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20 numbers and 8 primes (Posted on 2006-10-12) Difficulty: 4 of 5
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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