perplexus.info :: Numbers : 20 numbers and 8 primes

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"

Submitted by Steve Herman

Rating: 4.5000 (2 votes)

Solution: (Hide)

Thanks, Charlie, for solving this so thoroughly.

1) The first satisfying sequence is the 20 numbers starting with 260,813

2) The second satisfying sequence is the 20 numbers starting with 771,324

3) There are 20 satisfying sequences under 2*3*5*7*11*13*17*19 = 9,699,690. Any sequence which satisfies the condition can be turned into another satisfying sequence by adding 9,699,690.

4) Question 4, alas, was based on my bad solution. What is true and interesting is that the first number in the first sequence plus the last number in the 20th sequence = 9,699,690. The same is true of the 2nd and 19th sequence, and the 3rd and 18th, etc.

Bonus Question:
Contrary to suggestions in the question, 20 is not 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? 18 is the smallest n, and the first satisfying sequence are the 18 integers starting with 27,829.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
Puzzle Thoughts	K Sengupta	2023-06-10 00:02:55
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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