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 20 numbers and 8 primes (Posted on 2006-10-12)
I am looking for n consecutive integers such that (i) every number in the sequence is divisible by a prime <=n and (ii) every prime number <=n is a factor of at least two of the numbers. For example, consider n=3:
a) There are two primes less than or equal to 3. They are 2 and 3.
b) 6 7 8 does not work, in part because 7 is not evenly divisible by either 2 or 3
c) 8 9 10 does not work, even though all are divisible by 2 or 3, because 3 divides only one of them

There is some reason to believe that no sequence of positive integers works for n < 20. For n = 20:

1) What is the first sequence of 20 consecutive positive integers that works?
2) What is the second?
3) How often do they repeat after that?
4) What interesting number results if you add the first integer from one of the first two sequences to the last integer of the other?

By the way, this problem grew out of JLo's innocent perplexus problem "Six numbers and a prime"

 See The Solution Submitted by Steve Herman Rating: 4.5000 (2 votes)

Comments: ( You must be logged in to post comments.)
 Subject Author Date n=18 should be it JLo 2006-10-16 06:17:55 re: program exploration (spoiler?) -- extended sequence of starting integers Charlie 2006-10-14 14:02:33 re(3): Confessions and a new problem, thanks to Charlie Charlie 2006-10-13 11:46:21 re(2): Confessions and a new problem, thanks to Charlie brianjn 2006-10-12 22:04:14 re: Confessions and a new problem, thanks to Charlie Charlie 2006-10-12 15:48:47 Confessions and a new problem, thanks to Charlie Steve Herman 2006-10-12 14:30:36 program exploration (spoiler?) Charlie 2006-10-12 14:07:27
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