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-- The Riddle Thread
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quote:

Originally posted by Syphiux It cannot be seen, cannot be felt, Cannot be heard, cannot be smelt. It lies behind stars and under hills, And empty holes it fills. It comes first and follows after, End life, kills laughter. What is it?

c'mon, this is from "the hobbit," but it seems as if you're combining the time and darkness riddles.

quote:

Originally posted by Noisician no, the real answer is 301. but u still owe me the formula, since 301 is just the smallest solution of many possible ones edit. and by "possible" i mean the ones that make perfect sense)

You said "at least" so 301 is the right answer, you never said anything about all possible solutions.

(But I'm curious to see the formula, hehe)

quote:

c'mon, this is from "the hobbit," but it seems as if you're combining the time and darkness riddles.

It isnt combined, and heres another riddle:

Two sheep farmers meet out in the fields. The frist farmer said to the other:
-If I give you you of my sheep, then you will have twice as many sheep as i do.
Then the otehr farmer said:
-Thats right, but if I give you one of my sheep, we will have the same amout of sheep.

How many sheep did they have?

First one has 5, second one has 7

quote:

Originally posted by Omegasox You said "at least" so 301 is the right answer, you never said anything about all possible solutions. (But I'm curious to see the formula, hehe)

shut up, meatbag. it was obviously implied in my post

quote:

Originally posted by Syphiux It isnt combined, and heres another riddle: Two sheep farmers meet out in the fields. The frist farmer said to the other: -If I give you you of my sheep, then you will have twice as many sheep as i do. Then the otehr farmer said: -Thats right, but if I give you one of my sheep, we will have the same amout of sheep. How many sheep did they have?

x - 1st farmer, y - 2nd farmer

first condition: y+1=2(x-1)
second condition: x+1=y-1

when u solve it, u'll get x=5 and y=7

ok, using the numbers 1, 5, 5, and 5, get a result of 24 using only multiplication division addition and subtraction

quote:

Originally posted by piggy ok, using the numbers 1, 5, 5, and 5, get a result of 24 using only multiplication division addition and subtraction

5x(5-1/5)

heres one of my favourites:

As soon as you say what i am, i no longer exist. What am I?

silence ?

quote:

Originally posted by Noisician silence ?

oldie:

2 coins (US money) make up 30 cents. one of them is not a nickel.
what are the coins?

quote:

Originally posted by Inertia oldie: 2 coins (US money) make up 30 cents. one of them is not a nickel. what are the coins?

but the other one is a nickel. the first one is a quarter.

I got one that took me a while to solve.......

You have 9 machines (numbered one through nine) that produce screws. However, one of them is producing screws that are .01 grams overweight (a difference that you couldnt possibly recognize without an accurate scale)...... Now, you are given a very accurate scale, however, you can only use the scale once (only one load can be placed on the scale and it can't be touched/changed)....... How do you find out which machine is the one producing the overweight screws?

quote:

Originally posted by A_M_8 I got one that took me a while to solve....... You have 9 machines (numbered one through nine) that produce screws. However, one of them is producing screws that are .01 grams overweight (a difference that you couldnt possibly recognize without an accurate scale)...... Now, you are given a very accurate scale, however, you can only use the scale once (only one load can be placed on the scale and it can't be touched/changed)....... How do you find out which machine is the one producing the overweight screws?

that's a good one. here's what i think:

u need to take 1 screw produced by the 1st machine, 2 screws produced by the 2nd machine, ... , 9 screws produced by the 9th machine and put all of them on the scale. now, if all the machines produced the same screws, the ideal total weight would be (45x) grams, where x is the weight of one screw. by comparing the actual weight to the ideal one, u can figure out which machine produces defective screws because, in this case, the actual weight will vary, depending on which machine contributed to the overweight.

if the actual weight is (45x+0.01) grams - it's machine 1
if the actual weight is (45x+0.02) grams - it's machine 2
if the actual weight is (45x+0.03) grams - it's machine 3
...............................................
if the actual weight is (45x+0.09) grams - it's machine 9

make sense?

Family guy:
Ok, here's another riddle. A woman has two children. A homicidal murderer tells her she can only keep one. Which one does she let him kill?

quote:

Originally posted by moncster Family guy: Ok, here's another riddle. A woman has two children. A homicidal murderer tells her she can only keep one. Which one does she let him kill?

the ugly one.

quote:

Originally posted by Noisician just remembered one decent math problem/riddle i had to solve recently. there are a certain number of books lying on the floor that u need to pack in boxes via tying them up into packages. u first try to make some packages with only 4 books in each, but then u notice that, at the end, 1 book is left unpacked. u then try to make a number of packages with 5 books in each, which also leaves only 1 book left. so u make packages with 6 books in each, which, once again, yields one single book remaining on the floor. finally, tying up 7 books together leaves u no additional books whatsoever. my question is, *at least* how many books had to be lying on the floor? it's pretty easy, though

4a + 1 = n
5b + 1 = n
6c + 1 = n
7q = n

n is divisible by 7.
n - 1 divisible by 4,5 and 6. lcm(4,5,6) = 60. thus n-1 is divisible
by 60.
so need only consider those n, such that n-1 divisible by 60, and n itself is divisible by 7.
n - 1 = 60a
n = 7b.

hence: 60a + 1 = 7b. Now we can solve this using Euclid's GCD algorithm.

done.

quote:

Originally posted by Durafei

GCD algorithm isn't really a formula... there's a more formal and concise mathematical way of dealing with this question, but I forget what it is, lol. But I do remember that it was the answer to one of Flyboy's questions a couple of months ago.

quote:

Originally posted by DigiNut GCD algorithm isn't really a formula... there's a more formal and concise mathematical way of dealing with this question, but I forget what it is, lol. But I do remember that it was the answer to one of Flyboy's questions a couple of months ago.

Well, as far as I remember you need to apply GCD algorithm (well, actually the reverse of GCD algorithm) and you'll get the solutions.

but i've learnt that stuff over 3 years ago, and I can't remember exactly how it goes.

quote:

Originally posted by Noisician that's a good one. here's what i think: u need to take 1 screw produced by the 1st machine, 2 screws produced by the 2nd machine, ... , 9 screws produced by the 9th machine and put all of them on the scale. now, if all the machines produced the same screws, the ideal total weight would be (45x) grams, where x is the weight of one screw. by comparing the actual weight to the ideal one, u can figure out which machine produces defective screws because, in this case, the actual weight will vary, depending on which machine contributed to the overweight. if the actual weight is (45x+0.01) grams - it's machine 1 if the actual weight is (45x+0.02) grams - it's machine 2 if the actual weight is (45x+0.03) grams - it's machine 3 ............................................... if the actual weight is (45x+0.09) grams - it's machine 9 make sense?

did you know this one already or did u figure it out right now? bec. thats the fastest time someone has anwered that riddle for me

quote:

Originally posted by A_M_8 did you know this one already or did u figure it out right now? bec. thats the fastest time someone has anwered that riddle for me

nah. about a year ago, there was a semi-similar problem in one of my math classes. it gave me a general idea how to do yours.

anyway, here's another riddle that i think is pretty good (and hard)

it's a riddle about two mathematicians, mister a and mister b.

a certain person picked two natural numbers, each of which was greater than 1 and smaller than 100. to mister a, that person gave only the sum of those numbers. to mister b, the person gave only their product. they then were asked to find the numbers, but they were not allowed to tell each other what they initially knew about those numbers.

when they met, they had this short dialogue:

i'm afraid i can't deduce the right answer. there is not enough information given.
yes, i know that u can't possibly know the answer. i can even prove that.
in that case, i know what the numbers are, as u just unintentionally gave me a great hint.
but now i, too, know the answer, as what u just said to me was enough for me to pick the right numbers.

believe it or not, that dialogue should be enough to enable u to figure out the initial numbers, without even knowing their sum or product. so what were they?

Here's one I heard ages ago... Apparently this is another one that kids do well on but uni graduates rarely get.

What is this?

It is greater than God,
More evil than satan,
The impoverished have it,
The rich want for it,
If you eat it you will die.


PS - No googling... If you can't figure it out yourself, or you've heard it before don't post.

[no text]

quote:

Originally posted by astroboy Here's one I heard ages ago... Apparently this is another one that kids do well on but uni graduates rarely get. What is this? It is greater than God, More evil than satan, The impoverished have it, The rich want for it, If you eat it you will die. PS - No googling... If you can't figure it out yourself, or you've heard it before don't post.

The answer is "Nothing"



What will go up but never come back down?

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