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The problem is as follows:

A merchant has a kiosk and has one bag with $6.8$ kilograms of corn and a two pan scale and three weights. One measures $5$ kilograms, the other $2$ kilograms, and the other $1$ kilogram. In how many weighing trials at minimum could he serve an order of $2700$ grams of corn?

The choices given are:

  1. $4$ trials
  2. $3$ trials
  3. $1$ trial
  4. $2$ trials

I'm not sure exactly how to get to the right combination for this problem. It would be very nice if the answer could include the steps in a logical and very detailed way so I can understand.

To my best of ability I thought that I could arrange the weights as follows:

$5$, $2$, $1$, $6.8$

To get $2.7$ from these numbers is somewhat convoluted to me. Thus I need some help. What can I do?

The thing is either method which I've tried from subtracting this amount from $6.8$ does give me more than what I need, thus I am going in circles.

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    $\begingroup$ What you should do is play around with the numbers; what other numbers can you make? What kind of 'tricks' can you come up with to make new numbers? Don't try to get to 2.7 with the fewest steps right away. Just try to get there first. $\endgroup$
    – This site has become a dump.
    Mar 11, 2021 at 21:40
  • $\begingroup$ I can make a useless objection: it seems highly unlikely that you can get to exactly $2700 g$ unless the weight of each kernel of corn (or at least, a sufficient number of kernels of corn) is a precise rational multiple of one gram. $\endgroup$
    – Ross Presser
    Mar 12, 2021 at 15:14

5 Answers 5

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In don't think there is a logical step-by-step solution. Some trial and error seems inevitable to figure out which numbers can be constructed, and in how many steps. So I give you my almost unfiltered thought process:

Weigh off $5$kg. Then $1.8$kg remains. Weigh off another $1.8$kg against this. Now you have two $1.8$kg piles. Split a $1.8$kg into two $0.9$kg piles by weighing it against itself. Then $2.7=1.8+0.9$.

Is this minimal?

You can also put the $5$kg weight on the scale, and distribute all the corn such that the scale is balanced. Then the two sides have $0.9$kg and $5.9$kg. Use the $0.9$kg to weigh off $0.9$kg twice more, to get $2.7=0.9+0.9+0.9$.

Is this minimal?

You can also put $1$kg on one side of the scale, and $5$kg on the other side. Then distribute all the corn such that the scale is balanced. Then the two sides have $5.4$kg and $1.4$kg. Split the $5.4$kg in half by weighing it against itself to get $2.7$kg.

Is this minimal?

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  • $\begingroup$ Boo Hiss. I think that one of your solutions beat me. $\endgroup$
    – user2661923
    Mar 11, 2021 at 22:03
  • $\begingroup$ I missed the solution in two steps. Very clever! $\endgroup$
    – mau
    Mar 12, 2021 at 8:07
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    $\begingroup$ @Raffaele no need to yell. The weights are already provided. Put 1kg of metallic weight on one side, put 5kg of metallic weight on the other side, complete with corn until both sides have the same weight. You get 5.4kg of corn on one side, 1.4kg on the other. In one single step. $\endgroup$
    – Eric Duminil
    Mar 12, 2021 at 19:35
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    $\begingroup$ @Servaes I'm sorry for the late reply. But as the OP of this question your approach is the one which I like the most and beats down (to my understanding) all the other answers. I like the fact that you encouraged me to keep asking to myself if this is minimum. Honestly I agree with what you said, guessing and trial and error is somewhat necessary. It seems there isn't a straight method. Perhaps other than just using that you can split the large chunk those in two groups and weigh them to reduce the weighing trials. $\endgroup$
    – Chris Steinbeck Bell
    Mar 12, 2021 at 20:38
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    $\begingroup$ The fact that the initial bag has a declared weight is probably a clue that in some weighing trial, all of the corn will be used. Usually these problems (often about water) just say there's a sufficiently sized cistern or something. $\endgroup$
    – Ross Presser
    Mar 15, 2021 at 17:15
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Personally, I see no algorithmic alternative to trial and error.

Hint:

If you can weigh out 1.4 kg, then the remaining corn, when cut in half, will suffice.

If you cut the original corn in half, you get to 3.4.....

Not sure if there is alternative approach that involves fewer weighings.

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$6.8$ divided by two gives $3.4$. Divided by two again and you get $1.7$.

Add $1$kg and you are done.

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  • $\begingroup$ This is the best answer, because it doesn't require any receptacles but the scale and the bag. $\endgroup$
    – saulspatz
    Mar 11, 2021 at 21:54
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    $\begingroup$ @saulspatz Unfortunately it is not minimal. $\endgroup$
    – This site has become a dump.
    Mar 11, 2021 at 22:00
  • $\begingroup$ @Servaes When I wrote that, no one had come up with a way to do it in two weighings, and I thought it was impossible. Your last answer must be the best possible, since it uses no additional container and only two weighings. $\endgroup$
    – saulspatz
    Mar 11, 2021 at 22:05
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    $\begingroup$ @saulspatz I didn't think it was possible either, so I'm hesitant to suggest that it is in fact the best possible ;) $\endgroup$
    – This site has become a dump.
    Mar 11, 2021 at 22:08
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Servaes has given a solution in two weighings. I will show that it cannot be done in one weighing. It cannot be done by weighing some amount of corn in only one of the pans, for the corn weighed would be an integral number of kilos, and the number of kilos of corn remaining in the sack would be congruent to $.8$ modulo $1$. So, we must have corn in both pans, and unless we weigh all the corn, we have no way of knowing how much corn is in either pan.

Suppose one pan has $c$ kilos of corn, and $w_1$ kilos in weights, and the other pan had $6.8-c$ kilos of corn and $w_2$ kilos in weights. We have $$\begin{align}c+w_1&=6.8-c+w_2\\c&=3.4-\frac{w_2-w_1}2\end{align}$$

Since $w_1$ and $w_2$ must be integers, we see that either $c\equiv.4\pmod1$, or $c\equiv.9\pmod1$. In either case, we have $6.8-c\equiv c\pmod1$, and in no case do we have $2.7$ kilos of corn in one of the pans.

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A merchant has a kiosk and has one bag with 6.8 kilograms of corn and a two pan scale and three weights. One measures 5 kilograms, the other 2 kilograms, and the other 1 kilogram. In how many weighing trials at minimum could he serve an order of 2700 grams of corn?

One answer has already shown how it can be done in two weighing trials, and another has proven it cannot be done in one. So, my answer will lay out what's possible in each trial, for future reference:

  1. Starting from nothing, we can measure six exact kilogram quantities (or eight if our bag of corn was larger):
  • $1000$, $2000$, $5000$: place the corresponding weight on one pan, then fill the other pan with corn. Select the pan with corn.
  • $3000$: place the $1000$ and $2000$ weights together on one pan, then fill the other pan with corn. Select the pan with corn.
  • $4000$: place the $1000$ weight on one pan and the $5000$ weight on the other pan. Fill the $1000$ side with corn until balance. Select the pan with corn.
  • $6000$: place the $1000$ and $5000$ weights together on one pan, then fill the other pan with corn. Select the pan with corn.
  • $7000$ and $8000$ could also be computed if the bag of corn was larger.
  1. Starting with a quantity $q$, we can further make the following quantities:
  • $q/2$: place half of the quantity on each pan of the balance, then select one side and discard the other (or return to the bag).
  • $(q+1000)/2$: place the 1000g weight on one side, then arrange part of the starting quantity on each side until the scale is balanced. Select the portion from the side with the weight.
  • Similarly with any of the other kilogram amounts from section $1$.
  1. Starting with a quantity $q$ on one pan and sufficient corn left in the bag, we can further compute the following:
  • $q-1000$: Put the $1000$ weight on the second pan, fill that pan until balance, discard $q$ from the first pan and select what's in the second pan.
  • similarly for any other kilogram amount.
  1. If we have one additional container (or the table or floor1) available to store corn in, then we can add any of the above results to an existing result, by moving the result to the additional container, then computing the desired result, finally adding the second result to the additional container.

1The floor clearly won't work if the substance being measured is extremely fluid like water or flour, but corn should be possible, if a little tedious. Hopefully the floor is clean.

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