# A Question to Trouble Even the Best of You

Tags: trouble
 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.
 HW Helper P: 3,348 There is no such sequence.
 P: 367 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.
 HW Helper P: 3,348 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.
 Sci Advisor HW Helper PF Gold P: 12,016 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.
 P: 529 Yeah, is this a computing problem or a math problem? (Actually, it sounds more like a Google problem)
P: 14
 Quote by Gib Z 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++ ?
P: 14
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
P: 1,075
 Quote by Rhythmer 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..
P: 14
 Quote by d_leet 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:

$$X_n = X_{n-1} + ( (n-1) * 10 )$$

Starting from (n = 1), $$X_0 = 1$$

P: 1,075
 Quote by Rhythmer For all terms in the sequence: $$X_n = X_{n-1} + ( (n-1) * 10 )$$ Starting from (n = 1), $$X_0 = 1$$ 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.
 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!
 P: 998 The spacing also can't be 30: 7, 37, ..., 187 doesn't work because 187 = 11*17. But for any prime $p \neq 7$, we have $p \equiv 2k$ (mod 7) for some k, $1 \leq k \leq 6$. Since 30 mod 7 = 2 that means that p+30q is divisible by 7 for some q, 1
 HW Helper P: 3,348 Thats top stuff there Data.
P: 10
 Quote by Data The spacing also can't be 30: 7, 37, ..., 187 doesn't work because 187 = 11*17. But for any prime $p \neq 7$, we have $p \equiv 2k$ (mod 7) for some k, $1 \leq k \leq 6$. Since 30 mod 7 = 2 that means that p+30q is divisible by 7 for some q, 1
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.
 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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