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There is a broken two-pan arm scale, consists of different weighed pans and different arm lengths.

enter image description here

You are trying to weigh three different weights called X,Y,Z. You also know that the weights are even numbers:

  • When you place X and Y to the right hand side of the scale, 20g is needed to the left side to balance the scale.
  • When you place Y and Z to the left hand side of the scale, 18g is needed to the right side to balance the scale.
  • When you place all weights to the right hand side of the scale, the weight you need to put to the left hand side is 5g more than the weight you need to put to the right hand side when you place all weights to the left hand side of the scale.

Find the all weights.

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  • $\begingroup$ Just to be sure: When placing X and Y on the right hand side, the left hand side is empty and Z isn't used? $\endgroup$ – Lolgast Dec 8 '17 at 11:20
  • $\begingroup$ @Lolgast yes only X and Y, Z is out on the condition. $\endgroup$ – Oray Dec 8 '17 at 11:24
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    $\begingroup$ This seems like a basic simultaneous diophantine equations question - is there any trick or nicety that makes this a puzzle rather than a mathematical problem $\endgroup$ – Wen1now Dec 8 '17 at 11:26
  • $\begingroup$ When nothing is in the pans, does the scale balance? If not, is it known to have balanced with nothing in the pans with equal arm lengths (given that the scale is described as "broken")? $\endgroup$ – Gareth McCaughan Dec 8 '17 at 11:41
  • $\begingroup$ I asked about two different conditions; do you mean we don't know whether it balances in either of them? $\endgroup$ – Gareth McCaughan Dec 8 '17 at 11:42
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Firstly, to make life easier, we'll assume the pans themselves (and, by extension, arms) are weightless. That'll get rid of mess in our equations.

Suppose the ratio of the length of the left arm to the right is r. Then we get

When you place X and Y to the right hand side of the scale, 20g is needed to the left side to balance the scale.

So $20r=x+y$. Note that $20r$ is an integer.

When you place Y and Z to the left hand side of the scale, 18g is needed to the right side to balance the scale.

$(y+z)r=18$

When you place all weights to the right hand side of the scale, the weight you need to put to the left hand side is 5g more than the weight you need to put to the right hand side when you place all weights to the left hand side of the scale.

So $(x+y+z)=Wr,W-5=r(x+y+z)$

which means that $r^2(x+y+z)+5r=(x+y+z)$

Therefore $\triangle=25+4(x+y+z)^2$ is a square (as $r$ is rational). This quickly breaks as it implies $x=y=z=2$ which is bad. In fact the middle equations wasn't even necessary.

However I suspect that things get nasty if we remove the initial assumption. I'll work on it a bit more.

Weighted pans

I'll quickly write up the equations:

$20r=x+y+N,(y+z)r=18+n,(x+y+z)r^2+(5-n)r=(x+y+z+n)$

Somebody please point it out if I'm being stupid.

Anyway my python script finds

x  y  z  r    n
6  6  18 1.5  18.0                 bad (not all different)
18 4  10 0.66 -8.666666666666668   good
20 2  2  0.25 -17.0                bad (not all different)

as solutions, so I suspect that with three equations in five unknowns, even with the unique even integers restriction, that more solutions exist. If not, then proving uniqueness will be hard. At least that's what my intuition tells me.

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  • $\begingroup$ This was how I was originally doing it, until I read the comments on the question and realised that both the $a$ and $b$ (in the notation of my answer) are needed. $\endgroup$ – Rand al'Thor Dec 8 '17 at 12:49
  • $\begingroup$ @Sid The weighted pans part is done at the very end $\endgroup$ – Wen1now Dec 8 '17 at 23:37
  • $\begingroup$ @Wen1now different weights restriction exists in the question, could you consider that condition and find the answer accordingly? $\endgroup$ – Oray Dec 9 '17 at 8:26
  • $\begingroup$ I still don't believe it'll be (easily) possible to prove that there exists exactly one solution, if true. Otherwise big solutions exist as well... $\endgroup$ – Wen1now Dec 9 '17 at 10:05
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Partial answer

Unless my physics is wrong, the fact that the pans have different weights and the arms have different lengths means that there exist constants $a$ and $b$ such that the two pans balance with weight $P$ on the left and $Q$ on the right if and only if

$P=aQ+b$. (The $a$ describes the difference in arm length, and the $b$ describes the difference in pan weight.)

So the three data points you've given us can be written in equation form as follows:

  • $20=a(X+Y)+b$

  • $Y+Z=18a+b$

  • $X+Y+Z=aW+b$ and $W+5=a(X+Y+Z)+b$ for some $W$, or in other words $a(X+Y+Z)+b-5=\frac{X+Y+Z-b}{a}$.

Let us rewrite the third condition as follows:

$a(X+Y+Z)+b-5=\frac{X+Y+Z-b}{a}\Rightarrow(X+Y+Z)a^2+ab-5a=X+Y+Z-b\Rightarrow(X+Y+Z)(a^2-1)=5a-ab-b\Rightarrow X+Y+Z=\frac{\frac{5a}{a+1}-b}{a-1}$.

We know $X+Y+Z$ is an even integer (call it $K$), and that $a$ and $b$ should be rational. Solving as a quadratic equation in $a$, we find:

$Ka^2-K=5a-ab-b\Rightarrow Ka^2+(b-5)a+(b-K)=0\Rightarrow a=\frac{5-b\pm\sqrt{(b-5)^2-4K(b-K)}}{2K}$. So the number $(b-5)^2-4K(b-K)$ must be a perfect square.

I'm not sure how to proceed from here ... am I barking up the right tree?

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  • $\begingroup$ I find it amusing how much of the same notation you've used as I have in my own (unposted) scribblings :-). (Same a,b,W. I used S instead of K, though.) $\endgroup$ – Gareth McCaughan Dec 8 '17 at 12:05
  • $\begingroup$ @Gareth Mathematicians notate alike? :-) Even better, I was also about to use S instead of K, and then changed it to K to remind us that it's an integer! $\endgroup$ – Rand al'Thor Dec 8 '17 at 12:10
  • $\begingroup$ So it would seem. $\endgroup$ – Gareth McCaughan Dec 8 '17 at 12:11

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