A Two-pan Non-Equal Arm Scale

Question

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

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.

Answer 1

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.

Answer 2

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?