h2g2 The Hitchhiker's Guide to the Galaxy: Earth Edition

I think (BIG assumption) I can prove that it is possible to make the number 42 from any 6 numbers in a specific order PROVIDED at least 3 of them are non-zero.

I'll try to make it simple and I'll send the proof to you soon. Anyone who fancies trying to have a go themselves....Feel free.

TDiD

P.S. U176272 => ((1*7)*(6+2))-(7*2)=42

and

U000000 = invtan(0!)-(0!)-(0!)-(0!)+0+0=42

(Just thought I'd throw that in!)

Sounds interesting. The 3 numbers non zero bit intrigues me - unless thats for purposes of concatenation, you can turn any zero into a 1 by doing a ! to it. I'll be very interested to see this proof

Dittlyio

Right, here we go.

We have a user no', Uabcdef (where a,b,c,d,e,f are integer values and are independently congruent (If they want to be the same, they can be!!).

IF 2 or more NUMBERS ARE THE SAME

It isn't difficult to rearrange the formula's so you get a zero somewhere. after this it should also be obvious that 2 or more of the numbers MUST be 2 or 3 apart. This allows a three to be made in any situation. This allows a simple inverse tan to be achieved to create an answer. Even if the identical numbers are first and last it means there must be consecutive numbers in the middle which allow you to eliminate any unco-operative numbers!!!

(i.e

U146571

=>-1-4+6=1
=>5-7-1=-3

so intan(1)-3=42)


IF ALL THE NUMBERS ARE DIFFERENT

imply either form a 6 and a 7 (do I really need to prove that bit?) or use a similar method to the above to make a 1 and a 3. I think (I havn't found an example that doesn't work) there are no three different numbers that can't be combined to make either a 3 or a 1. This stems from the idea that the maximum difference between two single number integers is 8. If you add a third, the maximum diffrence goes down to 3. (either divide or subtract depending). Skillful use of brackets then allows you to create invtan(1)-3 or -3+invtan(1).

(i.e

U169247

=>1*6-9=-3
=>-2-4+7=1

-3+invtan(1)=42)

(Those two example were totaly random by the way!!!)

These proofs work in every case I have tried up to three congruent numbers. After that, I'm not sure if a proof can exist since it relies on numbers having a split pair (i.e being odd) and being adjacent to another number. After three, there is a chance this won't happen.

I'm sure you can see why the two proofs here work. They are pretty simple and only require you to be able to add up to 3. (True).

AFTER THIS WE ENTER UNKNOWN TERITORY!!!

If anyone can make a proof of three congruents and more (I'm working on it as well!) I'd love to hear it, It's killing me not being able to work it out! Well, at least I got a ball rolling!

TDiD

Feel free to ask questions now he seminar is over!

Right, small mistake in the above posting.

Everywhere I have written 0 I mean 1 or 0 since 1=0!

Everywhere I have written 1 I mean 1 or 0 since 1=0!

This means 0 is interchangable for a 1 in this proof although 1 IS NOT exchangable for a 0!!

Sorry about the mess.

TDiD

Dont have time to look at this properly right now, hope I'll get time later. There seem to be a fair few 'probably's in there, and as a mathmo I'd need more rigor to call it a proof...

...the other way to play this game would be to try to come up with a U# that cannot be 42ed...I'll have a think.

As a mathmo you obviously appreciate the value in such a proposition. If you can prove it wrong you show it's a blind alley.

The big maybe about the above idea (The big thing that stops it being a proof) is that I can't think of a decent starting point for trying to prove (With official maths rigour i.e induction or throughflow) that any 3 non congruent numbers can always be rearranged to 3 or 1. After that, if you can prove that any choice of 3 of the remaining 7 numbers can be arranged to the other (1 and 3 respectively) then the proof is as rigourous as the basics need to be. We'll deal with congruents later.

I tried using the empty set to produce collected pairs with and odd one left but I kept finding that it was equally possible that the order may affect the outcome. It becomes difficult and I believe a maths program on a computer may have to be employed (I should get a copy of MAPLE 6 tommorow!)

The proof I wrote was never meant to be the be all and end all. I'd appreciate any help you could give and I hope you agree that what I wrote should give a fairly stable background for the next stage of research. The hardest part of any proof is working out how to start!!!

Cheers for the feedback.

TDiD

P.S. (Being a second year maths student at Nottingham uni, I'm tempted to use this question as the beginning of a project on six letter variations within algebra. I have to do a pure maths project this year and I've put quite a lot of work into this already. I could just prove which numbers were and weren't possible and why. As a second year project, even that's pretty advanced!)

TDiD (again)

What exactly do you mean by 'congruent' here? I've heard the term used referring to numbers being congruent modulus something (eg 2 = 7 mod 5), and in congruent triangles...but I dont think you mean that.
Do you just mean that congruent is the same as equal?

Lemme get the basic method straight: make 1 and 3, invtan the 1 and take away 3. Ok, makes sense. Its the being sure that you can make 1 and 3.

How would you get that from 888888? (8+8+8)/8 and 8/8. Hmm. That sorts out all numbers the same (apart from 0, but youve done that already.

How about 788887? I can't see an easy way to get 3 and 1 from that. Though its easy enough to get 42 by other means.

I get the feeling its a nasty messy problem, and not really very nice mathematically - there'll be lots of messy little counterexamples that need to be sorted individually. Theres also things like the / operation, for which you can't use brackets to be able to swap the order of things. Good luck, though I'd make sure this sort of thing would be suitable for your project before you start on it properly

argh, the curse of the 8) - 8 ) ...

This is where the fun comes in I think. They'll be seven different proofs, depending on the numbers of identical numbers.

Congruence in this example is being used to indicate the same number. It isn't technicaly correct but similar and identical would cause problems in the understanding!! Then again, I suppose congruence has as well!!

To prove the second bit- All 3 number combinations of diffrent numbers can be made into 3 and 1, I'm going back to the good old fashioned method of writing them all out and working them all out. If I can do that, I'll have proved it for proof 1 (All numbers are different) and part of proofs 2,3and4 (where the first three numbers share one identical number with the second three!!, two identiacl numbers and three identical numbers respectively!!)

I'll make a guide entry of all the three number combinations at some point, hopefully next week!!!

Until then

Any help you can give on the consecutive number problem would be greatly appreciated.

TDiD

(It's fun, in a mathsy kind of way!!!)

Okay, I've started the page!

It's

A617988 - "42% Proof"

It's only just really started but will grow again on Monady of next week!!

Take a look, there's a copule you could really help me on!

TDiD,

(Making an effort)

Bloimey........

A whole area of Mathematics spawned......

I'll have to put a link in....

what the hell is this all about? you are all mental, stop being gimpy and drink copious beer

Aha - what makes you think we aren't already - I know I am.....