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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: ( Back to comment list | You must be logged in to post comments.)
 re: program exploration (spoiler?) -- extended sequence of starting integers | Comment 6 of 7 |
(In reply to program exploration (spoiler?) by Charlie)

`   260813   771324  1360457  1870968  1954131  2464642  3410063  3545723  3920574  4056234  5643437  5779097  6153948  6289608  7235029  7745540  7828703  8339214  8928347  9438858  9960503 10471014 11060147 11570658 11653821 12164332 13109753 13245413 13620264 13755924 15343127`

In each case, the first member of the sequence is shown.

 Posted by Charlie on 2006-10-14 14:02:33

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