This puzzle celebrates our puzzle solver Annie, who quietly reached over the 10 puzzle mark.

In this puzzle Annie is Manhattan's top private detective and she is being called to Tiffany's jewelry store to help investigate a fraud. Tiffany's manager informs Annie that they sold 10 sets of the Tiffany & Co Silver Utensils yesterday to 10 different people. Each set consisted of 24 pieces, all of them of the same weight. It came to their attention that one of the sets was a fake set that was supposed to be used as a window display. All the utensils in this set weigh 1/10th of the silver utensils. Tiffany's manager wanted to avoid a scandal and notification of all the buyers of a potential fraud. However, she realized that all the buyers should be called back to the store (with their utensil sets) in order to detect and replace the fake set. She asked Annie for the fastest way of detecting which of the 10 sets is the fake one in order to minimize the commotion and suspicion.

Annnie told her: "Reach out to all the 10 buyers and ask them to come to the store with the utensil sets they bought claiming that they forgot a butter knife that should be matched to their specific set. Once I have all the utensils here, all I need is to use your weight scale one time only to detect which set is fraudulent. You will need to present these buyers with 10 butter knives to leave them satisfied."

Do you think Annie was over-confident? How could she avoid weighting each of the utensil sets and use the scale only once to find the fake set?

Answers accepted all day long on Friday May 27th and Saturday May 28th, on our Family Puzzle Marathon. They will be hidden until Sunday morning (EST) and everyone who solved it will get a puzzle point. Please, explain your answer.

In this puzzle Annie is Manhattan's top private detective and she is being called to Tiffany's jewelry store to help investigate a fraud. Tiffany's manager informs Annie that they sold 10 sets of the Tiffany & Co Silver Utensils yesterday to 10 different people. Each set consisted of 24 pieces, all of them of the same weight. It came to their attention that one of the sets was a fake set that was supposed to be used as a window display. All the utensils in this set weigh 1/10th of the silver utensils. Tiffany's manager wanted to avoid a scandal and notification of all the buyers of a potential fraud. However, she realized that all the buyers should be called back to the store (with their utensil sets) in order to detect and replace the fake set. She asked Annie for the fastest way of detecting which of the 10 sets is the fake one in order to minimize the commotion and suspicion.

Annnie told her: "Reach out to all the 10 buyers and ask them to come to the store with the utensil sets they bought claiming that they forgot a butter knife that should be matched to their specific set. Once I have all the utensils here, all I need is to use your weight scale one time only to detect which set is fraudulent. You will need to present these buyers with 10 butter knives to leave them satisfied."

Do you think Annie was over-confident? How could she avoid weighting each of the utensil sets and use the scale only once to find the fake set?

Answers accepted all day long on Friday May 27th and Saturday May 28th, on our Family Puzzle Marathon. They will be hidden until Sunday morning (EST) and everyone who solved it will get a puzzle point. Please, explain your answer.

## 14 comments:

Wheigh together one piece from the first set, 2 pieces from the second, 3 from the third etc. 10 pieces from the 10th set.

If 1/10 is missing, the first set is fake, if 2/10 are missing, it's the second one, etc.

You can even be more economical by weighing only 9 sets. If nothing is missing, the set that has not been wheighed is the fake one. This in case one of the customers could not come at the same time as the others...

The silver is not supposed to be magnetic. A magnet test would work to discover the fake test before using the scale.

If Annie takes one piece from one set, two pieces from the second set, three pieces and so on, then using a fine scale, she can figure out how much less of the weight it should be (say its short 1/10th the weight) and this tells her that the fake set is from the first set.

If its short 2/10, its from the second set and so on.

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Weigh 1 piece from customer #1, 2 pieces from customer #2, 3 pieces from customer #3, etc., up to 9 pieces from customer #9.

Assume the weight of each piece is w.

Then if customer #1's set is fake, it weighs

(1/10 + 2 + 3 + ... + 8 + 9)*w = 44.1*w

if customer #2's set is fake, it weighs

(1 + 2/10 + 3 + 4 + ... + 8 + 9)*w = 43.2*w

if customer #3's set is fake, it weighs

(1 + 2 + 3/10 + 4 + ... + 8 + 9)*w = 42.3*w

and following this pattern:

if customer #4's set is fake, it weighs 41.4*w

if customer #5's set is fake, it weighs 40.5*w

if customer #6's set is fake, it weighs 39.6*4

if customer #7's set is fake, it weighs 38.7*w

if customer #8's set is fake, it weighs 37.8*w

if customer #9's set is fake, it weighs 36.9*w

For the last case, it's a little bit different since we weigh no fake pieces of silver:

if customer #10's set is fake, it weighs

(1 + 2 + 3 + ... + 8 + 9)*w = 45*w

She can do it with one weigh!

1) Get the 10 sets

2) Label them 1-10.

3) From set one, take one utensil and label it as coming from set one.

4) Take two utensils from set two and label them as coming from set two.

5) Repeat with 3 from 3, 4 from 4, 5 from 5, 6 from 6, 7 from 7, 8 from 8, 9 from 9, and 10 from 10, labeling each piece as to the set it game from.

6) Take the 55 labeled pieces (1+2+3+4+5+6+7+8+9+10) and weigh them.

7) You know what 55 real pieces should weigh; you will get a number less than this. The difference between what they should weigh and what you get will be a multiple of 1/10th of the weight of a real piece.

8) Divide the difference by 1/10th of the weight of a real piece; the resulting number from 1 to 10 will tell you which set is the fake!

Dennis

I don't see that it's a fraud in the legal sense; just a mistake. But that's not the point of the puzzle.

Nor do I see, yet, how "one time only" is going to discover, for sure, the lighter display sample, unless she gets lucky. Someone has already proved me wrong, betcha.

Whether she weighs the entire sets, or just the forks, or single forks, the process should be the same....

But if the display set really weighs only 1/10 of the genuine sets, the heft is going to be VERY easily noticed. And THEN I suppose she could use the scale, once, just to be sure.

Take one utensil from the first set, two from the second, three from the third, etc. ... then from the tenth. Weight them all together. If there were no fakes, the weight would be N*(N+1)/2 = 55 times the weight of one genuine piece. To detect which of the sets if fake, take the difference between the weight you get and 55, and divide by the weight of one fake (1/10th of genuine). You get a whole number, which is the number of the set that contains all fakes.

P.S. why is it that even disposable forks are called silverware? :-)

I think I'm stymied...but I do know that if one fork weighs 1/10 as much as another, anyone could tell the difference just by picking it up. Politely lifting a piece of the customer's flatware to "compare" it to the butter knife would answer the question without a scale. It would also allow you to bring in customers one at a time until the bad set was found, which might allow you to alert fewer customers and save some bread knives.

But I know that's just evading the math problem that has me stumped...

Aha!

We know the weight of a utensil. I'm going to call that "x".

If I take one piece from the first set, two from the second, three from the third, etc. until I get to taking ten from the tenth, I will have 55 pieces.

If they were all good, they should weigh 55x. If the piece from the first set was bad, it will weigh 54.1x (54x + 1 piece that weighs .1x). If the two pieces from the second set was bad, the weight will be 53.2x (53x + (2 * .1x)). If three pieces from set three are bad, then the weight is 52.3....all the way to the tenth set, which should weigh 46x.

So you would know which set was bad from that weighing. But honestly, it might be best to tell the mathematician that she's down a rabbit hole and pull over the common sense clerk who can tell the weight by feel.

We could take 1 piece from the 1st set, 2 pieces from the second, 3 from the 3rd, etc., but none from the 10th set. This makes 45 pieces total. The total weight should be 45 units. If it actually is 45 units, you know the 10th set is the display. If the total weight is 44.1 units, then the 1st set is the display. Since there is a different number of items from each set, we can see how short the combined weight is.

Having set many tables and washed much silverware, I am confident that one can distinguish between one utensil set and another that weighs 1/10 of the first one. Therefore just carrying the utensil sets to the back of the store will give some indication by comparison of which set is the fake. I'm sure Tiffany's knows (or can find out) how much a genuine set weighs. When the suspect utensil set is identified, it can then be verified as the fake by one weighing on the scale. This is not sophisticated or scientific but I think it'll get quick results!

No, she is not overconfident and this is why. You take 1 piece from the first set, 2 from the second set, 3 from the third set and so on. If the real set weighs 10 grams apiece then the fake set will weigh 1 gram apiece. You set your 55 pieces on the scale. the amount that it is off, the total should equal 550 but it weigh less than that. The total will be off by 9 grams per fake piece. e.g. the total if the total is 541 grams, there is one bad piece, meaning that set 1 is fake. If the total was 487 grams, that is 63 grams off. 63 being 9x7 means that there are 7 fake pieces making set 7 fake.

Ok, the trick is out.

Congratulations to everyone who either knew or figured it out. Kj provided a great step by step explanation. The rest of us will apply this trick next time when this puzzle comes in another reincarnation - as coin or diamond weighting problem.

A puzzle point for TyYann, Wang, kj, Ilya, Bean, SteveGoodman18, Lynnet.

Tom, Annie and anne-Marie - you each suggested some nice workaround solutions but I think they are not technically count as the correct solution. As you each are "celebrities" of our puzzle marathon with plenty of points, we will skip rewarding you for this one.

Have a great long weekend.

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