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Riddles!


Idontwantspam

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6?

No. :no:

I believe I mentioned you were short two prisoners at the beginning of this charade :whistle:

lol

I really can't claim to be the first or only person to have come up with the answer. I was writing up my post when Idontwantspam announced the error. At the time, I didn't even think I had the right answer given the constraints. :angel

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I should also ask "why?" 6 is not correct, but even if I gave you the answer to the riddle, the next question would be "why." Otherwise, it will only take a few guesses to guess the next digit. So if you think you know the answer, also tell how it's the right answer.

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Here's one:

There are nine iron balls exactly the same size (as in diameter, circumference, not weight), shape, color. However, only 1 weighs more than the rest. You have scales. With exactly 2 weighings, how can you determine which one weighs more than the rest.

* You cannot put the balls on the scales one by one and call it one weighing. Meaning, you can't put one, then another, then another on either scale. If you want to weight 3 balls, you must put all three at once on the scale you want.

* One weighing means, say for example, putting 3 balls at same time on right scale and 4 balls at same on on left scale.

Other do's or don'ts will be provided based on answer.

If you know the answer, do not tell us the answer, just say you know.

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What is the next number in the sequence 3 3 5 4 4 3 5 5 4 _

3. Because 'ten' has three characters.

@spacesurfer:

I know.

Good job, Mijzelf! :thumbup I really thought that would take longer to figure out. In case anyone didn't follow that short thread:

The word "one" contains 3 letters.

The word "two" contains 3 letters.

The word "three" contains 5 letters.

The word "four" contains 4 letters.

The word "five" contains 4 letters.

The word "six" contains 3 letters.

The word "seven" contains 5 letters.

The word "eight" contains 5 letters.

The word "nine" contains 4 letters.

The given sequence was 3 3 5 4 4 3 5 5 4.

The next word in the series is "ten" which contains 3 letters, so the next number is 3.

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I made a diagram of the 1024 bottles and 8 prisoners. I don't see why my pattern won't work.

1st hour has 8 vertical blocks

2nd hour has 8 horizontal blocks

3rd hour has 8 vertical columns for each prisoner

and the 4th hour has 2 horizontal rows per prisoner.

Potential of one dead prisoner and up to 4 dead.

The picture is here.

DL

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A new one. Three men go fishing, and hire a boat. The boatlessor charges them for $30, so each man pays $10. While the men are walking to the landing-stage, the lessor remembers that the price is $25, so he sends his boy with $5 restitution. The boy decides that dividing $5 in three men is difficult, so he puts $2 in his pocket, and gives each man $1.

Now each man has paid $9, $27 in total. The boy has $2, that makes $29. Where is the remaining dollar?

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I don't have perfectly clear how you want to do this, but it seems to me that you forget that a prisoner cannot survive the 2nd hour, when he already died the 1st.

It is not in the second hour I have the problem, it is in the third. I could end up with two pairs of unchecked blocks as opposed to just one. Ok, so I double the tasters per column in the 3rd hour and triple in the 4th. I am ready to get as convoluted as necessary at this point. :rolleyes:

Dead men trial two.

What is surprising is that, so far, I have only used 7 sips off each bottle.

DL

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