Sensibilly -T ™ scales were out of stock, so Gramazon filled the order with a lousy Lopside® from Irreturnable Irrationals International, Inc.73205080...  How irredeemably irritating.  Sigh, might as well unpack it anyway and just start weighing things.

One factory-calibrated lopsided beam balance
One unit weight,          weight =  1   unit
One disposable weighing pan,    weight =  p   units ( p is an irrational amount)
Another disposable weighing pan, weight =  p   units
Another disposable weighing pan, weight =  p   units
       (infinitely many identical pans)

The scale balances when nothing hangs from either beam’s hook.
The scale also balances with the unit weight on the left side and one pan on the right.
Any given weight that’s a positive integer W units heavy may be balanced by hanging one or more pans, placing W in one of those pans, and optionally hanging the unit weight as well.
Each pan is good for a single balancing only and is then wastefully discarded.
The unit weight may be used again and again.

Action plan
Day  1.   Balance W =  1  unit   (then discard all used pans).
Day  2.   Balance W =  2  units (and again discard the pans)
Day  3.   Balance W =  3  units (and discard pans)
Day 30. Balance W =  30  units (discard pans one last time)

Among positive irrational numbers, what value for p allows the fewest pans total to be used up during these 30 action-packed days?


All items may hang on either side, but W always needs a pan on the same side to hang in.

Although the balance’s left beam is depicted above as being longer than its right beam, that disparity would be reversed if p were between 0 and 1, which is allowed.


1 Answer 1


With a weight $L$ on its left arm and $R$ on its right, the balance balances exactly when $pL=R$.

Suppose we hang $L$ pans on the left and $R$ on the right. Take $e$ to be 0 if we don't use the unit weight and 1 if we do. There are four options for where we locate our weight $W$ and our unit weight.

(L,L) : $p(Lp+W+e)=Rp$ or $L p^2 + (W+e-R) p = 0$
(L,R) : $p(Lp+w)=Rp+e$ or $L p^2 + (W-R) p - e = 0$
(R,L) : $p(Lp+e)=Rp+W$ or $L p^2 + (e-R) p - W = 0$
(R,R) : $p(Lp)=Rp+W+e$ or $L p^2 - R p + W+e = 0$

We can ignore case (L,L) because it cannot be true if $p$ is irrational. For every positive integer $W$ at least one of the others must be true, for suitably chosen $L,R$.

So, clearly $p$ satisfies some quadratic equation $Ap^2+Bp+C=0$. In this case it satisfies another quadratic equation precisely when that equation is a multiple of $Ap^2+Bp+C=0$.

When can case (L,R) work? Well, that constant coefficient is always either 0 or -1, and in fact it can't be 0 if $p$ is irrational. So if case (L,R) works then we have $L=A,W-R=B,-1=C$. The number of pans used will be $L+R=A+W-B$. If this case always works then we must have $B\leq1$ and we'd better take $A=B=1$, so we need $W$ pans every time. This requires that $p=\frac{\sqrt5-1}2\simeq.618$. But of course there might be solutions where we use this only for some $W$, and in that case we might be able to use a larger $B$.

When can case (R,L) work? Well, then we will need to take $C=-1$ again since $W$ always has to be a multiple of $C$. Then we do best to take $A=B=1$ giving the same $p$ as we needed to make (L,R) work. In this case we have $L=W,R=W+e$ so we'd better always take $e=0$ (i.e., not use the unit weight) and then we take $2W$ pans. This is always worse than case (L,R) so we can ignore it.

When can case (R,R) work? Here we can do a bit better because we get to choose whether the constant coefficient is $W$ or $W+1$. So we can make our constant coefficient be 2, and then we can scale the coefficients by about $W/2$ instead of $W$. But there is a snag. Our quadratic needs to have real roots! This means that $B^2\geq4AC$ which if $C=2$ requires $B^2\geq8A\geq8$ so $|B|\geq3$. So the best we can do is $A=1,B=-3,C=2$. But this won't do because this quadratic equation has rational roots. Since we end up needing about $(|A|+|B|)W/2$ pans, this is going to end up worse than case (L,R) again.

Conclusion: we can never do better than case (L,R) which requires exactly $W$ pans to achieve balance with weight $W$. So we will need, in total,

$1+2+\cdots+30=31\times15=465$ pans.

And we should take


Perhaps "Gramazon" is so-called

in honour of the golden ratio.

Cautionary note: This is exactly the sort of thing in which I easily make mistakes, so I will be unsurprised if any part of the above is wrong. The overall approach is clearly right, though.

  • $\begingroup$ Nice interpretation of "Gramazon," Gareth! That one was unintentional but the skirt of the scale does contain something similar on purpose. $\endgroup$
    – humn
    Sep 23, 2017 at 3:44

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