A Question to Trouble Even the Best of You


by adityab88
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adityab88
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#1
Apr5-07, 10:37 AM
P: 10
Find the smallest possible arithmetic sequence consisting of seven primes.

For example, the smallest possible arithmetic sequence consisting of five primes is: 5, 11, 17, 23, 29.

By the way, by "smallest possible" I mean that the last term in the sequence must be the smallest possible of all such sequences.
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Gib Z
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#2
Apr6-07, 09:19 PM
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There is no such sequence.
Moo Of Doom
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#3
Apr6-07, 11:44 PM
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Gib Z, you might want to think about that statement some more. It has been proven that there are arithmetic sequences of primes of any finite length.

Gib Z
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#4
Apr7-07, 04:50 AM
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A Question to Trouble Even the Best of You


Shhh! lol

Ill reword it then, There is no such sequence that I can be bothered to find and that has any mathematical value. To find such a sequence all one needs is some programming knowledge and to be really bored.
arildno
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#5
Apr7-07, 05:00 AM
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What can readily be established is that any member of such a sequence must have the same last digit.

So there, I have narrowed it down immensely.
christianjb
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#6
Apr7-07, 05:37 AM
P: 530
Yeah, is this a computing problem or a math problem?

(Actually, it sounds more like a Google problem)
Gib Z
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#7
Apr7-07, 05:41 AM
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lol adityab88 you google recruit!!!!
Rhythmer
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#8
Apr7-07, 11:55 AM
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Quote Quote by Gib Z View Post
To find such a sequence all one needs is some programming knowledge and to be really bored.
What type of programming do we need to solve this sort of mathematical problems? C++ ?
Gib Z
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#9
Apr7-07, 07:04 PM
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No idea don't ask me lol.

If you want, start a thread here inquiring about this. But not C++ I know.
Rhythmer
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#10
Apr8-07, 12:19 AM
P: 14
Quote Quote by Gib Z View Post
No idea don't ask me lol.

If you want, start a thread here inquiring about this. But not C++ I know.
Unfortunately, C++ is the only programming language I can use.


Regarding adityab88's question, here is a solution I've just gotten:
1, 11, 31, 61, 101, 151, 211, 281
And it's the smallest possible because 361 is not a prime
d_leet
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#11
Apr8-07, 01:02 AM
P: 1,076
Quote Quote by Rhythmer View Post
Unfortunately, C++ is the only programming language I can use.


Regarding adityab88's question, here is a solution I've just gotten:
1, 11, 31, 61, 101, 151, 211, 281
And it's the smallest possible because 361 is not a prime
However this is not an arithmetic sequence because consecutive terms do not differ by one single constant.

ie. 11-1=10, whereas 31-11=20, 61-31=30 etc..
Rhythmer
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#12
Apr8-07, 02:08 AM
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Quote Quote by d_leet View Post
However this is not an arithmetic sequence because consecutive terms do not differ by one single constant.

ie. 11-1=10, whereas 31-11=20, 61-31=30 etc..
For all terms in the sequence:

[tex] X_n = X_{n-1} + ( (n-1) * 10 ) [/tex]

Starting from (n = 1), [tex] X_0 = 1 [/tex]



Is my solution too bad?
d_leet
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#13
Apr8-07, 02:18 AM
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Quote Quote by Rhythmer View Post
For all terms in the sequence:

[tex] X_n = X_{n-1} + ( (n-1) * 10 ) [/tex]

Starting from (n = 1), [tex] X_0 = 1 [/tex]



Is my solution too bad?
The only thing wrong with it is that it is not an arithmetic sequence.

An arithmetic sequence has terms of the form: an=a0+n*d

Where a0 is the first term, and d is the common difference between terms.
adityab88
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#14
Apr8-07, 03:05 AM
P: 10
hello ... ppl
i am quite sure you do not need programming equipment.
just your brains will suffice; i am sure of this becuase i saw this question on a maths paper, where computers were not allowed


this is all i have: the difference between the consecutive primes has to be a multiple of 30, i have shown this using some number theory. so, for example, the difference could be 60; a sequence like this (7, 67, 127, ...).

good luck and have fun!
Data
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#15
Apr8-07, 04:52 AM
P: 998
The spacing also can't be 30:

7, 37, ..., 187 doesn't work because 187 = 11*17. But for any prime [itex]p \neq 7[/itex], we have [itex] p \equiv 2k[/itex] (mod 7) for some k, [itex]1 \leq k \leq 6[/itex]. Since 30 mod 7 = 2 that means that p+30q is divisible by 7 for some q, 1<q<=6.

That can be generalized to remove some other spacings too; I'm tired and will figure it out in the morning!

Edit: OK, I lied about figuring it out later!

You can also eliminate all other spacings, except those which are products of 7, in cases for any sequence which does not start with 7. So the sequence has to start with 7, or the spacing is a multiple of 7 (ie. 210, 420, etc.).


A little bit of experimenting then yields the sequence: 7, 157, 307, 457, 607, 757, 907
Gib Z
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#16
Apr8-07, 11:44 PM
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Thats top stuff there Data.
adityab88
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#17
Apr12-07, 06:38 AM
P: 10
Quote Quote by Data View Post
The spacing also can't be 30:

7, 37, ..., 187 doesn't work because 187 = 11*17. But for any prime [itex]p \neq 7[/itex], we have [itex] p \equiv 2k[/itex] (mod 7) for some k, [itex]1 \leq k \leq 6[/itex]. Since 30 mod 7 = 2 that means that p+30q is divisible by 7 for some q, 1<q<=6.

That can be generalized to remove some other spacings too; I'm tired and will figure it out in the morning!

Edit: OK, I lied about figuring it out later!

You can also eliminate all other spacings, except those which are products of 7, in cases for any sequence which does not start with 7. So the sequence has to start with 7, or the spacing is a multiple of 7 (ie. 210, 420, etc.).


A little bit of experimenting then yields the sequence: 7, 157, 307, 457, 607, 757, 907
good stuff
how long did it take you to "experiment" and find the final sequence? cause i was looking for it, too, but my experimentation did not really work.
Data
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#18
Apr12-07, 01:12 PM
P: 998
About five seconds . It makes sense to start at 7; if you can find a sequence starting with 7 that has spacing smaller than 210 (and there's no smaller sequence starting with 7), then that's obviously the smallest one.

Of course, from what I've done there was no particular reason to confine myself to sequences starting with 7, so it would be fair to consider it a lucky coincidence.


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