Guest Posted February 16, 2010 Report Share Posted February 16, 2010 Using one zero (and no other numbers) and mathematical (including trigonometric and advanced) operations, how can you achieve a final result of 6 Quote Link to comment Share on other sites More sharing options...

0 Guest Posted February 16, 2010 Report Share Posted February 16, 2010 Define a function Z = a^x + b^x + c^x + d^x + e^x + f^x Where a,b,c,d,e,f are non-negative Then, Limit x -> 0 (Z) = 6 Quote Link to comment Share on other sites More sharing options...

0 Guest Posted February 16, 2010 Report Share Posted February 16, 2010 (Floor(sq root(antilog(cos (0)))))! Explanation: cos 0 = 1 antilog 1 = 10 sq root 10 = 3.16 Floor 3.16 = 3 Factorial 3 = 6 Quote Link to comment Share on other sites More sharing options...

0 Guest Posted February 16, 2010 Report Share Posted February 16, 2010 Assuming you can physically change the shape of the zero, you can cut it on the upper right hand side and bend the bottom half in until it touches the inner left hand side creating a 6. Quote Link to comment Share on other sites More sharing options...

0 Guest Posted February 16, 2010 Report Share Posted February 16, 2010 (Floor(sq root(antilog(cos (0)))))! Explanation: cos 0 = 1 antilog 1 = 10 sq root 10 = 3.16 Floor 3.16 = 3 Factorial 3 = 6 really good try but i dont think this is right...you got 6 from 0 but you didnt prove 0=6.... Quote Link to comment Share on other sites More sharing options...

0 Guest Posted February 17, 2010 Report Share Posted February 17, 2010 (edited) Unless it is a trick we would need an equation set up. As we are only allowed one number i dont see how this is possible. However I have managed to prove it just using lots 0s. 0 = 0! - 0! 0! - 0! = 0!^{2} - 0!^{2} 0! - 0! = (0! + 0!)(0! - 0!) 1 = 0! + 0! [divide both sides by (0! - 0!)] 1 - 0! = 0! + 0! -0! 0 = 0! 0 * (0!+0!+0!+0!+0!+0!) = 0! * (0!+0!+0!+0!+0!+0!) 0 = 6 Edited February 17, 2010 by psychic_mind Quote Link to comment Share on other sites More sharing options...

0 unreality Posted February 17, 2010 Report Share Posted February 17, 2010 Unless it is a trick we would need an equation set up. As we are only allowed one number i dont see how this is possible. However I have managed to prove it just using lots 0s. 0 = 0! - 0! 0! - 0! = 0!^{2} - 0!^{2} 0! - 0! = (0! + 0!)(0! - 0!) 1 = 0! + 0! [divide both sides by (0! - 0!)] 1 - 0! = 0! + 0! -0! 0 = 0! 0 * (0!+0!+0!+0!+0!+0!) = 0! * (0!+0!+0!+0!+0!+0!) 0 = 6 you divide by 0 ie, you cant divide both sides by (0! - 0!) but the whole point of these kind of false-proofs is to disguise that as well as possible and you did a pretty good job Quote Link to comment Share on other sites More sharing options...

0 Guest Posted February 17, 2010 Report Share Posted February 17, 2010 you divide by 0 ie, you cant divide both sides by (0! - 0!) but the whole point of these kind of false-proofs is to disguise that as well as possible and you did a pretty good job Yeah i know that error but since 0 does not equal 6 i cant build a completely sound proof otherwise the whole of maths would fall apart. Quote Link to comment Share on other sites More sharing options...

0 Guest Posted February 17, 2010 Report Share Posted February 17, 2010 I thought the question was to make an equation that produces 6 using just one 0. However, as the replies go, its more about proving 0 = 6 In that regard, here's another attempt a^{x} = a^{y}; then x=y Using this, 1^{0} = 1^{6} Then 0 = 6 For that matter, using this method, 0 can be proved equal to another other number Quote Link to comment Share on other sites More sharing options...

0 plainglazed Posted February 18, 2010 Report Share Posted February 18, 2010 I read the problem as DeeGee did iniitially - to achieve a result of 6 using one zero and mathematical operations. to achieve this without the debate of whether or not square root requires a two is: floor(arccos(-cos(0)))! = 6 The subtitle does make me wonder tho... Quote Link to comment Share on other sites More sharing options...

0 Guest Posted February 18, 2010 Report Share Posted February 18, 2010 I read the problem as DeeGee did iniitially - to achieve a result of 6 using one zero and mathematical operations. to achieve this without the debate of whether or not square root requires a two is: floor(arccos(-cos(0)))! = 6 The subtitle does make me wonder tho... Or: factorial(floor(log(-log(cos(acos(0)))))) = 6 Quote Link to comment Share on other sites More sharing options...

0 plainglazed Posted February 18, 2010 Report Share Posted February 18, 2010 Using one zero (and no other numbers) and mathematical (including trigonometric and advanced) operations, how can you achieve a final result of 6 - 10=6 (base 6) Quote Link to comment Share on other sites More sharing options...

0 Guest Posted February 18, 2010 Report Share Posted February 18, 2010 INT(arccos(-cos 0))! Quote Link to comment Share on other sites More sharing options...

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Using one zero (and no other numbers) and mathematical (including trigonometric and advanced) operations, how can you achieve a final result of 6

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