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This is a discussion on cancelled within the General Chat forums, part of the Knight Online (ko4life.com) category; 9...
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I'm sure some Dr of math will answer that one. Here's the new question. What is the mathematical equation to determine the number of pubic and chest hairs luts has combined? :P ready
(0/X) + (0/Y)
Just kidding lutz...
Now give us a decent question man
? it's not a math problem, it's more like a logic puzzle.yeah paddy 2nd i agree with u
lutz wtf is that shit
thats not a game its a fucking good will hunting math problem
no one here is fucking matt damone you noob
you dont even give hints or make it to admins and mods cant play
ffs delete the whole post i will re post a new game[/b]
If you want people to not be able to Google them, you can't make it trivia - trivia is all Google-able.
? it's not a math problem, it's more like a logic puzzle.
If you want people to not be able to Google them, you can't make it trivia - trivia is all Google-able.[/b]
delete this whole thread i will redo it? it's not a math problem, it's more like a logic puzzle.
If you want people to not be able to Google them, you can't make it trivia - trivia is all Google-able.[/b]
From the numbers 1 to 999, how many numbers are there where one number is equal to the average of the other two distinctive digits?
a + x +c /3 = x
o.O
I'm stuck with that xD
I agree..slutz is a noobie
From the numbers 1 to 999, how many numbers are there where one number is equal to the average of the other two distinctive digits?
If you wanted the answer... just make a chart like this:
1 - (0,2)
2 - (1,3) (0,4)
3 - (2,4) (1,5) (0,6)
4 - (3,5) (2,6) (1,7) (0,8)
5 - (4,6) (3,7) (2,8) (1,9)
6 - (5,7) (4,8) (3,9)
7 - (6,8) (5,9)
8 - (7,9)
9 - None
Since all integers must be distinct (that is, there can not be more than one of each) then you can just take the number of averages that are listed in the chart above (which is 20), and now that you know that there are 20, do the following:
Since the integer can be in the hundreds place, tens place, or ones place, and each set can be flipped around as to create two instances per time it is in the hundreds, tens or ones place, there are six possible permutations of the set above.
That is, let's take 213 for example.
213 is a valid "special number".
231 is too.
Then we move 2 to the ten's place.
123.
321.
Then we move 2 to the one's place.
132.
312.
So there are 6 each time we flip it around, so it's 120.
But we're not done. There are some special cases, and repetitions, such as the fact that 102 cannot be switched to 012, as that's not a 3 digit number. So since we take out all of our "special instances" (which is 24 instances) we are left with the answer:
96
ffs stfu now...dont try to cover up that you suck at games
..ffs stfu now...dont try to cover up that you suck at games[/b]
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