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Posts posted by Pickett


So, is this anagram completely jumbled? or is each word its own anagram? I would guess that it's completely jumbled. If that's the case, are the word lengths at least correct?
The topic is LAITY = ITALY???...and I noticed that EVINCE could be VENICE....but not sure

So, my dad told me this little problem quite a few years ago (when I was 8 or so...took me years to figure out how to solve it). It has always stuck with me because of that...so it's just a straightforward math problem, but it's a fun one:
There is an alley. In this alley there is a 20ft ladder and a 30ft ladder. These ladders cross over the alley from the ground to the opposite building. They cross each other 10ft above the ground. How wide is the alley?
Here's a simple drawing to represent this problem:

This one was a little tougher:
Second letters of the month names:
jAnuary, fEbruary, mArch, aPril, mAy, jUne, jUly, aUgust, sEptember, oCtober, nOvember, dEcember
So the next two are o,e

Cards:
Ace, Two, Three, Four, Five, Six, Seven, Eight, Nine, Ten, Jack, Queen, King
so, the next two are q, k
adding even numbers:
0, 2, 6, 12, 20, 30, 42, 56, 72
so the next two are 56, 72

(6 * 8)  2  √4 = 44 (could also be + √4 if using the negative root)
oops, sorry for the double post...browser messed up when submitting didn't think it posted.

(6 * 8)  2  √4 = 44 (could also be + √4 if using the negative root)

Well, since I solved your number sequence #4, the second sequence was pretty simple for me:
The next number is 14.01 (depending on rounding)
Sequence is: f(n) = f(n1) + P(n1)/pi (assume P(0) = 0, and f(0) = 1).
f(1) = 1
f(2) = f(1) + 2/pi = 1.6
f(3) = f(2) + 3/pi = 2.55
f(4) = f(3) + 5/pi = 4.14
f(5) = f(4) + 7/pi = 6.37
f(6) = f(5) + 11/pi = 9.87
f(7) = f(6) + 13/pi = 14.01

Well, finding the next number is really dependent on your rounding. it's almost easier just to give the formula used for the sequence:
f(n) = 2pi^{(n(n1)/2)}
OR
f(n) = f(n1) * pi^{n1}, where f(0) = 2 (rounded to 2 decimals)
Specifically, what I did to get the numbers close to yours, is I used pi = 3.14159...so using the second formula I got the following:
f(0) = 2
f(1) = f(0) * p^{0} = 2
f(2) = f(1) * pi^{1} = 2 * pi = 6.28
f(3) = f(2) * pi^{2} = 6.28 * pi^{2} = 61.98
f(4) = f(3) * pi^{3} = 61.98 * pi^{3} = 1921.77
f(5) = f(4) * pi^{4} = 1921.77 * pi^{4} = 187197.24 (this is probably just rounding "error")
f(6) = f(5) * pi^{5} = 187197.24 * pi^{5} = 57285798.44
or, just using the first formula, it's close enough (again depending on rounding...and obviously these are slightly different)
f(1) = 2pi^{(1(0)/2)} = 2pi^{0} = 2
f(2) = 2pi^{(2(1)/2)} = 2pi^{1} = 6.28
f(3) = 2pi^{(3(2)/2)} = 2pi^{3} = 62.01
f(4) = 2pi^{(4(3)/2)} = 2pi^{6} = 1922.77
f(5) = 2pi^{(5(4)/2)} = 2pi^{10} = 187294.51
f(6) = 2pi^{(6(5)/2)} = 2pi^{15} = 57315565.75
So, I would say the next number is about 57285798.44 or somewhere close to there...

This one appears to be the last digit of the fibonacci sequence:
1 2 3 5 8 13 21 34 55 89 144, 233, etc...
so, the next numbers would be 9, 4, 3...

NUNS: Naruto: Ultimate Ninja Storm

MIM: Mario is Missing
TTABBL: Tiny Toon Adventures: Buster Busts Loose
PPL: Pokemon Puzzle League
That only leaves one more:
NUNS

TCNB: Trauma Center: New Blood
PMTTYD: Paper Mario: The Thousand Year Door
GSTLA: Golden Sun: The Lost Age
ZAMN: Zombies Ate My Neighbors!
TOSDOTNW: Tales of Symphonia: Dawn of the New World
DHOD: Disgaea: Hour of Darkness

Phoenix Write Ace Attorney: Justice For All

The Legend of Dragoon

Super Mario Sunshine

Well, the first one is easy. (ab) is equal to 0, so when you divide by (ab), you're dividing by zero, which causes your error.I'm not sure about the second one, but I think the error for the third one lies in the infinite series. Basically you started with a finite series of zeros, then treated it as an infinte series later.
yes on the first one!
yes and no on the third one: Yes it has to do with the infinite series...but 0 does equal 0 + 0 + 0 + ... as an infinite series...so, there's something else wrong with it...

It's been a long time since calculus...an error with the derivative. the derivative of x^2 isn't 2xdx but 2dx. You're treating 2*dx as 2*x*dx.
I think???
nope, the deriviative of x^{2} is in fact 2x dx. getting somewhat closer though.

What is stronger than the Devil,more powerful than God,
a rich man needs it,
a poor man has it,
and if you eat it, you will die?
Not that hard but just something I hear a while back.
Nothing

I've seen all of these "proofs" being added recently for 2 = 0, 1 < 0, etc, etc...so I figured I'd add my two proofs for 2 = 1, and proof that magic exists...
I apologize if these have already been posted...I haven't seen them yet though, so I thought I would put them up. These aren't difficult, and I fully expect them to be solved very quickly, but they're fun.
This first one is the golden oldie (I learned this one quite a few years ago and still like it):
let a = b
a^{2} = ab (multiply both sides by a)
a^{2}  b^{2} = ab  b^{2}(subtract b^{2} from both sides)
(a + b)(a  b) = b(a  b) (factor)
(a + b) = b (divide both sides by (a  b) )
b + b = b (substitution)
2b = b
2 = 1
Then this following one uses calculus (fun one at first when learning calculus):
x = 1 + 1 + ... + 1 (x times)
x^{2} = x + x + ... + x (multiply through by x)
2x dx = (1 + 1 + ... + 1) dx (take deriviative of both sides)
2x = (1 + 1 + ... + 1)
2x = x
2 = 1
and my last one is to show that something TRULY can come from nothing:
0 = 0 + 0 + 0 + ...
0 = (1  1) + (1  1) + (1  1) + ... (since 0 = 1  1)
0 = 1 + (1 + 1) + (1 + 1) + (1 + 1) + ... (associative law of addition)
0 = 1 + 0 + 0 + 0 + ... (since 1 + 1 = 0)
0 = 1
So, as if by magic something appeared out of nothing! YAY!

Here are all of the answers that follow the example given:
3 ^{69258}/_{714}
81 ^{5643}/_{297}
81 ^{7524}/_{396}
82 ^{3546}/_{197}
91 ^{5742}/_{638}
91 ^{5823}/_{647}
91 ^{7524}/_{836}
94 ^{1578}/_{263}
96 ^{1428}/_{357}
96 ^{1752}/_{438}
96 ^{2148}/_{537}
I wrote a small Java program to find them all for me :c) Don't know if that's cheating, but I think it was a completely valid way to solve that riddle.
Great riddle, it was fun.

I figured this one out last night actually...
I actually got it after I had a drink or two at a bar
You can also start a similar sequence with 18 13 10 2...
Good sequence riddle. I liked it. I'm amazed I actually figured it out...

So, I don't know if there is enough information to solve this puzzle completely...however I got this far with it:
SIR GOOD:
1  Harl
2  Gort or Kal
3  Gort or Kal
4  Lok
5  Jol
SIR PURE:
1  Gort, Harl, or Jol
2  Kal
3  Gort, Harl, or Jol
4  Gort or Jol
5  Lok
SIR NOBLE:
1  Gort or Jol
2  Kal
3  Gort or Jol
4  Harl
5  Lok
SIR WISE:
1  Gort or Kal
2  Harl
3  Gort or Kal
4  Lok
5  Jol
SIR BLACK:
1  Gort
2  Harl
3  Lok
4  Kal
5  Jol
Basically Lok has 21 total points (duh)...Jol COULD have up to 22...Kal and Gort COULD have up to 14...and Harl COULD have up to 12. You can arbitrarily assign the points that are still in question to the squires and not violate any of the clues...

This is my first post on Brainden!!
My guess is RH stands for Robin Hood...so stealing from the rich to give to the poor...some somehow this company probably rigs online blackjack games to steal rich people's money and give it to poor people....
That's just my guess, could be way off.
in New Logic/Math Puzzles
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Well, he mentioned that there were still unanswered questions..and that figuring them out would be key. So, while I agree the capital letters are probably important, I think figuring out the questions is a good idea...
Those are some of the questions that I think may help in solving this...