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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.)
Confessions and a new problem, thanks to Charlie | Comment 2 of 8 |
Hmm...

CONFESSIONS
a) Charlie has uncovered more solutions than I expected.
b) My 3rd and 4th questions no longer make sense, given all of Charlie's solutions
c) Also, my assertion that "no sequence of positive integers works for n < 20" is incorrect.

NEW PROBLEM
What is the smallest value of n for which there are n consecutive positive 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?  Hint: It's less than 20.

KUDOS and THANKS
Nice work, Charlie!




  Posted by Steve Herman on 2006-10-12 14:30:36
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