I just wantto know how in the world do you plan on getting 4900 and 5100?

bobbym wrote:

Hi anna_gg;

You eliminated all the possible weights that are not multiples of 100. Is that okay?

Yes, because the problem says the grocer only sells in multiples of 100 grams.

bob bundy wrote:

LATER EDIT:  I am in the process of correcting this 'solution'.  I am optimistic that I'll get there some time (... length of time unspecified ...)

I am assuming the grocer has a limitless supply of weightless bags to put his derived flour weights into.

How about subscripts?

A weight that exists at the start will have zero as its subscript.

A weight derived at 1st weighing will have subscript 1.

I did have some others here, not multiples of 100.  I have now removed these. 
I do not think something like 4850 can later generate a multiple of 100 except by restoring the number that made it. 
This conjecture needs proving (or a counter example).

I believe this is a complete list of 1st weighings.

These 1st weighing amounts are put into bags and become additional 'weights'.

Now, what can be made at the 2nd weighing?

Theorem: (i) Where different flour bags are used  the subscript total to achieve a new weight must be strictly less than the subscript of the new weight. *

                (ii) but if the same bag is used in several stages the subscript total to achieve a new weight need not count this bag repeatedly 
to make this total strictly less than the subscript of the new weight. An example of this is shown later in achieving 1400_3 and 1600_3.

This theorem needs to be modified.  My thanks to anonynmstefy for pointing this out.

* so here, 2 > 1 + 0 + 0

Now those 'weights' can also be bagged up to use at the next stage.

3rd weighings

I cannot find a way to do 1400.

But anoninmstefy did:

I think that's it.  If you spot any more, please let me know.

Oh darn it!  I've just noticed that 5000.  That'll be 5000 against 5000.

So the list needs some additions.  ..... Tomorrow perhaps.  No better still, you've seen the method, I'll leave it as an exercise for the reader.

Additionally, for whatever amount, x, that can be measured out, the amount 10000 - x is also available for sale which doubles the list!

List so far:

I'm assuming the probability is easy to get from here.


Bob

Why not also
10000_1
5000_1
100_1
9900_1
200_1
9800_1
4900_1
5100_1
300_1
9700_1
500_1
9500_1

Last edited by anna_gg (2012-04-15 19:59:45)

These are "possible" alternative weighings, so why not use them?
To make 9500, for example, you take the total quantity of 10000 (no need to weigh it, as we already know it's 10000) and remove the quantity required to balance the 200 and 300 g weights together; so the remaining is 9500.
Same way with 9700 (we use only the 300g weight), 9800 (only the 200g weight), 9900 (we use both, one in each pan, to weight 100g, so the remaining is 9900).

Hi bob

See #20 for 2000 and as for 3600 you can take the first two measurements from #21 which makes 1800g and just double it in the last measurement to get 3600.

Same as all of them really.

You just pour a bit and see how the balance moves.

If you have put too much in one pan you move a bit across.

This is a maths puzzle;  not a lesson on running a grocer's shop!

Bob

Yes, sorry haven't had time to re-write the list as I've been helping zetafunc.

Suggestion:  Copy my list yourself into a new post, adding your values to it. I am confident you are right! 

What?  You want me to play the teacher game and mark all of your work before putting it up on the wall?! 

Huh!  not likely, I've retired! 

Bob

Hi bob

I've got an idea for 4700. Stay online while I post it.

Oh that's very clever.  the list of 'not yet done' values is getting smaller and smaller.  I wonder if it will shrink all the way down to the empty set.

I had hoped to pass on the job of up-dating the list but it looks like I'll have to do it ... but later as I'm still catching up on other posts.

Meanwhile, keep looking.  Only a few to go. 

Bob

4800:

4600:

4400: