Puzzle Corner

[superlion]

is the scale used in the solution?

[Jay]

Yes, the scale is used in the solution. When done correctly, the reading from the scale will indicate which drum contains the rivets that weigh less. The tough part of the puzzle is determining what to put on the scale to weigh. The drums are not sealed.

[superlion]

Well you want to put rivets on the scale to weigh them individually (rather than a whole drum at a time) because there aren't a set number per drum, right? Well, I still don't see how to use the scale and get it with one reading.

[Jay]

It is correct that there are not a set number of rivets per drum. (However, even if there were, the approach would still be the same.) It is also correct that rivets will be taken from drums and weighed. And it is also correct that we are trying to find the approach used that will allow those rivets to be weighed only once. (This is a classic puzzle because it is tough for most people to come up with the approach; yet the approach is perfectly logical.)

[superlion]

*wishes she were an engineer*

"feel" for the weight of the rivets then put the one that feels lightest on the scale to verify? (but you said the rivets feel the same :\ )

[Jay]

The rivets do indeed feel the same. So, no, you cannot pick the lighter one by feeling it. It is very unlikely that anyone can notice such a small weight difference.

And, no, being an engineer would not necessarily help. However, this puzzle (and probably all of the "story" types of puzzles) have a better chance of being solved with "right brain" creativity (which some engineers possess); while the "word" or "phrase" types of puzzles have a better chance of being solved with "left brain" creativity (which some writers possess).

What would happen if diifferent numbers of rivets were taken from drums?

[Capt.Rutlinger]

ok, you take let's say 10 rivets from the first drum then 9 from the second, then 8 from the third and so on,... until you take 1 from the last one

so the total amount of rivets is 10+9+8+7+6+5+4+3+2+1=55
now the total weight of this would be (if the rivets were all normal) 55*6= 330 ounces. Now since we know that there is one drum that has been filled with wrong rivets that weigh 5 ounces each when you read the scale you become another number (not 330) now do this: 330-x (x=the number you read on the scale); divide the solution you get by 5 and you get a number from 1 to 10 so now you know wich on is the drum with the wrong rivets cause when the final solution is let's say ten, the you've put 10 wrong rivets on the scale and since you've got them out of the first drum you now know that drum number 1 has been filled with wrong rivets.

BTW thanks jay for the hint

[Jay]

That is close enough. You actually do not have to do any dividing. It is not that the lighter rivets weigh 5 ounces; it is that they weigh 1 ounce less. And the technique can be made a little less confusing.

So here is the official answer:

Take one rivet from the first drum, two from the second drum, three from the third drum, ..., through ten rivets from the tenth drum. You will then have 55 rivets.

(As a side note, there are two different ways to determine how many rivets you have. You can either add 1+2+3+4+5+6+7+8+9+10. Or, if you are familiar with arithmetic progressions in algebra, you can use the formula, N x (N + 1) / 2 where N is 10. It does not matter, as long as you know there are 55 rivets.)

Weigh these 55 rivets together. It must be assumed that the 55 rivets can easily fit on the scale (which is usually true at places that make or use rivets). If they were all 6 ounce rivets, the total weight would be 330 [55 x 6 = 330] ounces. Subtract the actual weight from 330 and that is the drum with the 5 ounce rivets. For example, if drum one had the lighter rivets, the one rivet taken from there is 1 ounce less. (The total weight will be 329 ounces.) If drum two had the lighter rivets, two rivets were taken, each weighing 1 ounce less, for a total of 2 ounces less. (The total weight will be 328 ounces.) If drum three had the lighter rivets, three rivets were taken, each weighing 1 ounce less, for a total of 3 ounces less. (The total weight will be 327 ounces.)

Although this was just a puzzle, it is something that could exist in real life. In real life, you probably would be allowed to use the scale more than once. However, there is settling time in taking the readings. So this approach is actually much faster than using the scale up to 10 times.

[Jay]

And the new puzzle is... the same puzzle, well with one slight twist. But before that, a story.

You probably have had times in your life when you did something especially well, something that made you proud of yourself, possibly a moment of inspiration. Here is one of mine.

The previous problem was similar to a problem given in an early Columbo episode where the killer was a member of MENSA. (Columbo was a detective on a television show.) That problem involved bags of coins where a counterfeit coin had a different weight than a good coin. Only one bag contained legitimate coins and you had to pick which bag you wanted to keep after only one weighing. The above solution is the one given by Columbo at the end of the show, but I came up with a more flexible solution before Columbo gave his solution.

So back to the next puzzle. There are actually many solutions to the previous puzzle. The above solution is the most efficient when there is only one drum with 5 ounce rivets. It is the most efficient solution because it involves weighing the least number of rivets. However, it will not work if there is more than one drum with 5 ounce rivets. So this puzzle is to come up with a more flexible solution that allows any number of drums to have 5 ounce rivets. You have to know which drums have 6 ounce rivets and which drums have 5 ounce rivets after only one weighing.

This is tougher than the previous solution. But I want people to have a chance to think about it before I give a more flexible solution. The above solution could be found just with "right brain" creativity. But the more flexible solutions will be more obvious to either computer nerds or mathematicians.

[Capt.Rutlinger]

That is close enough. You actually do not have to do any dividing.

yeah I figured that out to yesterday eevening, but it was to late for me to get the pc back running

now for the next one does it involve chance calculations?

[Jay]

No, it does not involve chance calculations, Just like the previous solution, it involves taking specific numbers of rivets from each drum. However, the numbers will not be sequential like they were in the previous solution. Like the previous solution, we are looking for what those numbers are and how the reading should be interpreted. There are actually many solutions. (And I will accept any of them.) But there is one solution that is more efficient than the others.

[Capt.Rutlinger]

is it about using the last letter of the addition?

[Jay]

I do not understand the question.

[Capt.Rutlinger]

well if they would weight 10 ounces, the total amount of grams would be 550 and then the the last letter of the weight you read on the scale would be the amount of rivets you've taken, that weigh 9 grams. Anyway in the other solution is algebra involved?

[Jay]

I understand the question now. No,the solution does not involve only the last digit in the reading. All digits are important. (That is actually a HUGE clue, but it will only make sense when you know the solution.) But you might be on the right track, if you can think about all digits. Algebra is not involved in the solution.

When a correct solution is found, the one reading will tell you which drums have lighter rivets and which drums have heavier rivets. Any number of drums can have the lighter, 5 ounce rivets, as few as 0 drums and as many as all 10 drums.