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[W4veMantis]

heh, that's right. umm...don't worry about the black house. We'll just say black bricks.
Alright:
There are three birds sitting on a fence. You shoot one of them with a rifle. How many birds are still on the fence?

 

[Volrec]

There are many answers:
2: You shoot one, it dies
0; Shoot one, they all fly away

 

[W4veMantis]

Hmm...i suppose you're right about both of them. The anwser I was thinking about was 0. Birds usually fly away when they hear a loud noise, you know.
Anyway:
What is so fragile that when you say it's name, you break it?

 

[Kragen]

A secret

Spoiler

blabla 10 chars and stuff

 

[W4veMantis]

Nah, but nice try though.
The anwser is Silence. If you say silence...well you pretty much break the silence.
Another one:
Until I am measured, I am not known. Yet how you miss me, when I have flown. What am I?

 

[buddy9246]

water? ahh! make em easier!

 

[W4veMantis]

LOL, sorry bout that I'll post an easier one.
I'll let the other one stay a little longer. See if anybody can guess it.
Here:
What kind of storm is always in a rush?

 

[Platypus]

No puns. Hurry-cane is a pun.

How about, What can never clean anyone, but gets cleaner every day?


EDIT: Whoops, said it wrong. What can never be cleaned by anyone, but gets cleaner every day?

 

[W4veMantis]

oops, sorry bout the pun.
umm...i know for a fact that it's not my house.
hmm...

Oh, the anwser to my other one was time. You really don't know what time it is until you look at a clock or somthing, and sometime's time just fly's by too fast.

 

[Anime Phr3ak]

This is more of a math question than a riddle but whatever I guess

1+1+2+4=8 and 1*1*2*4=8

8 is both the sum and product of the numbers shown above. What is the smallest number of numbers that both add up to and multiply to get 2007?

How is this possible? Isnt 2007 a prime number?

 

[digitalmaster287]

2007 isnt a prime, it has factors if you looked for them. If you dont understand the question you hafta find the smallest list of factors to 2007 that also add to 2007, the same way 1,1,2,4 both add to and multiply together to get 8.

 

[Kalmetam]

First of all, 2 and 7 add up to nine and the is divisible by three so it isn't prime. Second the person who said 92 was correct. It was pi (you can guess the pattern from there) 3.14159265... Third I have to prove my answer for the other problem now... It will be a while 'cause I lost it lol.

 

[Kalmetam]

Look here people, here is my work for the days in a decade problem. First the first 10 levels of the Paschal triangle add up to 2045 as the sum. Now that is how many years in a decade. Now for the alphabet part. 1+2+3+4+5+6+7+8+9+10..... all the way up to 26 is 341 (maybe you misunderstood what I was asking) so that's how many days in a year, now for the final problem. 2045 x 341 = 697345.

Edit: I was a bit off my first time (ok, a lot, I admit it). I might've added wrong. But this is the answer I've tried it a couple of times.



Answering the 2007 Question now:

You just need to use pretty much some logic and dividing. So it can't be divisible by even numbers so I tried all odd numbers 1-10. So 1,3, 9 go into 2007. I think 9 works the best out of them so my answer is 1 (1765 times) + 223 + 9 = 2007. Ok, am I right?

 

[digitalmaster287]

Theres a smaller list of numbers that add up to 2007 try again. The riddle is to find the smallest possible list that adds up to and multiplies to get 2007

 

[Doraki]

the 1st ten levels of the Pascal triangle add up to (2^0 + 2^1 + 2^2 + .... + 2^9) = 2^10 - 1 = 1023

the alphabet part is 26 * 25 * 24 * 23 * .... * 2 * 1 (26 choices for the a letter ; 25 for the b letter.. etc...)
(Anyway 1+2+3+...+26= 26*27/2 = 13*27 = 351, not 341)

 

[Kalmetam]

2 to the 10th power is 1024 but I don't see where you got the minus 1 from. You needed to add the others too! (+1) so it is 1+4+8+16+32+64+128+256+512+1024 = 2045. For the alphabet one, you were right but I totally don't see how you got the 1023

 

[Schweppes]

I got a puzzle.

1
11
21
1211
111221

What's the next line?

 

[digitalmaster287]

Umm Kalmetam 2^0+2^1+2^2...+2^10=2047, not 2045. And Doraki got the minus 1 because 2^x+2^x-1+2^x-2...all the way to+ 2^0=2^(x+1)-1

So then the answer would be 2047*351 which is 718497

 

[#HBC | FrozeηFlame]

I got a puzzle.

1
11
21
1211
111221

What's the next line?

112221111 ?

 

[Kalmetam]

Well whatever, lets just focus on newer riddles.

Is it:

112112

or
112112112112

 

[Eor]

Ah, crafty. If I'm right, you have to write the digits out with letters to get the pattern, treating each one like it was diffrent.

So, 11 would be "One one". 21 would be "Two one", or "two ones" if it is describing the set before it.

It should be 312211, as 111221 had three ones, two twos, and one one.

 

[Kalmetam]

Wow, i think that is right... good riddle. I'll say a simple one.

I travel around the world but I'm only in one corner, what am I?

 

[St. Viers]

a stamp...

riddle dee dum, riddle dee dee
many leaves have I but not a tree
nor yet a bush, nor yet a flower
but instead a gateway to a happy hour

 

[len1]

a book you are

 

[digitalmaster287]

Since no one got my riddle, I'll just say the answer. It takes 1337 numbers. 669, 3, 1(1335 times) all add up to 2007 and multiply together to 2007.

 

[Kalmetam]

Oh... I noticed it was divisible by 3 but I didn't bother trying the problem that way, oh stupid me! Well err... I guess I should make a riddle (not extremely hard math and fighting questions)

What is the next set of numbers?

You might be thinking of some famous sequence thingy but I don't even care if this is that or not. Chances are it isn't but this still is a pattern so don't complain about that thinking i tried making an exact copy of a famous pattern and my problem makes no sense because my pattern does make sense. It isn't that hard.

0,1,1,2,3,5,8,13,21,34,55,81,136
0,0,1,1,3,5,9,17,31,57,105,193,355
0,0,0,1,1,2,4,8,15,30,53,98,181

Here's another question

What's the difference between the sets of letters?


B,F,G,J,L,N,P,Q,R,S,Z
A,C,D,E,K,M,T,U,V,W,Y
H,I
X
O

 

[buddy9246]

OMG its 8:00pm! dont make my brain hurt!!! ahh!!!

 

[Schweppes]

Ah, crafty. If I'm right, you have to write the digits out with letters to get the pattern, treating each one like it was diffrent.

So, 11 would be "One one". 21 would be "Two one", or "two ones" if it is describing the set before it.

It should be 312211, as 111221 had three ones, two twos, and one one.

Just to confirm, this is correct. :D

@Kalmetam's problem: Each letter in the first line has no symmetry, each in the second have one line of symmetry, each in the third have two, the fourth (X) has three (y-axis, x-axis, and y = x line), and the fifth line (O) has an infinite number of lines of symmetry.

Am I close? >_>

 

[digitalmaster287]

OK I have another math riddle:

what is the highest integer value of n where the following equation is true:

8n^1999 < 16^1500

 

[Eor]

I got

n<1.001040782129.......

 

[Kuriko]

That's not an integer though. I think n = 0.

 

[digitalmaster287]

Nope thats not it.

 

[Kuriko]

Mistake... never mind.

 

[digitalmaster287]

Eor, I think you misunderstand the problem.
Its :
8(n^1999) < 16^1500

NOT

(8n)^1999 < 16^1500

 

[Kuriko]

Yeah I realized that... lol.

Second try:
8n^1999 < 16^1500
n < (16^1500/8)^(1/1999)
n < 8
n = 7

 

[Sandy]

A man was to be sentenced, and the judge told him, "You may make a statement. If it is true, I'll sentence you to four years in prison. If it is false, I'll sentence you to six years in prison." After the man made his statement, the judge decided to let him go free. What did the man say?

 

[Eor]

Eor, I think you misunderstand the problem.
Its :
8(n^1999) < 16^1500

NOT

(8n)^1999 < 16^1500

Ah, yeah, I did mess up on that. My bad.

Back to the drawing boards.

A man was to be sentenced, and the judge told him, "You may make a statement. If it is true, I'll sentence you to four years in prison. If it is false, I'll sentence you to six years in prison." After the man made his statement, the judge decided to let him go free. What did the man say?

"I'll serve 6 years in prison"

 

[digitalmaster287]

Still wrong, Im not sure how you made the mistake tho. It was right up till your 2nd step. The mistake is between the 2nd and 3rd step.

 

[Eor]

I'm getting 0.0005002501...., but that is not an integer.

What I did

8(n^1999)<16^1500
n^1999<(16^1500)/8
n<((16^1500)/8)^(1/1999)
N<0.0005002501

If you hit enter twice, you get 8.

 

[Kuriko]

Is it? If my second step is right, I don't know what's wrong either. I'm getting 8 from that calculation... *shrugs*

@Eorlingas: 0.0005002501 = 1/1999