New coin challenge

the difference in weight is 10g/coin. 30gx150=4500g (real) 20gx150=3000g (fake) 1500g difference per box what do I do with the scale?

The idea is not to guess but to have positive proof.

Open each crate. Unpack all boxes. Blow up a balloon.

Tie balloon to box and immerse in very large see-through plastic container. The balloon and box should sink a certain amount. If balloon is still out of water reduce amount of air in ballloon or if sinks straight to bottom of container add more air to balloon. Mark spot on side of container at level of equalisation (the bouyancy of ballooon is equal to density of mass in the box). Test all boxes and separate the boxes that sink deeper compared to the boxes that didn't sink as much. The boxes that sunk deeper have all the real coins and the others have the fakes.

No...please...no more with the coins and the weighing...

I want to go home now.....

A little bit of logic wont hurt. It is just the sort activity required to get the "boring" weekend done with.

Remove the boxes from a create. Remove one coin from each box and mark the coin to which box it came from. Turn off the scale place all eight of the coins on it. Turn it back on. Note the weight reduction as you remove one coin at a time from the scale.

Boxes with the fake coins weigh 3000 grams each and the boxes with the good coins weigh 4500 grams each. This is a substantial weight difference that is easy judged by just manually "hefting" each box. The lighter boxes are easily identified using the qualitative human sense of weight differences because the weight difference is so large.

No scale is needed.

How many crates have we got?

The solution is similar to the previous coin problem except that this time
sampling of the coin is done thru a geometric progression rather than arithmetic.

Because of the limitations set by the scale, we'll be collecting coins (for weighing purposes) from seven (7)
out of eight (8) boxes in a geometrically progessing manner, i.e.

we'll take one coin for first box, two from the second box, four from the third, eight from
the fourth... and sixty-four from the seventh, a total of 127 coins (summation of 2^(x-1) where
x are box numbers from 1 thru 7).

Maximum mass possible using this arrangement will be 3740g (that is boxes 4 thru 7 contains genuine coins
while boxes 1 thru 3 contains fake ones) which is less than 4000g.

The trick then is to determine the deviation of the actual masses of the 127 coins from the mass
assuming all coins are genuine (127x30g = 3810g). If we divide this mass difference by 10g, then
we'll have an integer which is equal to the total number of coins from 3 or 4 boxes containing
fake coins (3 if we have sampled coins from 4 good boxes and 3 bad boxes, and 4 otherwise).
Remember we collected coins only for 7 boxes.

Hence if we got a mass of 3740g, which gives us a mass 70g less if all coins are genuine,
then we can conclude that boxes 1 thru 3 are all containing fake gold coins (total number of coins equals 70g/10g = 1+2+4),
and the eighth box is also a bad box.

...or if we have a mass difference of 890g (89 fake coins) then, boxes #1, #4, #5 and #7 are all bad boxes.
...or if we have a mass difference of 410g (41 fake coins) then, boxes #1, #4, #6 are all bad boxes, and the eighth
unsampled box is also a bad box.
...etc., etc.

Good mind practice! c:

Thats what I had in mind -

Of course, what's been ignored with this problem is the fact that real coins have a weight variation between samples anyway.

So the real coins won't all each be exactly 30g.

They will be something like 30g +/-.075%, probably with a normal distribution curve...

Meanwhile, the fakes will also have variations - maybe a greater percentage, as they will have been crafted in what is likely to be sub-standard manufacturing facilities.

My brain hurts.

This was intended as logic.

The South African Kruger Rand varies from 33.93 to 34 grams.

I don't see any problem determining fake coins even if we take into consideration this tolerance level (+/-0.075%).

With this tolerance level, we will only have worst case mass error of less than 3g for all of the 127 sampled coins (.075% x 30g x 127 coins), which is fairly less than half of 10g deviation of each fake coin from the genuine ones.

Thus, i'll go for this formulation even for practical purposes (assuming those scaling restrictions are possible, though i doubt they're not).

I would choose one coin from first box ,2 from second ,three from third, and so on up to 36 coins from the 8 boxes in the crate.

Weigh them. If they were pure gold would have weighed 1080 grams. Substract the real weigh from 1080 and find the possible combination of four numbers of which sum equals the above difference. These numbers would be multiples of 10 from 10 to 80.

Thus the individual numbers of which sum fit the above difference, represent the numbers of the faulty boxex in the crate.

I didnot check it,but assume there is only one possible combination.

Meir

I beg your pardon Meir but error in this arrangement can be realized in a snapshot.

Keep in mind that we have 4 bad boxes out of eight and not just one.

Collecting fake coins in this manner from boxes 1,4,7,8 or 2,3,7,8 for example, will give the same difference in mass. To site more,

...try combinations 1,2,5,8 or 1,2,6,7

...or combinations 1,3,5,8 or 1,2,6,8, and lots more!!

Take one coin from each box and put in a stack and place on the scale. The coins position in the stack corresponds to the location of the box it was taken from. Place the stack on the scale. Now start taking the coins off noticing if the new weight is either 20 or 30 grams less This is based on the scale freezing if adding coins. I am not adding coins, I am subtracting them

Weigh each crate, while the crate is on the scale note the total weight of the crate. Remove a box at a time and note weight drop on the scale and the total weight as each box is removed subtract the weight of the box from the previous total weight. Those dropping 4500 grams/box are real coins and those dropping 3000 grams /box are fake coins.

Since the scale only freezes if weight is added removal of weight will adjust on the scale.

Paul Dam

Dam@Damm.com

You might have forget that the scale can't carry that much weight. Not even a single box of genuine golds!!!

Cheers...

Its quite simple you are allowed to use the scale twice, taking the first four boxes take one coin from the first, two from the secound, three from the third, 4 from the fourth.

thus if all the coins are real their total weight is 400g

1x30=30

5x30=150

9x30=270

12x30=360

=810

if there is a weight difference of 10g then the first box is fake, a weight difference of 50g it is the secound box, a weight difference of 90 it is the third box etc. It it one of more of the boxes for example the first and secound there should be a weight difference of 60 etc etc

Its quite simple you are allowed to use the scale twice, taking the first four boxes take one coin from the first, 5 from the secound, 9 from the third, 12 from the fourth.

thus if all the coins are real their total weight is 810g

1x30=30

5x30=150

9x30=270

12x30=360

=810

if there is a weight difference of 10g then the first box is fake, a weight difference of 50g it is the secound box, a weight difference of 90 it is the third box etc. It it one of more of the boxes for example the first and secound there should be a weight difference of 60 etc etc

Oops!!!

"And you are allowed 1 (one) weighing per crate..."

I don't agree with being allowed to use the scale twice. "one weighing per crate"

In your case the scale would have to be left in the on position while taking away the previous lot. and will freeze as soon as new coins are added.

Ezekiel49 already supplied the correct solution in post 9.

Please return my scale before somebody breaks it.

Having just finished talking to MacGyver and seeing monk waiting I pulled out a 16.9 fl. oz bottle of water and pour a shot for me and monk. I drank mine he starts talking about his. I pull out my ruler and unlaced my shoes. I tie the water bottle to one end and make loop on the other end for the boxes. Depending on the weight of the box itself the water bottle should be heavier than each box of fake coins. I attach one box to the shoe lace and lift up the ruler in the middle and start sorting the boxes. If all the boxes are heavier than the water bottle I can pour monk's water back in the bottle or add an empty box to the water bottle side.Alas if I hadn't just finished talk to MacGyver I would I would have just compared the weight of the box in my left hand to the box in my right until I got them sorted as heavy and light. Then weighted one coin from each heavy to verify that they were gold. sigh I really don't want the bottle water will kill you maybe Monk will leave soon.

You have to set up a matrix. Take 1 coin from the 1st box, 2 from the second, 4 from the 3rd, 8 from the 4th, 16 from the 5th, 32 from the 6th, and 64 from the 7th. Mark off the 7 areas on the flat plate scale to put your 7 stacks of coins on so they don't get mixed up. You don't have to worry about the 8th crate because when you know what is in the first 7, you know what is in the 8th.

Realize that the puzzle stated that there are 4 boxes per crate that are fakes, that leaves 4 boxes per crate that are pure gold. Therefore there cannot be a combination of coins where there are 5 boxes with one type in a crate -- hence those possibilities can be eliminated. The matrix will look like this with the lightest possible shown:

array Weight

GGGLLLL 2610

GGLGLLL 2650

GLGGLLL 2670

LGGGLLL 2680

GGGGLLL 2690

That's the first 5 values of 72 total. G stands for gold and L for light of fake. The order L's and G's are in the array by box number -- 1, 2, 3, 4, 5, 6, 7.

The heaviest possible combination is:

LLLGGGG 3740

Which is below the scales maximum range of 4000. Every combination has a unique weight.