Imperial Hero International
General Category => General Talk => RPG Area => Topic started by: Sheremetev on August 30, 2010, 12:35:01 PM
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First topic in RPG area!
... though I think this is the place for all forum games.
So, rules are simple- the first person to guess the right answer posts the new teaser (in case you don't have one ready yet- you might give the right to someone who has already come up with one).
(http://a.imageshack.us/img839/6621/mathq.jpg)
Couldn't come up with something more original, sorry.
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:evil4:
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What, I told you I am out of ideas. Besides, messing with maths was not my idea. ;)
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The... what-root?
(Sorry, never heard the expression "circle root" in english before and I can't seem to get what exactly it is in math language... Google doesn't help either...)
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I guess ridle is made ok, and that pi m is not missing 2(square) on them.
Is it maybe 100 cm?
or 200 cm?
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I don't say the riddle is not correct. I say I don't know what circle square is. ???
Can someone put an image/link with its mathematical expression?
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Math lessons even here *xxx*
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I think it has to do with language "playing"... SQUARE->circle...
And since you gave dimensions m (meters) will remain m. So i guess the result will be πm.
Or πroot (at english proot...) funny sound...
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Circle root doesnt exist, atleast as far as i know. Square, cubic... exist.
Maybe its playing like for square - square root of 4 m2 = 200 cm. its square 2*2 m - surface area.
So if for circle you have surface area of pi m(i guess in this case it should write 2) - r2*pi=pi m so r2 =1 -- r= 1 m or 100 cm. But i dont know if thats solution. Or there is some word game in that ridle, because english isnt my native language, so i am not good in word games.
But for answer, we must wait creator of ridle.
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I will not give you the answer. Not yet. But I will give you a hint and +k to Tiger and Crassus, since they noticed both the catches here (and the analogy which solves half the riddle).
1 m is not the correct answer (nor 100 cm).
Hint: I never make mistakes (atleast at maths). The word game goes only as far as what a "circle root" is supposed to be. And the answer might be a little out of line.
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Oh lol, I didn't realise it was a riddle... I though it was a math problem. *xxx*
I'll be back if I have a solution. :)
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Ok, I'd say 50cm. :)
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Since Curion allready wrote pi m, i guess 100 pi m isnt solution? Logic behind is that circle root is first root, so it doesnt change anything, just change it into cm as you did in first example.
50 cm i solution if you think of pi m as perimiter(way arround circle - his length) 2 r pi = pi m - that crossed my mind first but i decided to go with P= r2 pi formula.
I am really curios, what the solution will be.
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Our problem here is to define "circle root". Since Sheremetev measured it in meters and not square meters I don't think it's some kind of surface we need to root but length (that's how I understood it at first at least). That's why I used the circle's perimeter formula instead of its surface.
Something still doesn't feel right though... I'll recharge myself and then try to find out what's not in the right place.
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I think i got something...
Since square have 2 dimensions and the root is the one dimension, if we apply the same to the circle. π (at meters) must be the circumference thus the root is one of it's "dimensions" the radians->R=0.5m... Cuorion's answer...
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Following one logic or another, Cuorion got the right answer. So, get creative and give us a new riddle. :)
P.S: Just to keep you occupied until and to satisfy my curiousity, did it cross your mind that the answer could be neither linear, nor planar? Imagine it was circular root (or smth), sticking to Tiger's first idea, what would you do with the measure unit part?
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Imagine it was circular root (or smth), sticking to Tiger's first idea, what would you do with the measure unit part?
Request a nobel prize if we managed to solve it? *hihi*
Anyway here's my teaser.
You toss a coin three times. First one was heads. What's the chance that it came tails at least once?
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75% or 3/4?
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Now if that was right, it would be too easy, wouldn't it? ;)
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For most people it wouldnt be to easy. I teach highscool kids same stuff and it needs a lot of explaining to understand it.
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I don't know, I always thought of conditional probabilities really intriguing.
You grasped the main idea, but there's a catch in there.
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Ok, right to the catch- how many sides does the coin have? :D
I assume that the only possible outcomes of each toss are heads and tails (equally possible- each 50%).
So, equally possible chains of events are:
H H H - 25%
H H T - 25%
H T H - 25%
H T T - 25%
All in all, 75%, as Tiger said.
If we assume this is the "dinosaur problem" from a more generalized point of view, chances are 50%- it will either fall tails atleast once or it will not. Just like the chances of seeing a living T-Rex (velociraptor for all xkcd fans)- you will either see one or not.
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Hint: What both of you did was "toss the coin twice" (since you both calculate 2² different outcomes).
EDIT: No dinosaurs were harmed during the testing of this teaser. :D
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If there are 3 tosses, then the solution is 87,5%.
Because if you toss 3 times and you know 1 outcome(first one was head), then it doesnt matter for tail in next 2 tosses. Thats why we calculate with 2 tosses.
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I never said though, that you make 2 tosses after you learn the first was head. ;)
Let's just say that there's another answer that I would consider more accurate. I think it's a matter of perspective, actually. If you think I caused too much confusion, I can always tell you what's in my mind and post another more... elegant mathematical riddle.
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After a bit of thought, I came to the conclusion that it's better if we stick to riddles/teasers that can have a definite answer. So let me write my point of view on the last matter and give a more solid riddle to solve.
On the previous problem we had two conditions to fulfill.
1) Get a head on the first toss.
2) Get a tail at least once.
We tossed 3 times, so all the possible outcomes would be 2³, or in detail,
Head Head Head
Head Head Tail
Head Tail Head
Head Tail Tail
Tail Head Head
Tail Head Tail
Tail Tail Head
Tail Tail Tail
Only 3 out of 8 fulfill both conditions. Head Head Tail, Head Tail Head and Head Tail Tail. So the chances of having a head first and then at least one tail is 3/8, or 37,5%.
Now for the more solid one. I want you to find the smallest number with the following properties.
When divided it by 2, the remainder is 1.
When divided it by 3, the remainder is 2.
When divided it by 4, the remainder is 3.
When divided it by 5, the remainder is 4.
When divided it by 6, the remainder is 5.
When divided it by 7, the remainder is 6.
When divided it by 8, the remainder is 7.
When divided it by 9, the remainder is 8.
When divided it by 10, the remainder is 9.
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No problem, its ridle, but was have meaning that you allready throw it. First should be head would be more clear.
2519 is the answer. If you have positive numbers in mind.
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2519 is the answer
That's right. :) Your turn!
Have you found it with the elegant or the brute way?
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Elegant - i wrote program in 4 lines in qbasic to give me answer. :D i know it exists nice way but i was to lazzy to go on foot.
I leave you the honor to make another question, if you wish.
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i wrote program in 4 lines in qbasic to give me answer.
Why, I call a basic programme as epicly smart.
So, I am to post another teaser?
It's already here. :-X
P.S.: For those who wonder, as we see on the table, X = a*n + n - 1 for every integral n(http://upload.wikimedia.org/math/7/b/3/7b327af63e7c8eee3ba86fe1a4c7cff0.png)[2,10]. This gives X + 1 = (a + 1) * n, so (X + 1) is the lesser common multiple (LCM) of the numbers 2, 3, 4, 5, 6, 7, 8, 9 and 10. Using the prime factorization method, we get X + 1 = 2³ * 3² * 5 * 7 = 2520.
Since X + 1 = 2520, X = 2519.
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Why, I call a basic programme as epicly smart.
So, I am to post another teaser?
It's already here. :-X
Hint: There is something hidden in this post. The teaser is to find it. P.S. is not included.
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Well, the quote option didn't help. I backtraced the quotation too, no success there either.
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Try observation and imagination.
P.S.: It is also related to math, in a way.
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Only thing i noticed is mistake in x-1=2520 it should be x+1 - you fix it before i wrote this. So obviusly, thats isnt solution.
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Could solution be :-2519?? because smiley is with x so we found x = 2519. but please dont laugh on stupid try. :D
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It's not a stupid try, although it was not intented. :)
Crassus also noted in phone that the smiley contains -X but as I said, the P.S. is not part of the post, considering the riddle.
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There shouldn't be an "as"?
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Well yeah, the first line of the post is a bit... strangely syntaxed.
And there's obviously a reason for this. :)
So, you located the row with the hidden math stuff. Come on! Find the surprise! *YAHOO*
EDIT: When I first made this post, the first row was writen in a bit different way. Then for some reason Internet Exporer crashed and I had to right it again. I didn't remember at that time the exact original phrase, so I had to make a new one, hiding the same stuff in it again. Right now I remembered the original. It was "Why, I find a basic programme so epicly smart...". I hope this helps a bit.
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Do you wish a code of my basic program?
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It must be epic, though (write it in purple).
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:D :D :D
It's an expression derived from greek actually. I don't know if it's used that way in english, but it has a good meaning.
(I'd love to look at the code of your program.)
Now what other hint can I give...
Let's say I want you to tell me what number you see in that line. (I'm almost giving it up with this... ??? )
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I want just one more hint- how much uzo do I need to see the number?
I see 4 in the first post connected to your last riddle.
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Not too much uzo. It can be done with vodka as well. :P
Let's say the number is not writen in Hindu-Arabic numeral system.
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Hint: The number is not inside the line, it is the whole line.
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One?
1one111one1
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*SCRATCH* *hmm*
(I'm a bit curious, how did you come up with something like "1one111one1"?)
P.S.: The hidden number is quite famous too.
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I saw Judith write that once. After my diamond-poem.
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It has to be like this :D
!!!11!1!11one!!11one!1!one!!!eleven!!111!one!!!1111eleven!one
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is the solution "e"?
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It could be, now that I think of it. But that's not what I had in mind, neither is the way you though of "e" the same as what you need to do to solve the riddle.
I'd also love to see Judith in this topic. *hihi*
P.S.: Except for imagination, the solution also involves counting.
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Hint: Even the punctuation marks of that line matter...
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Hint: Use the letters of the words.
(come on, I want to solve some teasers too!)
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The number is π... 3.14.
Why, I call a basic programme as epicly smart.
3, 1 4 1 5....
Each word's number of letters it's a number... If you haven't post this:
Hint: Even the punctuation marks of that line matter...
I would never be able to find it...
Soon i will post my teaser...
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Yataaaaaaaaaaa!!
That's right! The hidden number is pi.
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Here is my riddle (the answer is word):
Txdqwxp skbvlfv vxjjhvw wkdw rqh soxv rqh pdnhv d qxpehu iurp plqru lqilqlwh wr soxv lqilqlwh zlwk kljkhu srvvlelolwb wkdw qxpehu eh wzr. Lv lw fruuhfw?
P.S.the key is 3
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"Quantum physics suggest that one plus one makes a number from minor infinite to plus infinite with higher possibility that number be two. Is that correct?"
I am not sure if quantum physics have anything to do with adding 1+1. Because its mostly about small particles...But in quntum physics a lot of stuff were turned upside down. I think they somewhere wrote that 1 + 1 isnt two, it can be three. So i guess this could be true statement. But i dont remember this formulation.
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It was too easy to use one of the first codes ever...
Caesar rulez (even as salad).
Your turn.
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It was too easy to use one of the first codes ever...
Caesar rulez (even as salad).
Your turn.
I didnt remember that. When i saw mixed words, first thta caame on my mind is that they were swithced easily. Wehn i saw you also gave code, it was easy.
Here, very easy one. Its to late to find some complicated stuff
You must get number 50 with numbers 4,6,6 and 9. Ech number must be used once, you can use all mathematical operations(+,-,*,/) and ().
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49 + 6/6? ???
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Nope, you cant connect numbers like 49 - they must have operation between them.
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4+6+6+9=50-25
Each number is used once. *hihi*
I will try to find another (more appropriate) solution later.
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There is atleast 1 multiply(*) and () in solution.
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Do we have to use all numbers?
For example, is (9*6)-4 a solution?
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Well i found a solution with the use of all numbers!!!
(6*9)-4+(d6/dx)=50
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Yes, all numbers must be used. Or it would be to easy, with 3 numbers.
Additional help - solution uses fractions.
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Well i found a solution with the use of all numbers!!!
(6*9)-4+(d6/dx)=50
I was thinking of that. *hihi*
Let's find the most irrational solution now.
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(6*9)-4+(d6/dx)=50
d6/dx is soooo epic! :D
But it's not right, since you also used the number "x"... *hihi*
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Ok, I have one without the x AND it has * and () as said in the quest description (I sound like Red Mage, but I have a good reason- solution is claimed to use fractions, therefore a fractactical one will be even better):
(6*9 - 4) + [lg6]
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lets see
6*(........
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Ok, I think I might have the answer (though a little crazier than using the (http://img704.imageshack.us/img704/1918/solgt.jpg)
And, to understand what I mean- read this (http://en.wikipedia.org/wiki/Congruence_relation).
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Well, the solution was much simpler.
6*(9-4/6)
Who ever wish, can post next task. :)
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Simpe just doesn't work for me. And I came up with congruence solution about 02:00 AM today (people count sheeps when they want to fall asleep, but, as I said, simple just isn't my style).
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Simpe just doesn't work for me. And I came up with congruence solution about 02:00 AM today (people count sheeps when they want to fall asleep, but, as I said, simple just isn't my style).
Is this the next teaser-riddle?
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The ways of my mind are a riddle indeed, but no, there's no riddle in that. I will come up with smth later (unless someone hopefully posts one before me).
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I only asked coz you wrote
Simpe
and not
Simple
....