Legendary Pokémon

Why your answer is wrong:
1. Say the 20th prisoner can see 10 black hats and 9 red ones. The majority is black. The 19th guy has got a black one. Same as the majority, so the 20th guy says "Red". The 19th guy counts and sees 9 black hats and 9 red ones. Problem.
2. In any other case, the 20th uses the same way to tell the 19th his color. So, the 19th says the correct color and saves himself. What about the next ones? That way, the number of prisoners whose freedom you can guarantee is 10. Which is not as many prisoners as possible.

Also, the list isn't really needed.

It's 20 *κολοκύθι* prisoners.They can all kick their *κολοκύθι* if they revolute.so that's the way.

in other news
20th says Red if the no. of Red hats he see is Even. Says Black if its Odd.
Rest is simple.
If 19 sees no. of even Red hats in front of him and he heres Black from the 20th he knows he's got to be Red. if he heard Red, then he knows he got to be Black and so on ....

thus 19 people were saved min and I rule.

Correct. Link's majority/minority thing was close, but had some problems as I said. However, that way, the prisoners can easily count the number of hats in front of them and all say a reply based on what the 20th dude said.

Billaros won it and he may set the next for I am bored.

In the lands of Aldebaran, state of the akward lands, its people use a decimal system of numbering, giving the same meaning to the symbols +, x and =, but the digits 0,1,2,3,4,5,6,7,8,9 are used in a different row than the usual. To be more exact the symbol of each of the digits differs from usual (meaning that there's not a single digit with the same meaning in both systems). Below are some correct calculations in their numbering system, which, by COINCIDENCE, are right in our numbering system too:

4x7=28
5x7=35
4x6=24
1+4+6+7+7=25

If those people wrote "2 x 302 = ?", what number of their system should replace "?"?

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oh and don't mind rushing the answer.There won't be one for at least 5 days so take your time.

You all fail.Your brains are probably full of sand.Nm2 go ahead reveal the answer.You are allowed to.OH AND MIND THAT HE SOLVED THAT ON THE FIRST DAY IT WAS POSTED YOU ALL MORONS.

IMMA BUMPING THIS THREAD USING MAH LAZORZ.
I'll be posting the solution next weekend if nobody else gets it by then. One week for you guys to find it and prove that your level of intelligence can somehow come close to ours. For I is teh cleverzorz and you is teh stoopidzorz. So. Go find it. Use the maths skillz they've been teaching you at school all those years.

\O\ 460

Time's up noobs. Nobody got it so I win. Now, here's what your solution should look like:

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To begin with, let's set some variables for their numbers.
a=1, b=2, c=3, d=4, e=5, f=6, g=7, h=8, x=9, y=0 (with x and y because any maths problem respecting itself should contain x and y)
For example, a≠1 since it is equal to their 1, which is not equal to our 1. I hope I'm not making it too complex already.

Now, let's see what the problem gives us.
4 × 7 = 28
5 × 7 = 35
4 × 6 = 24
1 + 4 + 6 + 7 + 7 = 25

Which can be now transformed to:
d × g = 10b + h (1)
e × g = 10c + e (2)
d × f = 10b + d (3)
a+d+f+2g = 10b+e (4)
(From now on, I'll omit the "×" symbol so that you don't confuse it with the x variable. "d × g" is now "dg".)

Now the only way to move on is to assign random values to one of the variables. I've chosen b, after having tried with d and g first. b seems to be the correct choice, but I'm not sure, it may be solvable if you choose one of the others as well. So, let's go:

• b=0
  Equation (3) ⇒ df=d
  And since d≠0 (since b=0 and not two variables can have the same value), we get f=1.
  Equation (4) ⇒ a+d+1+2g=e
  The lowest possible value of "a+d+1+2g" is for g=2, d=3 and a=4.
  This means 4+3+1+4=e ⇒ e=12 which can't be right since e is between 0 and 9.
• b=1
  Equation (3) ⇒ df=10+d ⇒ d(f-1)=10
  Which means that either d=2 and f-1=5⇒f=6 which is wrong since f≠6 as we said in the beginning, or d=5 and f-1=2⇒f=3
  Equation (4) ⇒ a+5+3+2g=10+e ⇒ a+2g=2+e ⇒ 10g=10-5a+5e (5)
  Equation (1) ⇒ 5g=10+h ⇒ 10g=20+2h (6)
  According to (5) and (6), 10-5a+5e=20+2h ⇒ 2h=5(e-a-2)
  I'm bored to elaborate more on this, but the last equation will end in impossible equations in any case, so the whole thing was wrong which means b≠1.
• b=2, which is wrong since b≠2 according to the problem.
• b=3
  Equation (3) ⇒ df=30+d ⇒ d(f-1)=30
  Which means that either d=6 and f-1=5⇒f=6 which is wrong because they get the same value, or that d=5 and f-1=6⇒f=7
  Equation (1) ⇒ 5g=30+h ⇒ h=5(g-6)
  Since h<10, its only possible values are h=5 and h=0 (since the second part of the equation is multiplied by 5). But h≠5 since d=5. So, h=0, which means g-6=0⇒g=6
  Equation (4) ⇒ a+5+7+12=30+e ⇒ a=e+6 which means that a>6. But a≠7 since f=7 and a≠9 since then e=3 which is wrong since b=3. So, a=8 and e=2.
  Equation (2) ⇒ 12=10c+2 ⇒ c=1
  So, the only variables remaining are x and y and the only values remaining are 4 and 9. But x≠9 according to the problem, so x=4 and y=9
• In any other case (b=4, b=5, ... , b=9), you'll end in impossible results, which I'm bored to type.

So, to sum up. What they have as 1234567890, we've got it as 8315276049
Which means that their 2×302 equals our 3×193=579 and our 579 equals their 460. So the final result is 2×302=460.

Should you know Greek and are bored to read this, I've scanned my solution, click here for it.
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Fr3quency obviously googled the result so he didn't get it.
I'll post a new challenge when I find a cool, worthy one.
Also note that the above solution was not copy-pasted from some random site, I typed it myself, copying the solution I had in paper (the scan) which was in Greek. If someone doubts about it, I don't really care.

Ok, we've got a pentagram. A pentagram has five triangles (we do not count the triangles that have any other lines in them). By drawing just two more lines, you can construct a shape with exactly ten triangles, again not counting those with other lines in them. How can this be done?

Here's an image for you to work on:

Also, just for the sake of clarifying how triangles are counted. The pentagram you have to start with has five triangles, the ones in red:

Triangles like this one do not count since there are lines going through them:

Now, if you make your shape look like this after the first line, the triangles you get are those in red again:


It's not really that hard as the previous ones. I just hope you don't consider a pentagram to be a symbol of satan. If you do, ignore this challenge.

does this count?
edit: oh nevermind, there are nine tringles here anyway

edit:

Can we make new triangles outside the pentagram? If we can, then that's one of the answers. Although, I'm not sure that there are more than one..
[attachment=455]

Ok, to tell the truth, I'm not sure about how many solutions there are. I found this challenge on a certain site where the solutions to the problems aren't given so I cannot say how many possible ways exist for solving this. However, your solution is the only one I've found myself as well, so congrats, you win!

Next challenge when I find a good one.

whats the prize for winning

We are supposed to be a forum about Pokémon, so yeah. Nothing more to say.

(15-01-2009 09:36 PM)crobar Wrote: [ -> ]whats the prize for winning

Nothing. Just the pleasure of feeling superior to others.

What should we do???