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 Five Weights and a Broken Scale (Posted on 2004-08-02)
You have to sort five weights weighing 51g, 52g, 53g, 54g, and 55g. You have a balance scale with which you can compare the weights. But after solving so many sorting puzzles, it is starting to break down.

If the difference between two weights is greater than 1.5g, the scale will correctly determine which side is heavier. If the difference between the weights is less than 1.5g or equal, the scale will indicate the weights are equal.

Sort the weights in the smallest number of weighings.

 See The Solution Submitted by Brian Smith Rating: 3.5556 (9 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Re: 5 weighings--REALLY!!! | Comment 11 of 17 |

Charlie, great question! see my comments in bold italics<o:p></o:p>

Why not for example >,<,< or >,<,=?<o:p></o:p>

<o:p></o:p>

I am only citing >,>,> as the first case,  i was going to elaborate with all of the possibilities I cited above (from a-f). But I did not post all my solution, just for the first case. In all of the cases I got  maximum 5 weighings. If the first case is understood, it is easy with the others because they all go by the same reasoning<o:p></o:p>

<o:p> </o:p>

<o:p>And to your question: </o:p>Why not for example >,<,=?

>,<,=" is the same as case d) I cited in my first posting. The order does not matter, but the combination. I wrote: "d)      >,=,< ". It is the same if I wrote "d) >,<,=". What I mean is, " case d, if you get one >, one <, and one = in the first 3 weighings." It does not matter whether you got the '=' first or the '<' first, as long as you got >, < and =!<o:p></o:p>

<o:p> </o:p>

<o:p> </o:p>

Again to your question: Why not for example >,<,<?

My answer: You will never get >,<,<!

">,<,<" has one ">" and two "<'s"... This is IMPOSSIBLE to GET if you compare ANY weight with all the others. See this table:<o:p></o:p>

<o:p> </o:p>

WEIGHT       Result if compared with the 4 others     <o:p></o:p>

51:                      =,<,<,<<o:p></o:p>

52:                      =,=,<,<<o:p></o:p>

53:                     >,=,=,<<o:p></o:p>

54:                     >,>,=,=<o:p></o:p>

55:                      >,>,>,=<o:p></o:p>

<o:p> </o:p>

EXAMINE THE ABOVE AND YOU NEVER GET a result which has all >,<,<  in it!!! (Meaning, can never get a result which has one > and two >'s)

<o:p></o:p>

<o:p></o:p>

="II)                 <o:p></o:p>

A) suppose you got case a) >,>,>   <o:p></o:p>

then it follows that A is 55 since it is larger than the three others. It can't be 54 since the balance will display 54=53. This also leaves us to conclude that E is 54!"<o:p></o:p>

There are not just 3 others; there are 4 others, so A is not necessarily 55.<o:p></o:p>

<o:p></o:p>

-->That is why I said (above) " It can't be 54 since the balance will display 54=53"<o:p></o:p>

<o:p></o:p>

NOTE:<o:p></o:p>

There ARE 4 others: 51, 52, 53, 54<o:p></o:p>

If we weigh <o:p></o:p>

1) 51 vs 55: the scale will display '>'<o:p></o:p>

since 55>51   scale will display '>' since weight difference >1.5 g<o:p></o:p>

Also:<o:p></o:p>

2) 55>52   scale will display '>' since weight difference >1.5 g<o:p></o:p>

3) 55>53   scale will display '>' since weight difference >1.5 g<o:p></o:p>

If we weigh 55 with the FOURTH weight,<o:p></o:p>

4) 55=54   scale will display '=' since weight difference < 1.5g<o:p></o:p>

<o:p></o:p>

So, weighing 55 against the FOUR others, we got: >,>,>,=<o:p></o:p>

WE GOT 3 >'s or >,>,> (and one '='. But if we got three >'s in the first place, there is no need to weigh against the 4th one, we know it will be '='. WHY? Because if we first got  3 >'s, the difference for each with A must be greater than 1.5. But all the weights have a weight beside it that differs only by 1. Therefore, if we compare any weight against the FOUR others we get at least one '='. Since in this case since we have got three >'s already, then the remaining must be a '=' )

<o:p>If we got 3 >'s,  A cannot be 51,52,or 53 (see table), because for any of these (51,52,or53) we must get only a maximum of one '>' for 53, and zero '>' for 51 and 52</o:p>

<o:p></o:p>

Also, 'A' cannot be 54 since if we weigh 54 against the 4 others, it will get only two >'s: >,>,=,= .

<o:p></o:p>

54 > 51 since difference >1.5g<o:p></o:p>

54> 52 since difference > 1.5g<o:p></o:p>

54=53 since difference < 1.5 g<o:p></o:p>

54=55 since difference <1.5g<o:p></o:p>

<o:p></o:p>

It's impossible to get 3 >'s for 54! <o:p></o:p>

<o:p></o:p>

The only weight we CAN get at least 3 >'s if we weigh against all the 4 others is if the weight is 55! <o:p></o:p>

<o:p></o:p>

SO IF IN THE FIRST THREE WEIGHINGS WE GOT 3 >'S, WE ALREADY KNOW THAT A IS 55. <o:p></o:p>

<o:p></o:p>

If A is 55, then we know that the 4th weight is 54, even if we don't weight them together. Why? Because none of the first three gave us a '=' (since we got >,>,>), it must be that none of these 3  has a weight difference of only 1 compared to 55. But since all of them has a weight beside them that differs only by 1, then the remaining must be 54 because it is the ONLY weight that differs by 1 with 55.<o:p></o:p>

For this case, I elaborated on >,>,>. By the same reasoning it works with <,<,<.. Also with all the other cases but which are slightly different (if you got at least one '='). All will give a maximum 5 weighings!

<o:p></o:p>

(Whew! I hope I am able to explain it well.  It is harder to explain it than to actually solve it.) <o:p></o:p>

<o:p> </o:p>

Edited on September 12, 2004, 4:34 pm
 Posted by i_wish on 2004-09-11 13:05:01

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