riddles galore

[ConcreteSchlyrd]

Originally posted by Kim Eugene
Proj, you're missing a line somewhere. Unless what you say is false, that there are not 8 chalk lines.

My ass is dumb.

[Fit of Rage]

XII is twelve, not eleven.

[ConcreteSchlyrd]

Point well taken. The answer has something to do with roman and arabic numbers though. I know that much.

[Project]

whatever. 7 lines. the "riddle" remains the same

\ / | | = |

[ÆNIMA]

Originally posted by Troll King
You are given 10 sacks, each containing an unknown number of balls. (Yes, they are ball sacks, so get the giggles out of the way first.) All of the balls weigh 10 ounces each, except for the balls in 1 particular sack which weigh 9 ounces each. 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?) What is the minimum number of weighings needed to be sure which sack contains the 9 ounce balls?

(And by "to be sure", I mean not having to luck out and pulling out a 9 ounce ball on your first try.)

Ok, assume you have a scale that measures in ounces. Possibly, by chance....if you pick out the 9 oz ball-sack first, 1 weighing would be necessary. Because the number you would get would not end in a 0, like the sacks with 10 oz balls would. Of course, on the opposite...if the last sack you pick up is the 9 oz sack, then only 9 weighings would be necessary, you obviously wouldn't have to weigh the final one...

[Cpt.Guano!]

need to weigh the balls them 4 times max

i think if you used a balance scale you would only have to weigh them 3 times

for the line thing

move one of the vertical lines from the left side of the equation and put it like / over the = so it says V | =/= |

so \ / | is 6 in roman numerals...

so it says 6 != 1


?

[ÆNIMA]

Originally posted by Cpt.Guano!
need to weigh the balls them 4 times max

i think if you used a balance scale you would only have to weigh them 3 times

explain!!

[Cpt.Guano!]

take 5 balls and weigh them

if you get 49 then keep those 5 balls otherwise throw them out and take the other 5 that you didnt weigh.

take 3 balls and weigh them

if you get 29 then keep those 3 balls otherwise throw them out and take the other 2 balls that you didnt weigh.

if you have 3 balls then take 2 of those 3 balls and weigh them. if you get 19 then keep those 2 balls otherwise the one you didnt weigh is the correct ball

if you have 2 balls then weigh 1 of them. if it is 9 then you have the correct ball otherwise the other one is the correct one

[ÆNIMA]

I'm not sure if you noticed, but there are 10 different sacks. Each with an unknown number of balls...and only one contains 9 oz balls...I'm not sure if your answer fits this description:\ Frankly it's late, and I'm tired.

[Troll King]

Originally posted by ÆNIMA
Ok, assume you have a scale that measures in ounces. Possibly, by chance....if you pick out the 9 oz ball-sack first, 1 weighing would be necessary. Because the number you would get would not end in a 0, like the sacks with 10 oz balls would. Of course, on the opposite...if the last sack you pick up is the 9 oz sack, then only 9 weighings would be necessary, you obviously wouldn't have to weigh the final one...

You didn't read my caveat. Your method has a 1 in 10 chance of finding the 9 oz sack on 1 measurement. There is, however, no way to guarentee that this will happen. There is however a way using just 1 measurement that will guarentee you will find the sack with 9 ounce balls with 100 percent certainty.

[MaGi kOz]

Originally posted by Project
whatever. 7 lines. the "riddle" remains the same

\ / | | = |

umm

[TelC@t]

Originally posted by Troll King
We place inside a box 13 white marbles and 15 black marbles. We also have 28 black marbles outside the box.

We remove two marbles from the box. If they have a different colour, we put the white one back in the box. If they have the same colour, we put a black marble in the box. We continue doing this until there is only one marble left inside the box. What colour is it?

Kim got the answer ... but I am repeating the case if we can't see what's inside the box and just draw 2 balls at random ... it's just more interesting
---------------------------------------
Assuming:

W <--- white ball
B <--- Black ball

(W W) --> (B)
2 white balls are drawn, resulting in 1 black ball being placed in the box
(W B) --> (W)
1 white and 1 black are drawn, white ball is placed back into the box
----------------------------------------
Now considering ONLY the number of white balls:


If there is only 1 white ball left at the end:

(W B) --> W

For any other number of white balls, 2 scenrio possible:

(if 4 white balls left)

1) 2 white balls are drawn,

W W (W W) --> W W B

2) 1 white 1 black are drawn,

W W W (W B) --> W W W W

(if 5 white balls left)

1) 2 white balls are drawn,

W W W (W W) --> W W W B

2) 1 black 1 white are drawn,

W W W W (W B) --> W W W W W

It can be shown that after each draw, the number of white balls either stays the same or decreased by 2. It can NEVER decrease by 1.

Since we started with 13 white balls, at the end it will be 1 white ball left.

(W B) ---> W

So there will be 1 white ball left at the end.

Number of black balls is irrelevant, but it can be shown that before there is only 1 white ball left, 1 extra black ball will be resulted whenever the box runs out of black balls.

for example:

W (W W) --> W B

[TelC@t]

Originally posted by Troll King
You are given 10 sacks, each containing an unknown number of balls. (Yes, they are ball sacks, so get the giggles out of the way first.) All of the balls weigh 10 ounces each, except for the balls in 1 particular sack which weigh 9 ounces each. 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?) What is the minimum number of weighings needed to be sure which sack contains the 9 ounce balls?

Originally posted by Troll King
There is however a way using just 1 measurement that will guarentee you will find the sack with 9 ounce balls with 100 percent certainty.

You feel balls from each sack with your hands, and the 9 ounce ball will feel slightly lighter. To be certain, you will weight it on a scale to be 100% sure.

[Kim]

That method sounds retarded, and if it is the answer, SHAME.

My method was this. Put 5 sacks on your right side, and the other 5 sacks on your left, to ensure you don't get confused. Take one ball from each sack on the left, and put it on the left side of scale. Do the same with the sacks on the right, on the right side of scale. The side that weighs less contains the 9 ounce ball. So, lets say it was the left side that had it. So as of now, there is a 1/5 chance that the sack will be the one with 9 ounce balls.

From here, there are two scenarios.

1) Put two balls from any two sacks on the left side of the scale, and put two balls from any of the other 3 sacks on the right side. If in fact they balance out, that means that the remaining sack contains the 9 ounce balls.

2) Put two balls from any two sacks on the left side of the scale, and put two balls from any of the other 3 sacks on the right side. If the sacks aren't balanced, the lighter side is the one that contains the 9 ounce ball.

From scenario 2), there is a 1/2 chance that you will find the sack with the 9 ounce balls now. So just weigh it out again, and the lighter one is it.

[Sleepy Weasel]

Originally posted by TelC@t
You feel balls from each sack with your hands, and the 9 ounce ball will feel slightly lighter. To be certain, you will weight it on a scale to be 100% sure.

That doesn't give 100% certainty on one weighing though, that's just some bad logic. You won't be certain until after you weigh it, so if you picked wrong, you'll need to keep guessing. And obviously if you have to keep going if you picked wrong, that's not 100% accuracy.

I had the same plan as Kim... I don't think you can do better than 3 without some cheap trick.