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## Question

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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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

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(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

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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.

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(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....

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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 by psychic_mind
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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

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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.

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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

ax = ay; then x=y

Using this,

10 = 16

Then 0 = 6

For that matter, using this method, 0 can be proved equal to another other number

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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...

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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

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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 -

10=6 (base 6)

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INT(arccos(-cos 0))!

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