles in the light pan is light. Discard the marbles form the pan in the middle, then individually weigh 3 of the 4 marbles from the unequal pans. If the pans are light, normal and heavy, then it's easy. If one is heavy and the other two are equal, then the light marble is the one of the four you didn't weigh.

[Guest]

Then:

Do the same as what you just said but with 9

[superprismatic]

I can easily do 10.

[bushindo]

Minor improvement

I can do 12. I think that is near the limit. I am working on possible improvement.

Edited March 20, 2011 by bushindo

[bushindo]

Minor improvement

I can do 12. I think that is near the limit. I am working on possible improvement.

I think I can do 16. I'm about to board an airplane now, so I'll elaborate later.

[Guest]

I think I can do 16. I'm about to board an airplane now, so I'll elaborate later.

I agree with this number. I'll let bishindo post his solution first. I'll post mine later if we don't hear form him today.

[bushindo]

I agree with this number. I'll let bishindo post his solution first. I'll post mine later if we don't hear form him today.

I now can do 1 better than that

I can do 17. Some preliminary details first. Let's label the marbles 1 to 17. For the first weighting, put marbles 1-4 onto pan number 1. Put marbles 5-8 onto pan number 2, and then put marbles 9-12 onto pan number 3. We can represent this weighing with the following notation

     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14] [,15] [,16] [,17]

W1    1    1    1    1    2    2    2    2    3     3     3     3     0     0     0     0     0

where the row indeces represent the marble indeces, and the number beneath each row index represents which pan it goes in. We use 0 to indicate that a marble is not included in the weighing. The first weighing will result in 3 categories of outcome. The first is that all three pans will have the same height. The second outcome is that exactly two pans will weight the same, and the last outcome is that the three pans have 3 different heights. The first situation (the three pans all weight the same) is the most difficult, because it has the most potential marble configurations. We solve the first situation, and the remaining two will be trivial. So, suppose that the first weighting shows that all three pans have the same height. That means that the heavy and light marbles MUST occur together within marbles 1-4, or within marbles 5-8, or within marbles 9-12, or within marbles 13-17. We take advantage of that fact, and rearrange the marbles in the second weighing so that the heavy and light marbles are forced to be in different pans (unless the heavy and light marbles occur exactly at marbles 13 and 17). So, for the second weighing, assign the following marbles to the pans

     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14] [,15] [,16] [,17]

W1    1    1    1    1    2    2    2    2    3     3     3     3     0     0     0     0     0

W2    1    2    3    0    2    3    0    1    3     0     1     2     0     1     2     3     0

Depending on the result of weighing 2, we can determine 4 or 5 possible heavy/light pairs, and then we can narrow it down with the last weighing.

For instance, let's say that the 3 pans all weight in same in weighing 2, that means that marble 13 and marble 17 is the heavy/light pair, and we can determine which is which with the last weighing.

For another example, suppose that the the second weighing gives for following heights for pans 1,2, and 3 - (a, a, b), where b < a. This means that for the second weighing, the light marble must occur in pan 3, and the heavy marble must be in the unweighted group. Looking the the table above, we can work out that the heavy/light marbles must be one of the following pairs.

(13+, 16-)

(17+, 16-)

(4+ , 3-)

(7+ , 6-)

(10+, 9-),

where (10+, 9-), for example, means that marble 10 is the heavy marble, and marble 9 is the light one. Remember only one of those 5 pairs are true, and the rest all weight 10 units. We can use that fact to narrow those 5 down with the 3rd weighing. So for the last weighing, put marble 13 and marble 3 on pan 1, put marbles 17 and marble 6 on pan 2, and then put marble 9 and marble 1 (which is now known to be a *good* marble) on pan 3. Depending on the light/heaviness of the pans, we can now tell which configuration it is.

Edited March 21, 2011 by bushindo

[Guest]

take 9 of the marbles and put 3 in each pan. You will either get "all equal", or one different- 2 equal, or all different.

- If all are different, it's a simple matter to establish which of the light 3 is the light one, and which of the heavy 3 is the heavy one.

- If only one is different (say it's light), take two of the light three, omitting the other one, and weigh against the remaining 4 split into two pans. This will ID a pair that must be heavy, and either a pair that must be light, or else the omitted marble is light. Remove one marble from each pan to ID both heavy and light.

- Now the hard part: Suppose first weighing gives equal weights. Call the 9 marbles A1,A2,A3, B1,B2,B3, and C1,C2,C3. Re-group the marbles like this: A1B2C3, B1C2A3, C1A2B3 (second weighing).

- If still equal, the heavy and light are with the remaining 4. Pick 3 and weigh (3rd weighing), and the answer will emerge.

- If not, you will have a heavy and a light and a middle pan in the second weighing. Say it's H,L,M per the order A1B2C3, B1C2A3, C1A2B3. Thus, C1A2B3 are all normal, and can be ignored. Either (1) A1=H and A3=L, or (2) B2=H and B1=L, or (3) C3=H and C2=L.

- Set the A's aside, and weigh as follows (third weighing): B2, B1,C3. (Set C2 aside as well). If (2) as defined above you will get H,L,M and conversely. If (1) you will get M,M,M and conversely. If (3) you will get M,M,H and conversely.

[plainglazed]

First compare 6 marbles on each pan.
Let's call them a1,a2,a3,a4,a5,a6 vs b1,b2,b3,b4,b5,b6 vs c1,c2,c3,c4,c5,c6
One of three general outcomes can occur:

I. If one pan is high or low and the other two equal:
-lets say pan "a" was low meaning one of a1,a2,a3,a4,a5,a6 is heavy and the 19th marble must be light.
compare a1 vs a2 vs a3 to see which one might be heavy then, if those are equal,
compare a4 vs a5 vs a6 to find the heavy marble

II. If one pan is low indicating a heavy marble and one pan is high indicating a light marble.
-lets say pan "a" is low so one of a1,a2,a3,a4,a5,a6 is heavy and pan "b" is high so one of b1,b2,b3,b4,b5,b6 is light

compare a1,a2,X vs a3,a4,X vs b1,b2,b3 where X is a previously established "normal" marble

if all are equal, then a5 or a6 is heavy and either b4 or b5 or b6 is light so one of the following must be true:
a5H,b4L
a5H,b5L
a5H,b6L
a6H,b4L
a6H,b5L
a6H,b6L
compare a5 vs b4 vs b5
if they are are equal then a6 is heavy and b6 is light
if only a5 is heavy then b6 is light
if only b4 is light then a6 is heavy
if only b5 is light then a6 is heavy
or a5 is heavy and b4 is light
or a5 is heavy and b5 is light

if a1,a2,X (or a3,a4,X) is heavy and b1,b2,b3 is light then one of the following heavy,light combinations must be true:
a1H,b1L
a1H,b2L
a1H,b3L
a2H,b1L
a2H,b2L
a2H,b3L
compare a1 vs b1 vs b2
if they are are equal then a2 is heavy and b3 is light
if only a1 is heavy then b3 is light
if only b1 is light then a2 is heavy
if only b2 is light then a2 is heavy
or a1 is heavy and b1 is light
or a1 is heavy and b2 is light

if a1,a2,X (or a3,a4,X) is heavy and b1,b2,b3 is "normal" then either b4 or b5 or b6 is light so either:
a1H,b4L
a1H,b5L
a1H,b6L
a2H,b4L
a2H,b5L
a2H,b6L
compare a1 vs b4 vs b5 similarly to above

III. If all three pans are equal then BOTH the heavy and light marble are in one of
a1,a2,a3,a4,a5,a6 or b1,b2,b3,b4,b5,b6 or c1,c2,c3,c4,c5,c6.

compare a1,b1,c1,a4 vs a2,b2,c2,b4 vs a3,b3,c3,c4

if all are equal then the heavy and light marble are either
a1,a4
b2,b4
c3,c4
a5,a6
b5,b6
c5,c6
compare a1,b2,c3 vs b4,a6,b6 vs c4,b5,c5
if only a1,b2,c3 is light then a1L,a4H
if only a1,b2,c3 is heavy then a1H,a4L
if only b4,a6,b6 is light then a6L,a5H
if only b4,a6,b6 is heavy then a6H,a5L
if only c4,b5,c5 is light then c5L,c6H
if only c4,b5,c5 is heavy then c5H,c6L
if a1,b2,c3 is heavy and b4,a6,b6 is light then b2H,b4L
if a1,b2,c3 is light and b4,a6,b6 is heavy then b2L,b4H
if b4,a6,b6 is heavy and c4,b5,c5 is light then b6H,b5L
if b4,a6,b6 is light and c4,b5,c5 is heavy then b6L,b5H
if c4,b5,c5 is heavy and a1,b2,c3 is light then c3L,c4H
if c4,b5,c5 is light and a1,b2,c3 is heavy then c3H,c4L

if say a1,b1,c1,a4 is heavy and a2,b2,c2,b4 is light then one of the following heavy,light combinations must be true:
a1H,a2L
a4H,a2L
b1H,b2L
b1H,b4L
c1H,c2L
compare a2,b1 vs a4,b4 vs c1,c2
if all are equal then c1H,c2L
if only a2,b1 is light then a1H,a2L
if only a2,b1 is heavy then b1H,b2L
if a2,b1 is heavy and a4,b4 is light then b1H,b4L
if a2,b1 is light and a4,b4 is heavy then a4H,a2L

if only one pan is heavy or light, say a1,b1,c1,a4 is heavy then one of a5, a6, b5, b6, c5, c6 is light so the following combinations are possible:
a1H,a5L
a1H,a6L
a4H,a5L
a4H,a6L
b1H,b5L
b1H,b6L
c1H,c5L
c1H,c6L
compare a1,c5 vs b1,a6 vs c1,b5
if all are equal then a4H,a5L
if only a1,c5 is heavy then a1H,a5L
if only b1,a6 is light then a4H,a6L
if only b1,ah is heavy then b1H,b6L
if only c1,b5 is heavy then c1H,c6L
if a1,c5 is heavy and b1,a6 is light then a1H,a6L
if b1,a6 is heavy and c1,b5 is light then b1H,b5L
if c1,b5 is heavy and a1,c5 is light then c1H,c6L



am skeptical that I have not screwed something up but have been looking at it too long to find out where.

a case for 19?

[bushindo]

a case for 19?


First compare 6 marbles on each pan.

Let's call them a1,a2,a3,a4,a5,a6 vs b1,b2,b3,b4,b5,b6 vs c1,c2,c3,c4,c5,c6

One of three general outcomes can occur:


I.  If one pan is high or low and the other two equal:

-lets say pan "a" was low meaning one of a1,a2,a3,a4,a5,a6 is heavy and the 19th marble must be light.

compare a1 vs a2 vs a3 to see which one might be heavy then, if those are equal,

compare a4 vs a5 vs a6 to find the heavy marble


II. If one pan is low indicating a heavy marble and one pan is high indicating a light marble.

-lets say pan "a" is low so one of a1,a2,a3,a4,a5,a6 is heavy and pan "b" is high so one of b1,b2,b3,b4,b5,b6 is light


compare a1,a2,X vs a3,a4,X vs b1,b2,b3 where X is a previously established "normal" marble


if all are equal, then a5 or a6 is heavy and either b4 or b5 or b6 is light so one of the following must be true:

  a5H,b4L

  a5H,b5L

  a5H,b6L

  a6H,b4L

  a6H,b5L

  a6H,b6L

compare a5 vs b4 vs b5

  if they are are equal then a6 is heavy and b6 is light

  if only a5 is heavy then b6 is light

  if only b4 is light then a6 is heavy

  if only b5 is light then a6 is heavy

  or a5 is heavy and b4 is light

  or a5 is heavy and b5 is light


if a1,a2,X (or a3,a4,X) is heavy and b1,b2,b3 is light then one of the following heavy,light combinations must be true:

  a1H,b1L

  a1H,b2L

  a1H,b3L

  a2H,b1L

  a2H,b2L

  a2H,b3L

compare a1 vs b1 vs b2

  if they are are equal then a2 is heavy and b3 is light

  if only a1 is heavy then b3 is light

  if only b1 is light then a2 is heavy

  if only b2 is light then a2 is heavy

  or a1 is heavy and b1 is light

  or a1 is heavy and b2 is light


if a1,a2,X (or a3,a4,X) is heavy and b1,b2,b3 is "normal" then either b4 or b5 or b6 is light so either:

  a1H,b4L

  a1H,b5L

  a1H,b6L

  a2H,b4L

  a2H,b5L

  a2H,b6L

compare a1 vs b4 vs b5 similarly to above  


III. If all three pans are equal then BOTH the heavy and light marble are in one of 

	a1,a2,a3,a4,a5,a6 or b1,b2,b3,b4,b5,b6 or c1,c2,c3,c4,c5,c6.


compare a1,b1,c1,a4 vs a2,b2,c2,b4 vs a3,b3,c3,c4


if all are equal then the heavy and light marble are either

  a1,a4

  b2,b4

  c3,c4

  a5,a6

  b5,b6

  c5,c6

compare a1,b2,c3 vs b4,a6,b6 vs c4,b5,c5

  if only a1,b2,c3 is light then a1L,a4H

  if only a1,b2,c3 is heavy then a1H,a4L

  if only b4,a6,b6 is light then a6L,a5H

  if only b4,a6,b6 is heavy then a6H,a5L

  if only c4,b5,c5 is light then c5L,c6H

  if only c4,b5,c5 is heavy then c5H,c6L

  if a1,b2,c3 is heavy and b4,a6,b6 is light then b2H,b4L

  if a1,b2,c3 is light and b4,a6,b6 is heavy then b2L,b4H

  if b4,a6,b6 is heavy and c4,b5,c5 is light then b6H,b5L

  if b4,a6,b6 is light and c4,b5,c5 is heavy then b6L,b5H

  if c4,b5,c5 is heavy and a1,b2,c3 is light then c3L,c4H

  if c4,b5,c5 is light and a1,b2,c3 is heavy then c3H,c4L


if say a1,b1,c1,a4 is heavy and a2,b2,c2,b4 is light then one of the following heavy,light combinations must be true:

  a1H,a2L

  a4H,a2L

  b1H,b2L

  b1H,b4L

  c1H,c2L

compare a2,b1 vs a4,b4 vs c1,c2

  if all are equal then c1H,c2L

  if only a2,b1 is light then a1H,a2L

  if only a2,b1 is heavy then b1H,b2L

  if a2,b1 is heavy and a4,b4 is light then b1H,b4L

  if a2,b1 is light and a4,b4 is heavy then a4H,a2L 


if only one pan is heavy or light, say a1,b1,c1,a4 is heavy then one of a5, a6, b5, b6, c5, c6 is light so the following combinations are possible:

  a1H,a5L

  a1H,a6L

  a4H,a5L

  a4H,a6L

  b1H,b5L

  b1H,b6L

  c1H,c5L

  c1H,c6L

compare a1,c5 vs b1,a6 vs c1,b5

  if all are equal then a4H,a5L

  if only a1,c5 is heavy then a1H,a5L

  if only b1,a6 is light then a4H,a6L

  if only b1,ah is heavy then b1H,b6L

  if only c1,b5 is heavy then c1H,c6L

  if a1,c5 is heavy and b1,a6 is light then a1H,a6L

  if b1,a6 is heavy and c1,b5 is light then b1H,b5L

  if c1,b5 is heavy and a1,c5 is light then c1H,c6L




am skeptical that I have not screwed something up but have been looking at it too long to find out where.

The logic seems good to me. The first part of step III is particularly nice as it extracts maximum information out of the 3rd weighing. Great work, plainglazed!

Edited March 21, 2011 by bushindo

[Guest]

a case for 19?


First compare 6 marbles on each pan.

Let's call them a1,a2,a3,a4,a5,a6 vs b1,b2,b3,b4,b5,b6 vs c1,c2,c3,c4,c5,c6

One of three general outcomes can occur:


I.  If one pan is high or low and the other two equal:

-lets say pan "a" was low meaning one of a1,a2,a3,a4,a5,a6 is heavy and the 19th marble must be light.

compare a1 vs a2 vs a3 to see which one might be heavy then, if those are equal,

compare a4 vs a5 vs a6 to find the heavy marble


II. If one pan is low indicating a heavy marble and one pan is high indicating a light marble.

-lets say pan "a" is low so one of a1,a2,a3,a4,a5,a6 is heavy and pan "b" is high so one of b1,b2,b3,b4,b5,b6 is light


compare a1,a2,X vs a3,a4,X vs b1,b2,b3 where X is a previously established "normal" marble


if all are equal, then a5 or a6 is heavy and either b4 or b5 or b6 is light so one of the following must be true:

  a5H,b4L

  a5H,b5L

  a5H,b6L

  a6H,b4L

  a6H,b5L

  a6H,b6L

compare a5 vs b4 vs b5

  if they are are equal then a6 is heavy and b6 is light

  if only a5 is heavy then b6 is light

  if only b4 is light then a6 is heavy

  if only b5 is light then a6 is heavy

  or a5 is heavy and b4 is light

  or a5 is heavy and b5 is light


if a1,a2,X (or a3,a4,X) is heavy and b1,b2,b3 is light then one of the following heavy,light combinations must be true:

  a1H,b1L

  a1H,b2L

  a1H,b3L

  a2H,b1L

  a2H,b2L

  a2H,b3L

compare a1 vs b1 vs b2

  if they are are equal then a2 is heavy and b3 is light

  if only a1 is heavy then b3 is light

  if only b1 is light then a2 is heavy

  if only b2 is light then a2 is heavy

  or a1 is heavy and b1 is light

  or a1 is heavy and b2 is light


if a1,a2,X (or a3,a4,X) is heavy and b1,b2,b3 is "normal" then either b4 or b5 or b6 is light so either:

  a1H,b4L

  a1H,b5L

  a1H,b6L

  a2H,b4L

  a2H,b5L

  a2H,b6L

compare a1 vs b4 vs b5 similarly to above  


III. If all three pans are equal then BOTH the heavy and light marble are in one of 

	a1,a2,a3,a4,a5,a6 or b1,b2,b3,b4,b5,b6 or c1,c2,c3,c4,c5,c6.


compare a1,b1,c1,a4 vs a2,b2,c2,b4 vs a3,b3,c3,c4


if all are equal then the heavy and light marble are either

  a1,a4

  b2,b4

  c3,c4

  a5,a6

  b5,b6

  c5,c6

compare a1,b2,c3 vs b4,a6,b6 vs c4,b5,c5

  if only a1,b2,c3 is light then a1L,a4H

  if only a1,b2,c3 is heavy then a1H,a4L

  if only b4,a6,b6 is light then a6L,a5H

  if only b4,a6,b6 is heavy then a6H,a5L

  if only c4,b5,c5 is light then c5L,c6H

  if only c4,b5,c5 is heavy then c5H,c6L

  if a1,b2,c3 is heavy and b4,a6,b6 is light then b2H,b4L

  if a1,b2,c3 is light and b4,a6,b6 is heavy then b2L,b4H

  if b4,a6,b6 is heavy and c4,b5,c5 is light then b6H,b5L

  if b4,a6,b6 is light and c4,b5,c5 is heavy then b6L,b5H

  if c4,b5,c5 is heavy and a1,b2,c3 is light then c3L,c4H

  if c4,b5,c5 is light and a1,b2,c3 is heavy then c3H,c4L


if say a1,b1,c1,a4 is heavy and a2,b2,c2,b4 is light then one of the following heavy,light combinations must be true:

  a1H,a2L

  a4H,a2L

  b1H,b2L

  b1H,b4L

  c1H,c2L

compare a2,b1 vs a4,b4 vs c1,c2

  if all are equal then c1H,c2L

  if only a2,b1 is light then a1H,a2L

  if only a2,b1 is heavy then b1H,b2L

  if a2,b1 is heavy and a4,b4 is light then b1H,b4L

  if a2,b1 is light and a4,b4 is heavy then a4H,a2L 


if only one pan is heavy or light, say a1,b1,c1,a4 is heavy then one of a5, a6, b5, b6, c5, c6 is light so the following combinations are possible:

  a1H,a5L

  a1H,a6L

  a4H,a5L

  a4H,a6L

  b1H,b5L

  b1H,b6L

  c1H,c5L

  c1H,c6L

compare a1,c5 vs b1,a6 vs c1,b5

  if all are equal then a4H,a5L

  if only a1,c5 is heavy then a1H,a5L

  if only b1,a6 is light then a4H,a6L

  if only b1,ah is heavy then b1H,b6L

  if only c1,b5 is heavy then c1H,c6L

  if a1,c5 is heavy and b1,a6 is light then a1H,a6L

  if b1,a6 is heavy and c1,b5 is light then b1H,b5L

  if c1,b5 is heavy and a1,c5 is light then c1H,c6L




am skeptical that I have not screwed something up but have been looking at it too long to find out where.

This thread took a few days to get going but eventually didn't disappoint. Nicely done plainglazed. About a week ago, I thought of this triple balance scale concept and how quickly you can glean a massive amount of info after a few weighings. 16 seemed like such a simple and efficient answer, but i'm not surprised that this forum blew that away. Glad I didn't spend a long time writing out my solution for 16 now. LOL

These kinds of problems are always tougher to write out than to think about.

[bushindo]

This thread took a few days to get going but eventually didn't disappoint. Nicely done plainglazed. About a week ago, I thought of this triple balance scale concept and how quickly you can glean a massive amount of info after a few weighings. 16 seemed like such a simple and efficient answer, but i'm not surprised that this forum blew that away. Glad I didn't spend a long time writing out my solution for 16 now. LOL

These kinds of problems are always tougher to write out than to think about.

How about 21?

Here's a case for 21.

Let's say that we label the marbles 1-21. In the first weighting (W1), assign each marble to a pan using the following scheme.

     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14] [,15] [,16] [,17] [,18] [,19] [,20] [,21]

W1    1    1    1    1    1    2    2    2    2     2     3     3     3     3     3     0     0     0     0     0     0

where the row indices represent the marble number, and the number below each row index indicates the pan number (1,2, or 3) that it should go to. A 0 indicates that the corresponding marble will be unweighted. There are three categories of outcomes. The first is that all three pans will have the same height. The second outcome is that exactly two pans will have the same height, and the last outcome is that the three pans have 3 different heights. We only solve the most difficult case where the three pans have the same height (this is because this case has the most potential marble configurations). The other two cases can be solved straightforwardly. So, let's say that in the first weighing, the three pans have the same height. Assign the second weighing (W2) as follows

     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14] [,15] [,16] [,17] [,18] [,19] [,20] [,21]

W1    1    1    1    1    1    2    2    2    2     2     3     3     3     3     3     0     0     0     0     0     0

W2    1    2    3    0    1    2    3    0    1     2     3     0     1     2     3     0     1     2     3     0     0

To understand what is going on, it may help to think that we have *four* pans, real pans number 1, 2, 3, and a virtual pan called 0 (this simply consists of all marbles that are unweighted). In the second weighing, we essentially take 3 marbles from each pan and distribute them to the other 3.

Again, we only examine the most difficult case, which is where the all three pans again have the same height. In this situation, we can see that there are 12 possibilities

(16+, 20-)

(16+, 21-)

(20+, 21-)

(16-, 20+)

(16-, 21+)

(20-, 21+)

(1+ , 5-)

(6+ , 10-)

(11+, 15-)

(1- , 5+)

(6- , 10+)

(11-, 15+)

where (11-, 15+), for instance, denotes that marble 11 is light, and marble 15 is heavy. Note that only 1 of these 12 possibilities is actually true, and the rest are all normal marbles. We can narrow these 12 possibilities down to 1 with the last weighing. So, for W3, put marble 16 and marble 1 on pan 1, put marble 20 and marble 6 on pan 2, and put marble 21 and marble 11 on pan 3.

If weighing 3 results in three different pan heights, then we know that the light/heavy pair must occur within the top 6 possibilities in the list above. If the weighing results in exactly 2 same heights, then we know that the light/heavy pair must occur within the bottom 6 possibilities. Depending on the exact pan-configurations, it is fairly easy single out exactly which pair it is.

In this outline we only looked at the most difficult class of outcomes, and showed that those are solvable. The remaining outcomes are not as hard, and we'll omit them here.

[Guest]

1.Put 6 marbles in each pan and weigh them (first time weighing), lets number them 1-6 in pan1, 7-12 in pan2 and 13-18 in pan3, put marble 19 aside for time being:

Thre Possible Outcomes: 1. All three pans are equal

2. All three pans are unequal

3. Two pans are equal and third one is lighter or heavier

Case 1) Take two marbles from each pan and put aside, say 1,2 from pan1, 7,8 from pan2 and 13,14 from pan3.

Now in pan 1 we have 3,4,5,6 pan2 9,10,11,12 and in pan3 15,16,17,18 numbered marbles

Lets change marbles as: Put two marbles from one pan to other two pans (one in each),say from pan1 put 3 in pan2 and 4 in pan3. Likewise...we will have:

In pan1: 5,6,9,15

pan2: 3,11,12,16

pan3: 4,10,17,18

Now weigh the balance (2nd weighing):

Here we will have 3 possible outcomes :

a) All three pans are equal

b) All three Pans are unequal

c) Two pans equal and 3rd one is heavier or lighter

Case a) As all pans are equal this time as well, so both the faulty marbles belongs to the same pan or are outside the pan which we have kept aside before 2nd weighing. Faulty marbles cannot be the one which we have moved from one pan to other (ie 3,4, 10,11, 15,16) otherwise all pans can't be equal.

So we remains with six set of marbles, each set have two marbles.

One of these sets have both faulty marbles.

These sets are (1,2), (7,8), (13,14), (5,6), (11,12), (17,18)

Now from any three sets put one marble aside from ach set...say 1, 7 and 13

Take any three sets, take one marble from each of three selected sets and put into the other set such that no two sets are getting and puting marble from/in same.

Lets take first three sets and suffles as above rule:

We will have (1,14), (2,7) and (8,13)

Now take one marble from each of the remaining three sets and put into the suffled three sets (one in each set)...

Lets say take 5, 11 and 17 from remaining sets and put one marble each into the suffled three sets... this way we will have three sets having three marbles each:

(1,14, 5), (2,7, 11) and (8,13, 17)

Lets weigh these three sets on the balance ( third weighing)

As we have suffled the marbles in the sets so both marbles can't be in the same pan and atleast one faulty marble is there in one of the pans.

So all pans can't be equal.

We will have two possibilities here:

X)One pan is heavier or lighter other two are equal

Y)All pans are unequal

Case X) It means only one faulty marble is there in the pan, and it is one among 5, 11 and 17, so see which of these marble is there in the heavier/leghter pan.

And if pan is heavier then this marble is also heavier and if pan is lighter so is the marble.

Now see the corresponding set from which we have taken this fauly marble, other marble from that set is heavier/lighter.

If 5 is the heavier then 6 is lighter...and vice-versa...

//end of solution

Case Y) It means both faulty marbles are in pan and it is from the first taken three sets (in our case (1,2), (7,8), (13,14))

Now as we have suffled such that both marbles from only one set will there in the heavier and lighter pan. So marbles from that set are lighter/heavier as per the pans.

For ex: we have weigh (1,14, 5), (2,7, 11) and (8,13, 17)

if (1,14, 5) is heavier and (8,13, 17) is lighter then only set of (13,14) is there in these two pans...and 14 is heavier and 13 is lighter.

//end of solution

Now case b) All three Pans are unequal

It means one faulty marble (heavier) is in heavier and other faulty marble (lighter) is in lighter pan.

Lets say pan1 is heavier and pan3 is lighter.

pan1: 5,6,9,15

pan2: 3,11,12,16

pan3: 4,10,17,18

It insure that the marbles in pan2 can't be faulty. Remove them from pan2.

Take two marbles which were in other pan (pan2) in first weigh from these pans ie 9 from pan1 and 10 from pan3. Keep them aside (may be used in further analysis)

Now take one marble from pan1 and pan3 which were in same pans in first weigh

(here 5 or 6 from pan1 and 17 or 18 from pan3) and weigh them with remaining marbles

Lets take 5 and 17 and put in pan2. Now we will have

Pan1: (6,15)

pan2: (5,17)

pan3: (4,18)

In pan1 and pan3 interchage one marble from same set in fist weighing.

Ex: interchange 4 and 6 or interchange 15 and 18.

we will have:

Pan1: (6,18)

pan2: (5,17)

pan3: (4,15)

Now weigh them (3rd weighing):

It can be possible that the marbles we removed (9, 10) can both be faulty or both are not faulty. Because one of them can't be faulty as these are the only two marbles remains from pan2 in the first weigh (and from that weigh all the faulty marbles are in same pan).

So there are two possible results of this weighing:

i) All the pans are equal.

j) All pans are unequal

Case i)

No two pans have marbles from the same set as in first weigh. So both faulty marbles can't be in same pan. it means faulty marbles are 9 and 10 and lighter and heavier as per the 2nd weigh results.

Case j)

We can find out lighter and heavier marble from heavier and lighter pan now.

if the results are same in 2nd weighing (ie pan1 is heavy and pan3 is light)

then the marbles which we havn't interchanged are heavier and lighter.

In this case 6 is heavy and 4 is light.

if the results are opposite as in 2nd weighing (ie pan3 is heavy and pan1 is light) it means the faulty marbles are those which we have interchanged:

In this case 18 is heavy and 15 is light.

other possible cases are:

- pan2 heavier and pan1 is lighter

- pan2 lighter and pan1 is heavier

- pan2 heavier and pan3 is lighter

- pan2 lighter and pan3 is heavier

if the pan2 is heavier it means we have put the faulty marble from the heavier pan in from 2nd weighing.

In this case marble number 5 from pan1 which was heavier in 2nd weighing, so it is the faulty heavier marble and lighter will be from the same set as in 1st weighing and can be picked from pan1 or pan3 depending upon wheather pan1 is lighter or pan3 is lighter.

Ex: if pan1 is lighter then marble number 6 will be lighter.

//End of solution

c) Two pans equal and 3rd one is heavier or lighter

It means One faulty marble is there in heavier/lighter set and one faulty marble is from the marbles which we have kept aside from each set of 1st weihing, ie (1,2) , (7,8) , (13,14).

We can ignore the marbles from two ewual pans.Now take the marbles from heavier/lighter pan in 2nd wighing. Among these marbles two are from the same set as in first weighing, take one out of these two marbles and keep aside (will be used in further analysis):

Say if the from the set of 2nd weighing:

pan1: 5,6,9,15

pan2: 3,11,12,16

pan3: 4,10,17,18

pan3 is the lighter/heavier pan. Now take 17 or 18 marble and keep aside.

Lets say we keep 18 marble aside.

Now we have 4,10,17 each from different set. put one marble from these three marbles in each pan.

Now from the remaining marbles ie (1,2) , (7,8) , (13,14) keep the marbles in the pan where there the marble from the same set of 1st weighing is not present.

Lets say we keep as:

Pan1: 4

pan2: 10

pan3: 17

Now put marble 1 and 2 in pan2 and pan3 as there is no marble from the same set as in 1st weighing:

Likewise we will have:

Pan1: 4,7,13

pan2: 10,2,14

pan3: 17,8,1

or similar combination where we can swap marble 1 and 2, or/and 7 and 8 or/and 13 and 14.

Now weigh them:

There can be two possible results:

o) All pans are unequal

p) two pans are equal and other one is lighter/heavier

Case o)

As we have chose pan3 lighter/heavier in 2nd weighing.

Lets say it was lighter. As we have distributed the three marbles in differnet pans so the pan which is lighter is beacuse of one of the marbles 4,10 or 17.

So we can say that marble is lighter.

Ex. if pan3 is lighter, it is because of marble 17 as the pan which contains this marble is 2nd weiging was lighter.

So we can say marble 17 is lighter. Now the heavier marble must be from the same set (1st weighing).

So, if pan1 is heavier then marble 13 is heavier.

and if pan2 is heavier then marble 14 is heavier.

case p)

This can be only possible if the marble which we have kept aside (marble 18 in this case) is faulty (heavier/lighter).

Only pan1 and pan2 can be heavier/lighter in this case. beacuse marble 17 can't be faulty otherwise we could not have landed here. If both 17 and 18 would be faulty then result of 2nd weigh would have been all equal.

As other faulty marble must be from the same set (1st weighing) so the marble 13 or 14 could be the other faulty marble.

So, if pan1 is lighter/heavier then marble 13 is lighter/heavier and other marble is already identified which is marble 18.

if pan1 is lighter then marble 13 is lighter and marble 18 heavier and vice-verca.

And if pan2 is lighter then marble 14 is lighter and marble 18 heavier and vice-verca.

//end of solution

Case 2) All three pans are unequal

We can ingnore six marbles from the medium pan. Lets say pan1 is lighter and pan2 is heavier then we can ignore pan3 marbles.

Take two marbles from pan1, another two marbles from pan1 and three marbles from pan2.

We will have:

Pan1: 1,2

pan2: 3,4

pan3: 7,8,9

We will use remaining marbles two from 1st set (5,6) and three from 2nd set (10,11,12) for further analysis.

Now ou

lets weigh them( 2nd weighing):

There are 3 possible results:

l) All pans are equal

m) All pans unequal

n) Two pans are equal and one is heavier/lighter

case l) It means faulty marbles are one from (5,6) set and other from (10,11,12) which we kept aside before this weighing.

case m)

It will also give us two sets (1,2) or (3,4) and (7,8,9)

case n) It will also give us a similar two sets.

For all three cases (case l, case m and case n), as we have two sets one of two marbles and one of three marbles...it can be solved as....

For example: We have following sets (1,2) and (7,8,9)

keep from marble from each set say 1 and 9 aside and weigh remaing three marbles:

In this case:

pan1: 2

pan2: 7

pan3: 8

We can easily distinguish heavier and lighter marbles....depeding upon results......

//End of solution

Case 3) Two pans are equal and third one is lighter or heavier

It means marble 19 which we kept aside is faulty and heavier or lighter depends upon the above result.

Remaining one faulty marble among six marbles can be found easily as we have two weighing left...

///End of solution

[plainglazed]

Do I hear 22?

First compare 6 marbles on each pan.
Let’s call them a1,a2,a3,a4,a5,a6 vs b1,b2,b3,b4,b5,b6 vs c1,c2,c3,c4,c5,c6 with d1,d2,d3,d4 not yet weighed

When a1,a2,a3,a4,a5,a6 balances with b1,b2,b3,b4,b5,b6 and c1,c2,c3,c4,c5,c6 the light and heavy marbles must BOTH be in either a1-a6, b1-b6, c1-c6, or d1-d4

compare a1,a2,b1,b2,c1,d1 vs a3,a4,b3,c2,c3,d2 vs a5,b4,b5,c4,c5,d3 with a6,b6,c6,d4 left out of this weighing

if again all are equal the heavy and light marbles must be in one of the following:
a1,a2
b1,b2
a3,a4
c2,c3
b4,b5
c4,c5
compare a1,b1,a3 vs b2,c3,b5 vs a4,b4,c4
if only a1,b1,a3 is light then a1L,a2H
if only a1,b1,a3 is heavy then a1H,a2L
if only b2,c3,b5 is light then c2H,c3L
if only b2,c3,b5 is heavy then c2L,c3H
if only a4,b4,c4 is light then c4L,c5H
if only a4,b4,c4 is heavy then c4H,c5L
if a1,b1,a3 is heavy and b2,c3,b5 is light then b1H,b2L
if a1,b1,a3 is light and b2,c3,b5 is heavy then b1L,b2H
if b2,c3,b5 is heavy and a4,b4,c4 is light then b4L,b5H
if b2,c3,b5 is light and a4,b4,c4 is heavy then b4H,b5L
if a4,b4,c4 is heavy and a1,b1,a3 is light then a3L,a4H
if a4,b4,c4 is light and a1,b1,a3 is heavy then a3H,a4L


In bushindo notation:

[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14] [,15] [,16] [,17] [,18] [,19] [,20] [,21] [,22]
W1 1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3 0 0 0 0
W2 1 1 2 2 3 0 1 1 2 3 3 0 1 2 2 3 3 0 1 2 3 0


(1+,2-)
(1-,2+)
(3+,4-)
(3-,4+)
(7+,8-)
(7-,8+)
(10+,11-)
(10-,11+)
(14+,15-)
(14-,15+)
(16+,17-)
(16-,17+)

thus far I have skipped all other outcomes except the case where all are twice equal. suspect that would hold up for this case but am checking...

[Guest]

yes, i guess I got a solution for 22, will post tomorrow......

[Guest]

[spoiler='Solution for 22 marbles....']

Solution for 22 marble...may be optimized further....

1.Put 6 marbles in each pan and weigh them (first time weighing), lets number them 1-6 in pan1, 7-12 in pan2 and 13-18 in pan3, put extra four marbles aside for time being

Lets name them 19, 20, 21, 22:

Three Possible Outcomes:

1. All three pans are equal

2. All three pans are unequal

3. Two pans are equal and third one is lighter or heavier

Case 1) Take two marbles from each pan and put aside, say 1,2 from pan1, 7,8 from pan2 and 13,14 from pan3.

Now in pan 1 we have 3,4,5,6 pan2 9,10,11,12 and in pan3 15,16,17,18 numbered marbles

Lets change marbles as: Put two marbles from one pan to other two pans (one in each),say from pan1 put 3 in pan2 and 4 in pan3. Likewise...we will have:

In pan1: 5,6,9,15

pan2: 3,11,12,16

pan3: 4,10,17,18

Now add marbles numbers 19, 20, 21 in each of pans.

We have now:

In

pan1: 5,6,9,15, 19

pan2: 3,11,12,16, 20

pan3: 4,10,17,18, 21

Now weigh the balance (2nd weighing):

Here we will have 3 possible outcomes :

a) All three pans are equal

b) All three Pans are unequal

c) Two pans equal and 3rd one is heavier or lighter

Case a)

As all pans are equal this time as well, we can conclude that the faulty marbles are not among the 19, 20, 21 as we have added these marbles in different pans. We can ignore these marbles.

so both the faulty marbles belongs to the same pan or are outside the pan which we have kept aside before 2nd weighing. Faulty marbles cannot be the one which we have moved from one pan to other (ie 3,4, 10,11, 15,16) otherwise all pans can't be equal.

So we remains with six set of marbles, each set have two marbles.

One of these sets have both faulty marbles.

These sets are (1,2), (7,8), (13,14), (5,6), (11,12), (17,18)

Take any three sets, take one marble from each of three selected sets and put into the other set such that no two sets are getting and puting marble from/in same.

Lets take first three sets and suffles as above rule:

We will have (1,14), (2,7) and (8,13)

Now take one marble from each of the remaining three sets and put into the suffled three sets (one in each set)...

Lets say take 5, 11 and 17 from remaining sets and put one marble each into the suffled three sets... this way we will have three sets having three marbles each:

(1,14, 5), (2,7, 11) and (8,13, 17)

pan1: (1,14, 5)

pan2: (2,7, 11)

pan3: (8,13, 17)

Lets weigh these three sets on the balance ( third weighing)

As we have suffled the marbles in the sets so both marbles can't be in the same pan and atleast one faulty marble is there in one of the pans.

So all pans can't be equal.

We will have two possibilities here:

X)One pan is heavier or lighter other two are equal

Y)All pans are unequal

Case X) It means only one faulty marble is there in the pan, and it is one among 5, 11 and 17, so see which of these marble is there in the heavier/leghter pan.

And if pan is heavier then this marble is also heavier and if pan is lighter so is the marble.

Now see the corresponding set from which we have taken this fauly marble, other marble from that set is heavier/lighter.

If 5 is the heavier then 6 is lighter...and vice-versa...

Case Y) It means both faulty marbles are in pan and it is from the first taken three sets (in our case (1,2), (7,8), (13,14))

Now as we have suffled such that both marbles from only one set will there in the heavier and lighter pan. So marbles from that set are lighter/heavier as per the pans.

For ex: we have weigh (1,14, 5), (2,7, 11) and (8,13, 17)

if (1,14, 5) is heavier and (8,13, 17) is lighter then only set of (13,14) is there in these two pans...and 14 is heavier and 13 is lighter.

//end of solutionNow case b) All three Pans are unequal

It means one faulty marble (heavier) is in heavier and other faulty marble (lighter) is in lighter pan.Lets say pan1 is heavier and pan3 is lighter.

pan1: 5,6,9,15, 19

pan2: 3,11,12,16, 20

pan3: 4,10,17,18, 21It insure that the marbles in pan2 can't be faulty. Remove them from pan2.

The possible solutions are:

-s1) Marbles 5 or 6 is heavier and 17 or 18 is lighter

-s2) 9 is heavier and 10 is lighter

-s3) 19 is heavier and 21 is lighter

-s4) 15 is heavier and 4 is lighter

Now take one marble from possibly heavier and one from lighter marble from s1, s2 or s3, lets say we take 9 (possibly heavier) and 21 (possibly lighter) and put in pan1.

Take one marble from s1 (where there are two candidates for one solution) from lighter marbles and one from heavier marbles and put in pan2 and pan3.

We have:

Pan1: (9, 21)

pan2: 5

pan3: 17

Put one marble which we are sure is not faulty in pan2 and pan3. As we know all other marbles which are not part of s1, s2, s3 and s4 are not faulty, lets rename then E1, E2 for simplicity.

Now we will have:

Pan1: (9, 21)

pan2: (5, E1)

pan3: (17, E2)

Weigh them (third weighing):

There are three possible results:

i) All pans are equal

j) All pans are unequal

k) Two pans are equal and third is heavier/lighter

Case i) If all pans are equal then faulty marbles are from s1 which we havn't put in pans.

In our case 6 is heavier and 18 is lighter.

Case j) faulty Marbles can be foud easily here...

Case k) faulty Marbles can be foud easily here... so not elaborating

//End of solution

Case c)

Two pans equal and 3rd one is heavier or lighter

Lets say pan1 is heavier and pan3 is lighter.

pan1: 5,6,9,15, 19

pan2: 3,11,12,16, 20

pan3: 4,10,17,18, 21

It means One faulty marble is there in heavier/lighter set and one faulty marble is from the marbles which we have kept aside from each set of 1st weighing, ie (1,2) , (7,8) , (13,14) and 22

Lets say pan3 is heavier/lighter pan.

As the result of first weighing was equal for all pans, it means faulty marble is in the same set (1-6) or (7-12) or (13-18) or (19-22)

So we have following possible solutions if we have pan3 heavier:

-p1)21 is lighter and 22 is heavier

-p2)4 is lighter and 1 or 2 is heavier

-p3)10 is lighter and 7 or 8 is heavier

-p4)17 or 18 is lighter and 13 or 14 is heavier

Now distribute the possible faulty marbles as:

pan1: one of the heavier possible marble from p2 or p3 or p4

pan2: one of the heavier possible marble from p2 or p3 or p4 but not from same as in pan1

pan1: one heavier and one lighter possible marble from p2 or p3 or p4

This way lets say we chose like this:

pan1: 1

pan2: 13

pan3: 18 and 7

Now put one ball (which we know is not faulty) each in pan1 and pan2, lets name them E1 and E2:

Now we have:

pan1: 1, E1

pan2: 13, E2

pan3: 18 and 7

Lets weigh them (third weighing)

Three possible results:

r1) All are equal

r2) All are unequal

r3) Two are equal and third pan is lighter/heavier

We can easily conclude from these results that which marble is faulty and heavier/lighter.

For example: case r) we know that 22 (heavier) and 21 (lighter) are the solution...

//end of solution

Case 2) All three pans are unequal

We can ingnore six marbles from the medium pan.Also ignore 19,20,21,22 marbles. Lets say pan1 is lighter and pan2 is heavier then we can ignore pan3 marbles.

Take two marbles from pan1, another two marbles from pan1 and three marbles from pan2.

We will have:

Pan1: 1,2

pan2: 3,4

pan3: 7,8,9

We will use remaining marbles two from 1st set (5,6) and three from 2nd set (10,11,12) for further analysis.

Now ou

lets weigh them( 2nd weighing):

There are 3 possible results:

l) All pans are equal

m) All pans unequal

n) Two pans are equal and one is heavier/lighter

case l)

It means faulty marbles are one from (5,6) set and other from (10,11,12) which we kept aside before this weighing.

case m)

It will also give us two sets (1,2) or (3,4) and (7,8,9)

case n) It will also give us a similar two sets.

For all three cases (case l, case m and case n), as we have two sets one of two marbles and one of three marbles...it can be solved as....

For example: We have following sets (1,2) and (7,8,9)

keep from marble from each set say 1 and 9 aside and weigh remaing three marbles:

In this case:

pan1: 2

pan2: 7

pan3: 8

We can easily distinguish heavier and lighter marbles....depeding upon results......

//End of solution

Case 3) Two pans are equal and third one is lighter or heavier

It means set of marbles (19,20,21,22) have one faulty marble, heavier or lighter depends upon the above result.

Remaining one faulty marble among six marbles can be found easily as we have two weighing left...

2nd weighing distribute like this:

Pan1: 1,2,3

Pan2: 4,5,E1

Pan3: 19,20,6

In every case it will give us 5 possible faulty marbles which can be easily solved in third weighing...

///End of solution

Edited March 24, 2011 by Panwar

[Guest]

I have a question, if you can have more than one marble in each pan at a time, and you can have both the 9 and 11 together, isn't it impossible to insure that you can find them? If you made the problem 11 and 8 it would be possible but 11 and 9 in the same pan is equivalent to 10 and 10.