You are given three meter scales A, B, and C are shown in picture given below. The scales are however, quite unique in their properties as they have different numbers of division per meter. You are asked to measure the length of the rod using the scales given to you. What is the range of the length of the rod as measured by the scales? Round off your answer up to two decimal places. You may use a scale only once.

enter image description here

Source: Question is designed by me.

  • 1
    $\begingroup$ We can assume it's exactly 42.2cm (that's not rounded at all)? $\endgroup$
    – msh210
    Jun 7 at 21:39
  • 2
    $\begingroup$ Lucky Bass that this was not rejected as a math problem and not a puzzle:) $\endgroup$
    – Moti
    Jun 8 at 6:23
  • $\begingroup$ Can I use one of the scales multiple times and measure out say 17 divisions of scale A? $\endgroup$
    – quarague
    Jun 8 at 11:13
  • $\begingroup$ In a comment to my answer somewhere far below (can't figure out how to link directly to the comment), OP says they have designed this puzzle themself. The VTC reason now reads "missing proper attribution", so I'm voting to reopen. (Then I may vote to re-close for it being a maths problem, since OP doesn't like the maths-puzzle-like answer ;-) ) $\endgroup$
    – Bass
    Jun 9 at 14:16
  • 1
    $\begingroup$ @I'm Nobody: You posted the question. What does it mean "I believe the actual intended answer has already been posted below"? Do you know the anwser to your question or not? And will ever an answer be flagged as correct or not? $\endgroup$ Jun 14 at 19:22

3 Answers 3


If I understood the question correctly, I guess the closest you can get is:

42.27 cm

by using for example:

3 divisions of B + 3 divisions of C - 1 division of A: $\frac{3}{11} + \frac{3}{12} - \frac{1}{10} = 0.422727..$

The lower bound:


For example via:

5 divisions of B + 2 divisions of C - 2 divisions of A: $\frac{5}{11} + \frac{2}{12} - \frac{2}{10} = 0.421212..$

But maybe there are smarter ways to get closer.

Well, I have some ideas

touching/placing the rod and each scale only once and without extra marks,

but I haven't yet found a way to prove that there might be closer bounds compared to the values given above.

OK, here is an attempt for a smaller deviation.


Admittedly, mathematically ok, but for practical purposes...

at least the rod and the scales are touched/placed only once and no extra marks are necessary. I know that even smaller deviations are possible with the same principle but I don't know what the absolute minimum is. I found 42.2107 but I haven't made a drawing for that.

enter image description here


If my script is correct (actually, you can't find this by hand), the lower bound is:


enter image description here

and the upper bound:


enter image description here

Maybe someone can confirm or contradict.

  • $\begingroup$ Nice, in my approach I only admitted the rod inside the triangle but not outside. I just forgot these cases... $\endgroup$ Jun 16 at 0:59

Approximate 0.422 as good as possible by a * 1/10 + b * 1/11 +c * 1/12, |a|<=10, |b|<=11 and |c|<=12
From above the best possible is for example
4 * 1/10 + 3 * 1/11 - 3 * 1/12 = 93/220
From below the best is for example
3 * 1/10 + 5 * 1/11 - 4 * 1/12 = 139/330
So the length of the rod is in the intervall (42.12, 42.27) cm

As proposed by theozh in two dimensions we can do much better...

For a lower bound of 42.197 cm build a triangle with the three scales with 9 units from scale A, scale B and 6 units from scale C. See picture for the position where to put the rod.
enter image description here

For the upper bound of 42.201 cm build a triangle with 7 units from scale A, scale B and 4 units of scale C. See the drawing where to put the rod.
enter image description here

For triangles the results are optimal.

  • $\begingroup$ Thank you for checking the ultimate limits with this method. I was about to write a little program to check all combinations... I assume you did it already? $\endgroup$
    – theozh
    Jun 9 at 4:53
  • $\begingroup$ @theozh: I checked all triangles. That the complete scale B is used in both solutions and one end of the rod sits in a corner seems pure coincidence. But it would be nice if my solution could be verified by you. $\endgroup$ Jun 9 at 8:21
  • $\begingroup$ if my script hopefully doesn't contain a bug, I have found narrower bounds. See my updated answer. $\endgroup$
    – theozh
    Jun 15 at 21:44

You can get the accuracy down to

exact length

by using only a single ruler.

The trick is to

use the rod itself as a measurement tool: Start by placing the two unnecessary rulers on a table, parallel to each other, a rod length apart. Then use the 1/10 metre ruler to measure "forwards" from one ruler in 10 centimetre units until you get past the other ruler. Then, measure backwards a single rod length, so your "spot" is again between the two parallel rulers. Continue measuring back and forth like this, always keeping count of the units measured in both directions.

Because the length of the rod (42.2 cm) and the measurement unit (10cm) are rational multiples of each other, eventually a marking on the ruler is guaranteed to exactly coincide with the target ruler we placed at the beginning.

When this happens, it is simple to calculate the exact length of the rod with basic algebra. In our case it will happen after measuring

211 units of 10 centimetres forward, and 49 rod lengths backwards,


$$211 \times 10 \text{ cm} = 50 \times 42.2 \text{cm}$$

The reason for the increased accuracy is of course that

* measuring a longer distance with the same sized ruler will give a smaller proportional error, and
* doing it one rod length at a time lets us stop at the minimal error,
* which will be zero unless the two lengths are irrational wrt. each other.

NB. You don't actually need any divisions on the ruler for this method to work, the actual length of the measurement unit is (almost) irrelevant. For practical purposes though, the 10cm unit is better than the 1 meter: with several units marked on a single ruler, you'll only ever need to move the rod to a spot that is clearly marked.

  • 1
    $\begingroup$ Brilliant solution, however there is always room for error when you mark the rod and move it backwards for further measurement. And for every marking, the error piles up, and the range of length of rod may become too large. I designed this question with least count and error analysis in mind. $\endgroup$
    – I'm Nobody
    Jun 8 at 12:16
  • $\begingroup$ You only need to mark the initial end (either one will be enough, really) of the rod; after that you only move either the rod or the ruler, always so that the end of the rod exactly coincides with a mark on the ruler. Also, if we're to ignore the tag's promise of a mathematics puzzle, and say that there's implicitly always room for error, then we should be using even more measurements, not fewer: random errors tend to cancel each other out instead of piling up. $\endgroup$
    – Bass
    Jun 8 at 12:56
  • $\begingroup$ @I'mNobody: Even if there's an uncertainty induced each time you mark a ruler and move it, it's still possible that you'll get a more accurate answer if the induced uncertainties are small enough. To accurately answer the question as stated, you'd need to know the size of that induced uncertainty. Alternately, if you wanted to outlaw the technique described here, you could just require that "you may not make markings on the rulers." $\endgroup$ Jun 8 at 14:47
  • $\begingroup$ @MichaelSeifert re "if you wanted to outlaw the technique described here, you could just require that 'you may not make markings on the rulers.'": I don't think you need markings on the rulers for this method. You can put the rod end to end a bunch of times, marking each time, and the ruler end to end a bunch of times, marking each time, until one of the first set of marks matches one of the second. $\endgroup$
    – msh210
    Jun 9 at 8:58

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