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Posts posted by Yoruichi-san

  1. That's not much of a contribution - some pzls don't hit the spot - many many more do. This was fine, just yo were sniffing at the red herring it seems (as was I)

    No need to snap, Cackle - pop off to another part of the forum and enjoy the ones you like - and add some too if you like - OH! and pls read the bit up the top, there's a bit in there about etiquette .... Other thatn that.. Hi and welcome to the den!

    I'm sorry...it wasn't suppose to be a red herring. It was suppose to be a hint...like I could of used some random names like Billy, Larry, and Joe or person 1, person 2, person 3 or something. I chose to use Snap, Crackle, and Pop because they are well-known cartoon characters, and the fact that they are cartoon characters was suppose to be a hint.

    I was trying to come up with something unconventional...I mean there are lots of "who am I" riddles and number patterns and stuff and a lot of them are really good, but I was hoping to create something new and original (plus that guarantees it hasn't been posted before...). Sorry if you don't like it. Hope you'll like my next one better...

  2. It takes immense genius to represent, simply and sincerely, what we see right in front of us.

    --Edmond Duranty

    Well, that's what differential equations are for (and why I say that they are the only useful form of math). ;)

    You can always write a differential equation describing what we observe. Solving that differential equation is another matter though...

    Thanks a lot for the words of praise! :)

  3. Different planes intersecting a cone? Any of those shapes can be made by slicing a cone (as well as an ellipse and possibly/probably some other shapes as well).

    Yep, basically, cuz...

    ...therefore they are two-dimensional. In this riddle, they live in two-dimensional planes that are intersecting a cone. If we put the cone on a table, then:

    Snap lives in a plane parallel to the table.

    Crackle lives in a plane perpendicular to the table but not through the tip.

    Pop lives in a plane perpendicular to the table and through the tip.

    Tony the Tiger lives in a plane that is Tangent to the Tip.

    Mmm...cereal and milk!

  4. So my friend made a comment that the problem with number patterns is that you can make them arbitrarily difficult...well, these get pretty difficult, although not too complex (i.e., the algorithms are short, but may be hard to find!).

    Find the next numbers and the algorithm for the following:









    GAME OVER....Do you wish to continue? ;P

  5. Oh, and after figuring that out, I get the first one too...

    Right, I forgot to include the pattern again, so here it is

    a(n)=n*(number of letters in the word for "n"+n)

    I.E. The second multipliers for: ONE is 3+1=4

    TWO is 3+2=5

    THREE is 5+3=8

    FOUR is 4+4=8

    FIVE is 4+5=9

    SIX is 3+6=9

    SEVEN is 5+7=12

    EIGHT is 5+8=13

    NINE is 4+9=13

    TEN is 3+10=13

    so the next ones should be:

    ELEVEN is 6+11=17

    TWELVE is 6+12=18

    THIRTEEN is 8+13=21

    So for n=11,12,13, a(n) should be 11*17=187,12*18=216, 13*21=273

  6. I think I got the second pattern:

    Oh, and this is the pattern:

    a(n)=n*(number of letters in the word for "n"+1)

    i.e. O-N-E and T-W-O both have 3 letters, so the second multiplier is 3+1=4. T-H-R-E-E has 5 letters, so the second multiplier is 5+1=6.

    E-L-E-V-E-N and T-W-E-L-V-E both have 6 letters, so the eleventh and twelfth numbers in the sequence should be 11*(6+1)=77 and 12*(6+1)=84. T-H-I-R-T-E-E-N has 8 letters, so the thirteenth number should be 13*(8+1)=117.

  7. aw, I thought I had it for a second, but one doesn't fit. Anyway, here was my first thought:

    A cone of some sort- from the top the point looks like a period, from the side the cone looks like an upside-down V, and from the bottom it's a circle or an O. I'm not sure about the U part.

    So they're living in an ice-cream cone? ;)


    You're on the right track...good work!

  8. Any chance of another hint here? I'm scundered.

    You have to figure out the question before you can figure out the answer.

    Each is separate from the others.

    Hmmm...I hope you guys aren't implying that I'm a monkey...then I would be offended, since I'm actually a rat (at least according to the Chinese zodiac). ;P

  9. With help from my friend Leroy...

    1. 1/4

    No piece can be larger than 1/2, and no piece can be smaller than the difference between the other two. So, calling the location of the two cuts x and y, which are both randomly distributed over the interval [0.1], we can draw a square which represents the possible x and y. Then divide this square into fourths with lines at x=1/2 and y=1/2. The condition that no piece can be greater than 1/2 rules out the square x>1/2,y>1/2, and the square x<1/2,y<1/2. Now we are left with two fourths.

    The last condition becomes |x-y|<1/2, i.e., the maximum difference is when one piece is almost 1/2 and the other piece is infinitesimally small. Rearranging, this condition is mathematically equivalent to y<x+1/2 and y>x-1/2 which cuts away half of the two remaining squares, leaving two triangles of area 1/8 each. So total probability is 2*1/8=1/4.

    2. 1/2

    For a coin toss, there are 2 possible outcomes (heads or tails). For n coin tosses, each toss can be heads or tails, so the number of different possible outcomes is 2^n. For two people (A and B) each tossing n coins, there are (2^n)*(2^n) = 2^(2n) possible outcomes.

    There are three possible cases considering the number of heads of A vs. B. A>B, A=B, A<B. Since the probability of heads is constant for both, the number of cases of A>B equals the number of cases of A<B, lets call this x. Call the number of cases of A=B y. So there are x+y+x=2^(2n) total possible outcomes.

    Now we add the n+1th coin toss.Now we have a total of (2^(n+1)*2^n)=2^(2n+1) possible outcomes. For A to have more heads, then if we get heads for the n+1th toss, we need originally A>B or A=B, so x+y cases. If it is tails, we need the original count to be A>B, so x cases. So A gets more heads in a total of (x+y)+x=2^(2n) cases.

    So the probability of A getting more heads is 2^(2n)/2^(2n+1)=1/2.

  10. Um, well from my experience riding bus routes...

    She on a loop that is on one side of both of them, and Mr. Downtown is closer to the loop, like:


    where she lives on the O and they are both on the -. Mr. Downtown is closer more west then Mr. Uptown.

  11. Coooll Dude> you stole my hypothesis.

    Haha, well, I'm pretty sure others have stolen this hypothesis before both of us...in fact, I'm sure there has been like at least a couple hundred science fiction stories which include it... ;P

  12. An electron, I have no idea what an anti bottom quark is *makes note to research it* but here's my reasoning:

    I get excited by the light,

    Electrons are exited by light

    I jump from state to state,

    Light causes them to jump the the next level (not sure if the terminology is correct)

    No one can be exactly like me,

    No two can have the same 4 quantum numbers

    My brothers I do hate,

    Electrons repel each other

    If when I turn one way you turn the other,

    A reference to the spin

    You can be my mate,

    there are two electrons per level

    You can try to guess what I'll do next,

    But I don't believe in fate.

    You can't know both were an electron is and what it it doing, therefore you can't predict what it will do.

    Who am I?

    Nicely done! Only thing I'd add:

    The last line actually refers to the Heissenberg uncertainty principle, which basically says we can not know the exact location and trajectory of a particle, which leads to the result that *we cannot predict the future*! (which I happen to think is the most important result of quantum physics and has repercussions on philosophy and the way we should look at the world...)

    Here's the line-by-line breakdown I was thinking of when I wrote the poem:


    Transitions between energy levels (states)

    Pauli exclusion principle

    Coulomb's law (coulombic repulsion)

    Electron spin

    Electron pairs having opposite spin in molecular orbits (MO theory)

    Schrodinger's equation/wavefunctions (probability density functions)

    Heissenberg uncertainty principle

  13. 1. Share lease of pastry over front

    2. Boring way to go of ate lunch

    3. Organic diary of squeal to the support proposal

    4. Important from pessimistic pizza to loser of omen of former over also latin gods hug

    If they ask "can you put that in plain engish?", show them this post...;P

    Look at my signature.

  14. Ooo...could it be...

    Death or the grim reaper (shinigami

    ;))? Taking "twilight" to refer to the "twilight of life"...and death can't be seen (unless you happen to possess a death note ;P)
  15. I like oranfry's method a lot. But in case any one cares, here is how to do it the conventional way with combinatorics:

    Okay, so as said before, there are 3 cases to consider:

    Case I: No repeating letters. (i.e. ABC)

    Case II: 1 repeating letter (i.e. AAB)

    Case III: 2 repeating letters (i.e. AAA)

    The total # of combinations is the sum of these three cases. Here is how I look at them:

    Case I: No repeating letters means that all three letters are different. Therefore, there are 26 choices for the first letter, 25 choices for the second, and 24 choices for the third. So there are a total of 26*25*24=15600 three-letter words of case I. Note that order *does* matter here, i.e. ABC is different from ACB. This is a permutation, not a combination. Put in mathematical terms, this is 26P3 (26 permutation 3).

    Case II: This is the tricky one. The way I think about it is that each of these has two unique letters, and the third letter is a repeat of one of the two.

    So first I figure out the number of two-letter combinations (i.e. AB), which is 26C2=26*25/2=325.

    Now for each of these two-letter combinations, there are 2 choices for the third letter (i.e. we can have a repeating A or B) and 3 places to put the non-repeating letter, so we can make 2*3=6 different three-letter words, (i.e. AAB, ABA, BAA, BBA, BAB, ABB).

    So there are a total of 6*325=1950 total words for case II.

    Case III: This is an easy one. Only one letter to choose, so there are 26 (or 26C1) choices for the letter. 26 words for case III.

    So there are a total of 15600+1950+26=17576 possible words of length three.

  16. so whats the pattern? i dont get it

    Each number can be divided into 2 numbers, between 1-26, which corresponds to the sequential letter in the alphabet.

    Each pair of letter is a state abbreviation, starting with AK (alaska). Basically, it is a list of the ten largest states (by area). The next two in the sequence are the 11th and 12th largest state.

    Hence the state pun. ;)

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