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solution gomez puzzle

 
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rimbaudblake
Icarian Member



PostPosted: Wed Jun 26, 2002 10:08 pm    Post subject: 1 Reply with quote

i think that i have the solution but i dont know to whom need to send it.
The next numbers are 37, 47, 58, 72, 87, 105

any doubt rimbaud_7@hotmail.com
thanks
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Dragon Phoenix
Judge Doom



PostPosted: Thu Jun 27, 2002 1:08 am    Post subject: 2 Reply with quote

Welcome to the GL. There is no need to send anyone your answers, but they can be posted in this forum. Please check however whether there is already an active thread for the puzzle you want to discuss. In this case, it is already here.

I am closing this thread to avoid double discussions.
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Dragon Phoenix
Judge Doom



PostPosted: Thu Jun 27, 2002 11:02 pm    Post subject: 3 Reply with quote

And it's open again - I did not realise that the 'Gomez-puzzle' had more than one component. Apologies and thanks Sofis for pointing it out.
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CrystyB
Misunderstood Guy



PostPosted: Fri Jun 28, 2002 12:55 am    Post subject: 4 Reply with quote

link and question: 2, 9, 15, 18, 19, 21, 24, 29, ?, ?, ?, ?, ?.

As for checking the above answer, i need a nutcracker!
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Gomez
candid chimera



PostPosted: Fri Jun 28, 2002 3:33 pm    Post subject: 5 Reply with quote

No need to send it to anyone rimbaud. Just post the solution & an explanation here.

For what it's worth, your answer is different from the official solution. That isn't to say your solution is invalid.



[This message has been edited by Gomez (edited 06-28-2002 11:35 AM).]
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Coldcube
Guest



PostPosted: Sat Jun 29, 2002 1:14 am    Post subject: 6 Reply with quote

its 30 right? 29 are 30?
it justt make things harder
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Gifthorse
Daedalian Member



PostPosted: Sun Jun 30, 2002 4:14 am    Post subject: 7 Reply with quote


This is probably wrong but I think the solution is . . .

dividing with primes

2,9,15,18,19,21,24,29. . . . . .27, 30, 39, 27, 30


How did I come up with this guess

We start with the Number 2

(9/2)= (4 Rem 1) = (5) x 3 = 15 then we let the next number be 18

(19/3)= (6 Rem 1) = (7) x 3 = 21 then we let the next number be 24

So when we do 29 we get. . . . .

(29/5) = (5 Rem 4) = (9) x 3= 27 then we let the next number be 30

This is where I get stuck, because I have to assume the next number to be 39


If thats true then when we devide by the next prime, which is 7, we get. . . . . .

(39/7) = (5 Rem 4) = 9 x 3 = 27 then we let the next number be 30


Its interesting to note that if we continue along with the sequence using only primes we get....

27,30, 39 ,27,30, 49, 27,30, 59, 33,36, 69, 15, 18, 79, 21,24, 89, 69,72,

99, 45,48

I MAY BE TOTALLY WRONG for instance my explanation doesn't explain the number 2 and its relevance although I did use it as my starting point for dividing

There is also the issue regarding transitions from a number to the numbers that end in 9.

Example (#)-> 9 (#) -> 19 (#) -> 29 I don't think my solution explains this part at all


I hope someone can shed more light on this. Also, what about normal counting numbers or using exponents.

Please reply
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Gifthorse
Daedalian Member



PostPosted: Sun Jun 30, 2002 4:18 am    Post subject: 8 Reply with quote

Somehow the sequence portion for 79 and 89 got erased or I forgot it.

Just know that the 79 and 89 portion is missing.

I appoligize
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Gifthorse
Daedalian Member



PostPosted: Sun Jun 30, 2002 4:19 am    Post subject: 9 Reply with quote

Oh wait no It did not.

I must be going crazy sorry,. Ok no more useless posting for me.
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Gifthorse
Daedalian Member



PostPosted: Sun Jun 30, 2002 4:23 am    Post subject: 10 Reply with quote

YO 27,30,39,27,20
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Gifthorse
Daedalian Member



PostPosted: Sun Jun 30, 2002 4:32 am    Post subject: 11 Reply with quote

How to explain the Number 2

I did the following (9/2) (19/3) (29/5) So. . .. . .

Its only follows the . . .we do this (2/3)= (0 Rem 0) = 0 x 3 = 0

If the equation gives zero this is probably might be the sequences' way of telling us this is definitely the beginning of the sequence
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The Unknown
Guest



PostPosted: Thu Jul 04, 2002 10:43 am    Post subject: 12 Reply with quote

i see that it is this thread now here it is again then:

Well, board, I agree with your solution for the most part, but. I am not sure if it originally started out with 2 maybe they started out with 0+2=(2)+7=(9) but if even they did not, you forgot to put a 7 in too as it is the first thing you add into to get the next number so. Here is what i think it is, I never seem to be able to explain things in word form, so i will show you the equation. Note the numbers in () are the numbers from the sequence and the next five numbers:
(2)+7=(9)+7-1=(15)+ <<<this si where it gets a little tricky, it can either be 6 divided by 2 or 6 minus 3 but i will go with six minus three>>> (2)+7=(9)+7-1=(15)+6-3=(18)+3-2=(19)+1+1=(21)+2+1=(24)+3+2=(29) <<<so now assumeing thatyou start off with 7 it would be>>>
(2)+7=(9)+7-1=(15)+6-3=(18)+3-2=(19)+1+1=(21)+2+1=(24)+3+2=(29)+7=
(36)+7-1=(42)+6-3=(45)+3-2=(46)+1+1=(48)+2+1=(51)+3+2=(56)
i went on to 7 for fun, but the next five numbers are 36,42,45,46,48
mainly for the fact that that is the pattern the first 8 numbers went in and if they are repeating that is it. Or you can have fun and flip the signs to their opposite. or start from the back and go to the begining. But that is it 36,42,45,46,48 in my opinion is the right answer : P. Sorry if i posted in the wrong forum. If so can you point me in the right one?

\


P.S the hardest puzzle i have found so far is the Albert Einstien one
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CrystyB
Misunderstood Guy



PostPosted: Thu Jul 04, 2002 1:41 pm    Post subject: 13 Reply with quote

there is no wrong forum. Most of us who follow on a puzzle usually read all about it. Meaning all 10 topics if there are as many...
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