Annoying IQ Test
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Thought the same, but it probably should read more like f(1) = 5, f(2) = 125 ... and in the end we are supposed to figure out what exactly f is and calculate f(5)
I'm invincible, I can't be vinced
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If that's really an equivalence on integers, I'm pretty sure it can be proven that x = y for all x and y that are integers. edit: actually, I'm not sure anymore. If this equivalence is assumed, do the usual integer axioms still hold? Is math still possible?
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Someone ask me this IQ Test and I got really annoyed when I found out the answer. Here it is: If 1 = 5, and 2 = 125, and 3 = 450, and 4 = 1250, then 5 = ?
Normally an IQ will provide you a multiple choice of answers. Else I will be playing my
leppie's functioncard! -
Normally an IQ will provide you a multiple choice of answers. Else I will be playing my
leppie's functioncard! -
Someone ask me this IQ Test and I got really annoyed when I found out the answer. Here it is: If 1 = 5, and 2 = 125, and 3 = 450, and 4 = 1250, then 5 = ?
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Why not 50000000? :-D
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler. -- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong. -- Iain Clarke
[My articles] -
Why not 50000000? :-D
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler. -- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong. -- Iain Clarke
[My articles] -
If that's really an equivalence on integers, I'm pretty sure it can be proven that x = y for all x and y that are integers. edit: actually, I'm not sure anymore. If this equivalence is assumed, do the usual integer axioms still hold? Is math still possible?
I think I'm onto the right way, but don't have it quite right yet: if there really is a function f(x), so that f(1) = 5, f(2) = 125, f(3) = 450 and f(4) = 1250 then this may help f(1) = 5 = 5^1 = 5^x f(2) = 125 = 5^3 = 5 ^ (x+1) = 5 * 5^x f(3) = 450 = 3.6 * 5^x f(4) = 1250 = 2 * 5^x At least I have found a way to show that all values are multiples of 5^x. If we bring this into the right form, we probably will quickly see the solution. Edit My current guess for f(5) would be around 5^x = 5^5 = 3125
I'm invincible, I can't be vinced
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Someone ask me this IQ Test and I got really annoyed when I found out the answer. Here it is: If 1 = 5, and 2 = 125, and 3 = 450, and 4 = 1250, then 5 = ?
2795 :)
f(x) = round(-179.9813978 + 307.4704136*x + -167.4867127*x*x + 44.99822836*x*x*x) -
I think I'm onto the right way, but don't have it quite right yet: if there really is a function f(x), so that f(1) = 5, f(2) = 125, f(3) = 450 and f(4) = 1250 then this may help f(1) = 5 = 5^1 = 5^x f(2) = 125 = 5^3 = 5 ^ (x+1) = 5 * 5^x f(3) = 450 = 3.6 * 5^x f(4) = 1250 = 2 * 5^x At least I have found a way to show that all values are multiples of 5^x. If we bring this into the right form, we probably will quickly see the solution. Edit My current guess for f(5) would be around 5^x = 5^5 = 3125
I'm invincible, I can't be vinced
I went brute force ;p See my latest answer.
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2795 :)
f(x) = round(-179.9813978 + 307.4704136*x + -167.4867127*x*x + 44.99822836*x*x*x)50000000 :) algo:
switch(in)
{
case 1:
return 5;
case 2:
return 125;
case 3:
return 450;
case 4:
return 1250;
default:
return 50000000;
}:laugh:
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler. -- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong. -- Iain Clarke
[My articles] -
2795 :)
f(x) = round(-179.9813978 + 307.4704136*x + -167.4867127*x*x + 44.99822836*x*x*x)Wolfram Alpha, anyone? Or is Mathematica your bitch?
Software Zen:
delete this; -
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Wolfram Alpha, anyone? Or is Mathematica your bitch?
Software Zen:
delete this;Gary Wheeler wrote:
Wolfram Alpha, anyone? Or is Mathematica your bitch?
Neither. Used http://creativemachines.cornell.edu/eureqa[^] :) Trying to find an 'integer' solution though. But 4 points is a little too little to infer a pattern :(
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I went brute force ;p See my latest answer.
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Do me one favor: Please try out the factors of 1, 5, 3.6 and 2 I got for X = 1, 2, 3 ,4 in this math software. What factor would be for x = 5?
I'm invincible, I can't be vinced
Not quite sure how to input that into the software :(
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Not quite sure how to input that into the software :(
input the points (1,1), (2,5), (3, 3.6) and (4, 2). I would like to know the value for (5, ?). I guessed that value to be 1 and then f(5) would be 3125. If your program can extrapolate a more precise value, we will se how good your brute force result and mine match.
I'm invincible, I can't be vinced
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Someone ask me this IQ Test and I got really annoyed when I found out the answer. Here it is: If 1 = 5, and 2 = 125, and 3 = 450, and 4 = 1250, then 5 = ?
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Someone ask me this IQ Test and I got really annoyed when I found out the answer. Here it is: If 1 = 5, and 2 = 125, and 3 = 450, and 4 = 1250, then 5 = ?
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Someone ask me this IQ Test and I got really annoyed when I found out the answer. Here it is: If 1 = 5, and 2 = 125, and 3 = 450, and 4 = 1250, then 5 = ?
