How to get $2700$ grams of corn from a bag of $6.8$ kilograms? 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:

*

*$4$ trials

*$3$ trials

*$1$ trial

*$2$ 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.
 A: 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.
A: $6.8$ divided by two gives $3.4$. Divided by two again and you get $1.7$.
Add $1$kg and you are done.
A: 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.
A: 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?
A: 
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:

*

*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.



*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 $1$.



*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.



*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.
