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You are in a room with 3 digital weight scales and a big book. The digital scales are small like the ones you see in small stores. They are identical in all aspects. They all have a READ button which when pushed gives you a weight reading for 1 second. Readings are in whole pounds, not fractions.

One scale is accurate.

One scale always gives 1 pound less reading.

One scale always gives 1 pound high reading.

By taking only TWO weight readings determine which scale is accurate, which one gives lower reading and which one gives higher reading. No, you cannot stand on the scale!

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  • 2
    $\begingroup$ Welcome to Puzzling! $\endgroup$ – Deusovi Feb 15 '17 at 22:38
  • $\begingroup$ I've found this: quora.com/… but I have no clue how the same answer could be applied here. If it is the same answer(somewhat like @stack readers solution), this puzzle is rather underwhelming to me. $\endgroup$ – Loko Feb 16 '17 at 12:58
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    $\begingroup$ (1) The scales are not “identical in all aspects” if they give different results. I guess you mean that they are identical except for calibration. (I guess you mean to say this (ROT13: gurl nyy jrvtu gur fnzr), but that would be too big a hint.) (2) If you mean that we’re not allowed to weigh anything that we brought with us (including ourselves), you should say that. (3) You say that scale readings are in whole pounds, not fractions. Is there anything in the room whose weight is not a whole number of pounds? How do the scales respond to this? $\endgroup$ – Peregrine Rook Feb 16 '17 at 16:50
  • $\begingroup$ ISTM that I have seen this puzzle (or, at least, one very similar to it, or at least, one with a similar answer) in the past six to eight weeks, but I’m not sure where. $\endgroup$ – Peregrine Rook Feb 16 '17 at 16:51
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    $\begingroup$ I believe this puzzle is underconstrained, mostly because of the "Readings are in whole pounds" rule. Can the "correct" solution deal with a book (or scale) that weighs 0.1 lbs? $\endgroup$ – trentcl Feb 16 '17 at 18:39
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Since all of the weight scales are identical in all aspects, their weight should be the same too:

Let the weight of a scale be X

Now weigh scales B and C on scale A, the reading will be 2X-1, 2X, or 2X+1 - I'll refer to this reading as R1

Next, weigh scale C on scale B, the reading will be X-1, X or X+1 - and this reading as R2

Take R1 - (2 x R2)

if the answer is:
-1 : scale A reads 1 pound less, scale B is accurate, scale C reads 1 pound high
-3 : scale A reads 1 pound less, scale B reads 1 pound high, scale C is accurate
+2 : scale A is accurate, scale B reads 1 pound less, scale C reads 1 pound high
-2 : scale A is accurate, scale B reads 1 pound high, scale C reads 1 pound less
+3 : scale A reads 1 pound high, scale B reads 1 pound less, scale C is accurate
+1 : scale A reads 1 pound high, scale B is accurate, scale C reads 1 pound less

Not going to bore you with the math, but not rocket science once you think out of the box

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    $\begingroup$ Yeah, I had the idea of weighing scales with other scales too just now, but I wasn't immediately sure how to make it work. This solution should be the one OP is thinking of. $\endgroup$ – edderiofer Feb 16 '17 at 4:22
  • $\begingroup$ Very close to my idea also involving weighing the scales but also the book. Had it working and ready for typing. +1 $\endgroup$ – Mordechai Feb 16 '17 at 5:01
  • $\begingroup$ This solution makes (and depends on) an assumption that I questioned in a comment on the question. That said, I suspect that this is the ‘’correct’’ answer. $\endgroup$ – Peregrine Rook Feb 16 '17 at 16:58
  • $\begingroup$ Does not work if a scale may weigh less than half a pound (depending on the rounding method assumed) $\endgroup$ – trentcl Feb 16 '17 at 18:33
  • $\begingroup$ @edderiofer what do you think of the answer I just posted $\endgroup$ – Kevin Feb 17 '17 at 17:15
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The solution is simple:

Just press the "READ" button on two of the scales without putting anything on them. They'll say either -1, 0, or 1. That tells you their offsets. Then you can just figure out the other one from what's missing.

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    $\begingroup$ @DeepakMahulikar Well that condition's pretty bleh. $\endgroup$ – edderiofer Feb 16 '17 at 4:12
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    $\begingroup$ They weigh to the nearest whole pound, plus/minus their inaccuracy. Weigh a penny. $\endgroup$ – Rubio Feb 16 '17 at 6:59
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    $\begingroup$ @DeepakMahulikar You're weighing whatever bits of dust may have settled on the scale. $\endgroup$ – Sneftel Feb 16 '17 at 20:48
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    $\begingroup$ @DeepakMahulikar you should call that out in the main post - this is the best answer given and works will all the rules you've set. $\endgroup$ – Buffalo5ix Feb 16 '17 at 22:32
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    $\begingroup$ Yes, this works at the moment. If you want to avoid this solution, a more plausible condition than "you must weigh something" would be to have the scales unable to display a negative weight, so the middle scale displays 0 when weighing 1lb or less. $\endgroup$ – Especially Lime Feb 17 '17 at 8:58
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Tear one page from the book.

Weigh it on scale 1. There are 3 possible readings:

1. Zero: Scale 1 is the accurate scale and we are done.
2. +1 or -1: Now we know that scale 1 is not the accurate scale. Weigh the page on scale 2. This will give 2 possible readings:
    1. Zero: Scale 2 is the accurate scale and we are done.
    2. -1 or +1: (The opposite of what we got on scale 1.) Scale 3 is the accurate scale and we are done.

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    $\begingroup$ But... the book! $\endgroup$ – boboquack Feb 16 '17 at 3:11
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    $\begingroup$ @boboquack: Use one of the blank pages in the back! $\endgroup$ – Deusovi Feb 16 '17 at 3:12
  • $\begingroup$ @Deusovi: Assuming there is one. I've had plenty of 'environmentally friendly' books where they just put bonus content to bring the number of pages to a multiple of the minimum correct number of pages. $\endgroup$ – boboquack Feb 16 '17 at 3:18
  • $\begingroup$ @boboquack I know, right? I have a big collection of over 1000 books at home myself. It made me cringe just writing that line, but such is the puzzle. :| I guess I could "assume" that the book is one of those hardcover ones, that have a paper cover around the cardboard, in which case, just use that paper cover. :) $\endgroup$ – Masked Man Feb 16 '17 at 3:29
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This is a hopefully easier to understand explanation of what I think some of the other answers are trying to get at.

First, use scale one to weigh scale two and three at the same time. If the reading is an even number, then that is the accurate scale. If it is an odd number, it is one of the inaccurate scales.

Next, use scale two to weigh scale one and three at the same time. If one was the accurate scale, then two will either show one pound less or one pound more and you will know which one is which. If scale one was inaccurate, then scale two will either show an even number (meaning it is the accurate scale), or an odd number (meaning that it is the other inaccurate scale).

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Here is a tentative solution.

Weight the book on scale A. For example, you get 21 pounds.
You then know that the correct weight must be 20,21 or 22.
Place the 3 identical scales next to each other to form a triangle and place the book in the middle so to split the weight in 3 then press the weight button on scale B.
If the weight is 6, you got 3 possibilities.
4,5,6 (14)
5,6,7 (18)
6,7,8 (21) = must be the correct answer
If the weight is 7, you got 3 possibilities.
5,6,7 (18)
6,7,8 (21) = must be the correct answer
7,8,9 (24)
If the weight is 8, you got 3 possibilities.
6,7,8 (21) = must be the correct answer
7,8,9 (24)
8,9,10 (27)
Since scale A gave you 21 it must be the accurate one,
If scale B gave you 6, it must be the "- 1 pound" scale.
If scale B gave you 8, it must be the "+ 1 pound" scale.
This method seems to work for any outcome.

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  • $\begingroup$ But what if you get 7 pounds? What if you get 20 pounds? I don't see how this accounts for all cases. $\endgroup$ – boboquack Feb 16 '17 at 2:01
  • $\begingroup$ @boboquack If you insist, I will add some more data. $\endgroup$ – stack reader Feb 16 '17 at 2:06
  • $\begingroup$ Because what you're relying on is a rounding error: you're assuming the weight must be divisible by 3. And 21, 7 gives you both scales being accurate, contradiction. $\endgroup$ – boboquack Feb 16 '17 at 2:08
  • $\begingroup$ @boboquack if it is dividible by 3, it works, if not, it will get round up(ceil I assume). If the total got round up, the 3 1/3 will also get rounded up. So it should come back to the same. 20.1 / 3 -> 6.7 -> 7 $\endgroup$ – stack reader Feb 16 '17 at 2:12
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    $\begingroup$ Consider the case where you got $25$, $8$. The book could weight $24$, and then A would be $+1$ and B $=0$, or the book could weight $25$, then A would be $=0$ and B $-1$ because $\left\lceil\frac{25}{3}\right\rceil=9$. $\endgroup$ – boboquack Feb 16 '17 at 2:18
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so here goes my answer, though just a guess of a guest:

I think, first you should weigh two of the weighing scale (A and B) in the remaining weighing scale (C), then divide the weight reading (A+B) into 2 ((A+B)/2). If the quotient((A+B)/2) has a remainder, it (C) is either the +1 or -1 pound scale, else it is the accurate scale.

If C is the accurate scale, weigh one of the scales (A or B), and if it is greater than the quotient((A+B)/2) then it is the +1 pound scale, or if it is less than the quotient((A+B)/2) then it is the -1 pound scale.

If C scale is either +1 or -1 scale because of the remainder, then weigh C and A together. Divide the weight into 2, and if it has a remainder, then B is either +1 or -1 scale, else it is the accurate scale. If the (A+B) is greater than (C+A), then B is the +1 scale, else C is the +1 scale.

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If this is allowed

1st Answer (wrong)
1. Put that big book on 2 weighing scales, both will show different results
2. Then swap out 1 of them for the other weighing scale and get that reading
3. the middle 1 is the correct 1

2nd Answer
1. Let's say the book is 10 pounds (solution does not work if book's weight is not whole pound and weight is not divisible by 2), and the weighing scales can weigh accurately but simply round to nearest whole pound
2. Putting the book on 2 weighing scales would show:
(5, 6), (4, 5), (4, 6)
3. If it is the 3rd reading, you know the other weighing scale is the correct one as this reading has a disparity of 2
4. If it is the first 2 reading, (possibly out of the boundary of the question) Quickly pull away 1 of the scales and if the remaining scale did not double, the 1 you pulled is the correct 1, else if it did double then the remaining 1 is correct (do all this under 1 second)

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    $\begingroup$ Unfortunately, this would involve 3 weighings, one on each scale. $\endgroup$ – boboquack Feb 16 '17 at 8:28
  • $\begingroup$ @boboquack right..., missed that $\endgroup$ – Stein121 Feb 16 '17 at 8:30
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    $\begingroup$ By the way, welcome to Puzzling! Interesting to note I had a similar idea, too, at first (placing the book straddled across all three scales), but which also doesn't work. $\endgroup$ – boboquack Feb 16 '17 at 8:33
  • $\begingroup$ Hmm... I think when you press the READ button it takes a solitary reading without recalculation, unfortunately. $\endgroup$ – boboquack Feb 16 '17 at 9:02
  • $\begingroup$ Oh well reading @just browsing answer, it makes more sense $\endgroup$ – Stein121 Feb 16 '17 at 9:05
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Lay the book evenly across the 3 scales so each scale bears a third of the weight. Pick one scale and use a READ. Now remove the book and place it on ONLY 1 of the 2 scales you haven't read yet and READ.

Now, multiply the first READ weight by 3.

If the result is < 3 (number will change based on divisibility by 3) units rom the second reading, then the first scale you chose is the correct scale. From there you can figure out the other 2 by whether the second reading is greater or less than your result.

If the result is > 3 units away from the second reading, you have picked the two incorrect scales and can work out which is which by whichever is higher/lower, leaving the unread scale to be the correct one.

If the result is exactly 3 units away from the second reading, then the second scale you read is the correct scale, and you can work out the two others by whether the result (first reading * 3) is less than or greater than the correct one.

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  • $\begingroup$ The second reading is with the entire book on a single scale, so in the case you describe you would have weight 8 (times 3 = 24) and then weight 25 (weighing entire book on ONLY the +1 scale). The < 3 difference means the first scale you read is correct, and since the second is more than the correct scale, you know it's +1. Following same steps, second scenario you gave should result in 8 (times 3 = 24) then 27. $\endgroup$ – Syps Feb 18 '17 at 21:57
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Scales A, B and C. All weigh same (they are identical so they all have same mass) There are six possible combinations: A accurate B Negative C Positive A accurate B Positive C Negative A Positive B Accurate C Negative A Negative B Accurate C Positive A positive B negative C Accurate A negative B positive C Accurate Put scale B or C on A : Take FIRST Reading x Put A AND C (both) on B : Take Second Reading y For all the six combinations above there are six DIFFERENT equations for x and Y y=2x-1 y=2x+1 y=2x-2 y=2x+2 y=2(x-1)-1 y=2(x+1)+1 Try a number

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