riddles galore

[Troll King]

I think you guys deserve a gold star for reading for paying such close attention to the part of the question where I said:

"Unlike similar puzzles, you have a scale that actually gives weight measurements (Honestly, why would anyone use a balance scale for a job like this?)"


As for the question about the black and white marbles, yes, you guys got the right answer, though you made the solution much more complex than it should be.

The easy, simple answer is that white balls can only be removed from the box in pairs. Since there is an odd number of white balls, there will always be one white ball left inside the box.


Now, back to the other question, there is indeed a way to do it with less than 3 measures, Sleeps. No "trick" about it, just plain old logic. I'll wait a little bit and post the answer later this evening if no one gets it by then, unless someone requests more time.

[syzygy]

this is really hurting my mind, but all i can think of is this:

place all the bags on the scale and remove them one at a time. the bag that makes the scale read a number ending in 0, or the remaining bag, is the one with 9-ounce balls in it.

flaw: this can only work provided the amount of balls in the bag with 9-ounce balls is not a multiple of ten.

[TelC@t]

Don't you mean put one ball from each bag on the scale ... so initially there are 10 balls on the scale, then remove 1 each time and the one that reduces the scale by 9 ounce is the ball from 9 ounce bag?

Of course the balls and sacks are marked.


Yeah I agree this one sounds less retartded than my 1st answer.

[Troll King]

Do we give up?

[ÆNIMA]

I do, and I speak for everyone, on all occasions.

[Troll King]

Okay, here's the solution. This only takes 1 weighing, and involves no trickery; just plain old mathematics.

Select 1 sack and remove 1 ball from it. From a second sack, remove 2 balls. Take 3 out of the third sack, and so forth until you take out 10 balls from the tenth and final sack. You should now have:

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55 balls selected from the 10 sacks.

Put these balls together onto the scale and measure their weight. If all the sacks contained balls weighing 10 ounces, the weight of these 55 balls would be:

55 * 10 = 550 ounces.

However, 1 sack contains balls that weigh 9 ounces each. Let's say for example if the first sack contains the lighter balls. Since we took just 1 ball from that sack, we'd have 1 ball weighing 9 ounces and 54 weighing 10 ounces. The scale would thus read:

1 * 9 + 54 * 10 = 549 ounces.

Likewise, if it was the second sack that contained the lighter balls, then those 55 balls we took would include 2 weighing 9 ounces and 53 weighing 10 ounces. That would give us a weight of:

2 * 9 + 53 * 10 = 548 ounces.

Following this progression, if the 10th sack contained the lighter balls, that would mean that our 55 ball sample would contain 10 balls weighing 9 ounces and 45 weighing 10. That would give us a measure of:

10 * 9 + 45 * 10 = 540 ounces.

What does this mean? Well if you look at the numbers, you would find that if you take the measurement and subtract it from 550 (the weight if all the balls weighed 10 ounces) you would get a number from 1 to 10. That number would correspond to the number of light balls you placed onto the scale, and therefore would tell you which sack contained the lighter balls.

Weight:--------Sack #:

549----------------1
548----------------2
547----------------3
546----------------4
545----------------5
544----------------6
543----------------7
542----------------8
541----------------9
540---------------10

[MaGi kOz]

i was thinking about something like that but i didn't know how to put it into words.

[Project]

square root of 1 = 1

[MaGi kOz]

Originally posted by Project
square root of 1 = 1

ah damn you're right.

[syzygy]

sorry to be a stickle in ur behind, tk, but seeing as we didn't know how many balls were in each bag, what if there weren't enough balls to take out of, say, the tenth bag?

i must say, tho, that was the best riddle so far.

all hail tk.

[Troll King]

Ballsacks pt deux

Okay, we have the same situation with ten sacks containing either 10 ounce balls or 9 ounce balls. The difference this time is that we don't know how many sacks contain 9 ounce balls or how many contain 10 ounce balls. For Syzygy's sake, each sack contains at least 1000 identical balls, and you still have your trusty scale that gives precise weight measurements, not one of those balance scales in more cliched questions.

How can we identify which sacks contain 9 ounce balls with the fewest number of measurements?


Yes the solution follows a similar track to the last one, with one vital change.

[TelC@t]

Still 1 measurement.

Just pick

1 ball from the 1st bag

2 from the 2nd

4 3rd

8 4th

16 5th

32 6th

64 7th

128 8th

256 9th

512 10th

basically you need to pick the sum of all the numbers of balls from the previous bags, plus 1, from the next bag.

Also, there are more than 1 way to solve this problem, but my way will result in the minimum number of balls that have to be measured.

[syzygy]

Re: Ballsacks pt deux

Originally posted by Troll King
For Syzygy's sake,

do NOT spell my name with a capital letter. that is all.

[3Stripes]

heres one:

there is one light bulb upstairs and 3 switchs downstairs, one of the switchs turns on the light. how can u find out which switch controls the bulb, for sure, only going upstairs one time?

[Troll King]

Flip two switches, leave them on for a bit, then flip the second one down.

Go upstairs.

If the light is on, then the switch you left up is the one. If the light isn't on, it must be one of the other two. Since you flipped the second one on for a bit, if it was the correct switch the bulb would still be warm. If the bulb is cool, the third switch must be the one.