Humour a little riddle...

Actually, this one's been solved.

Jedimdo told me over MSN. It was a cellphone and his keys.

But, I'll pass. No riddle from me.

 

actually this riddle page that jedimdo posted is fun!!!!! can't past the 8 lv though...

 

Right! The riddle has been solved by both pardal and Shinryu. So, any of them can post the next riddle.

And think that there are 80 levels! That means fun for a while. If you don't cheat, though.

 

i've posted this riddle earlyer but i haven't got an answer! try to answer this one...

 

my guess:

u weigh 4 balls on one side and 4 balls on the other. remove 4 balls which are more heavy and leave 4 which are less heavy. out of 4 ball which are left place 2 on 1st plate and 2 on second plate. from the heavier plate remove 1 ball and then remove 1 ball from the lighter plate. If the plates will be of the same balance that means that the ball that u just removed is the lightest. if they are not the same balance, then the ball on the higher plate will be the lightest.

 

thats not correct cause thats 3 chances, you only have 2 chances to height the balls...

 

You weigh one ball on one side and another ball on the other side...i don't know...I'm just guessing...

 

ummmmmmmmmmmm? ?? ???? ???? pass thats a little on the math side not my strong point

 

Look for the smallest one?

Find something better to do with you time?

Put 4 on each side and look for the lighter side and take a guess?

 

yeah , thats what i would probably do.

 

But the smallest one isn't noticeably lighter, why would it appear to be smaller?

Hmm... time for the riddle master (Me ) to have a go...

Let's see. 8 balls, two chances to weigh them, and you need to figure out which is the lightest one...

It seems to me to be mostly luck that would let you solve it. Let's see...

If you weigh six of the eight balls, and the scale is balanced, then you know, automatically, that one of the other two is the one you want, and it's a simple matter of weighing those two to find it. But, if they aren't balanced, that leaves... Ah-hah!

Okay. Weigh six of the eight balls. If the scale is balanced at that point, put those six aside, and weigh the other two. One of them is the one you want. However, if it's not balanced, take all three balls away from the heavier side, and put them with the other two. You have three balls left, and one of them is the lightest one. Weigh two of them, and if the scale isn't balanced, then you have the lightest one on the scale, otherwise, if the scale is balanced, you have it put aside (away from the other 5), so you know which one it is, anyway.

Am I right?

(Note, I left the entire thought process I went through in the post, so you can see how I figured it out... sort of)

Now, I need to think up another one.

[EDIT]Here we go:

My first is a number, my second another,
And each, I assure you, will rhyme with the other.
My first you will find is one-fifth of my second,
And truly my whole a long period reckoned.
Yet my first and my second(nay, think not I cozen),
When added together will make but two dozen.

How many am I?

(Okay, there was a different riddle up, but I got a notice back saying the guy's mailbox was full, so I couldn't confirm the answer. So, here's a different one)

 

sorry shinryu but nop... but if you follow a similar way of thinking you'll get there..


[EDIT]
to your riddle...
the first number is 4 and the second is 20????? they are = 24 that is two dozens and 4 is 1/5 of 20?????

 

No? But it works... :sad:

And, you're very close with my riddle. You've got to think more of language, and less of numbers. 4 and 20 don't rhyme. Or do they?

 

:dizzy2: Now I'm lost...sorry...if you give me a riddle like that (with math & rhyme)...then I'll be lost forever!!! :dizzy2:

 

sorry shinryu! i don't know what i drink that day to say that your answer to my riddle was wrong but i had something very strong to drink indeed...lol
your answer is correct. i was the one who was wrong sorry...

can't figure out your riddle anyway....

 

It's okay. I'm not mad or anything. Well, it's been long enough, anyway. The answer to mine was four score.

Anyone else have a riddle, or shall I post another?

 

lol... i would never have guessed it.

here's another...

in a chess board, place 8 pawns in the squares in a way that none of them cross it's view with each other, not in an horizontal way, not in a vertically way and not in a diagonal way. have fun

 

Saying the queen instead of a pawn would have been more corect. If you go by the rules of chest. With 8X8 squares. Little hard on the eyes but it works.

But here it is. The O is where they sit.

XXXOXXXX
XOXXXXXX
XXXXXXXO
XXXXXOXX
OXXXXXXX
XXOXXXXX
XXXXOXXX
XXXXXXOX

Does anyone have one to put up? I am to lazy to think of one.

Well about the ball one. they way it is worded sounds funny. First you height it. Which would make one or more bigger then the others. But you need to know what one weighs less. So by that one is smaller or one is larger then the rest.

Four balls on each scale take the lesser. So half them that would be two balls on each scale take the lesser again. 2 balls left. Can't feel it with your hands then you would have to look at them. The smaller one should be the lesser one unless the others have the same height then it is the bigger one. Unless this one was changed and not worded correctily then I am at a lost.

 

hehehe... the chess one is correct... i think?! it's a bit hard to see cause it's confusing but i think it's correct.

sorry for the error in the riddle of the balls, one of them is suposed to weight less then the others and not height less... lol
but anyway that riddle was already solved by shinryu....
if you don't want to post another one here i go again...

[EDIT]

ups!! already posted that one. forget it... i'll look for another.

 

i think that this riddle was already solved by someone.