# balance scale problem for 13 (not 12) items

The 12-item balance scale puzzle is very familiar. The object is to find the lone non-standard item (if one exists) out of a group of 12 seemingly identical items, using a balance scale and a maximum of three weighings. Did you know that it is possible to accomplish the same outcome given a set of 13 seemingly identical items? I've scoured the web for a discussion of this problem/solution and have not found one. Am I the only one that has solved this problem?

BTW: If you allow four (4) weighings on a balance scale, how many seemingly identical items can you analyze and be assured of finding the lone non-standard item within the group?

If there is sufficient interest, I will publish the answers in a future post.

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Sounds awesome, I've never heard of this kind of problem before. –  Simon Hayward Nov 27 '12 at 21:03
I just finisher (re) working the 4 "weigh" problem. It was trickier than I remembered. Back to the 3 "weigh" problem... –  tom Nov 28 '12 at 20:49
I just finished (re) working the 4 "weigh" problem. It was trickier than I remembered. Back to the 3 "weigh" problem...keep in mind that a balance scale weighing yields three (3) unique outcomes. Consequently, three properly planned weighings will yield 3**3 (or 27) unique outcomes. With 13 items, any one could be heavy, any one could be light, or they could all be the same (27 outcomes). Still, there is a logistical trick needed to make it all work, and I'm not ready to reveal it just yet. –  tom Nov 28 '12 at 21:01
You are not the only one who has solved the 13 problem. In the 12 problem, you need not only to find the odd one out, but determine whether it is light or heavy. Unlike your question, you are given that there is exactly one non-standard item. These details are important. [This link] has a solution if you are given whether the odd one is heavy or light. One extra bit of information gives one extra bit on the outcome information. They are both good puzzles. –  Ross Millikan Jun 18 '13 at 3:28

Solution for finding the odd marble from a set of thirteen (13) seemingly identical items

1ST WEIGHING - 5 ON 5, 2ND WEIGHING 3 ON 3, 3RD WEIGHING 1 ON 1. "S" REPRESENTS A MARBLE KNOWN TO BE OF STANDARD WEIGHT, BUT IS NOT ONE OF THE 13. USE OF THIS ITEM IS CRITICAL TO THE SOLVING THE PROBLEM

"/" => LEFT SIDE IS HEAVIER "\" => RIGHT SIDE IS HEAVIER "-" => BOTH SIDES ARE EQUAL

12345/6789S (OUTCOME 1A)

126/347

1/2 => 1H
1\2 => 2H
1-2 => 7L

126\347

3/4 => 3H
3\4 => 4H
3-4 => 6L

126-347

8/9 => 9L
8\9 => 8L
8-9 => 5H


12345\6789S (OUTCOME 1B)

126\347

1/2 => 2L
1\2 => 1L
1-2 =>7H

126/347

3/4 => 4L
3\4 => 3L
3-4 => 6H

126-347

8/9 => 8H
8\9 => 9H
8-9 => 5L


12345-6789S (OUTCOME 1C)

1011/12S (you can use "S" or any of 1 through 9)

10/11 => 10H
10\11 => 11H
10-11 =>12L

1011\12S

10/11 => 11L
10\11 => 10L
10-11 => 12H

1011 - 12S

13/S => 13H
13\S => 13L
13-S => ALL IDENTICAL

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Your question didn't say you were given anything else besides the 13 objects, otherwise you might as well have given the 13 objects with the odd one out in a nice little wrapping that marks it out. Incidentally it is not hard to prove that it is impossible if you only have the 13 objects and an ordinary weighing balance. –  user21820 Dec 29 '13 at 7:38

You can do it in three pre-determined weighings, by numbering the coins from 1 to 13 in the style 1, -2, 3, -4, 5, &c.

  1    001       6  M10     11   11M
2    0M1       7  1M1     12   MM0
3    010       8  M01     13   111
4    0MM       9  100
5    1MM      10  M0M     KG   M1M


You then weigh according to these rules. For each column, put coins numbered 1 in the left pan, and M in the right pan. If the '1' pan goes down, write '1', if the M pan goes down, write 'M'.

After three weighings, you should have a three-place sequence, like 01M. This abc gives 9a+3b+c, where M=-1. If it is an even number, reverse the sign, and that's the coin that is not fair, and whether it's overweight or underweight. So our 01M gives 9.0 + 3.1 + 1.-1 = 2, being even gives -2, so the coin numbered '2' is underweight.

This is different to the usual answer, since we always have seven coins on each pan, (you can't do it with simply six, because at any stage, you are testing for a trinary digit, and it needs here 9 coins, which is odd.

When you do the same for four or five weighings, the process is the same, but the known good is always opposite the half-point, ie 1M1M or M1M1M. You can get to 40 or 121 coins on 4 and 5 weighings, but if you are not allowed to add a known good, then you need to drop the half-coin (7 or 20 or 61) from the list.

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