perplexus.info :: Sequences : Two cases

Two cases (Posted on 2023-11-10)

Could one find a sequence of 14 consecutive positive integers such that:

(I) each of the members of such sequence will be divisible by a prime below 13?
(II) Same question for 21 members, each being divisible by a prime below 17?

Provide two independent answers for (I) and (II)

No Solution Yet Submitted by Ady TZIDON

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d1 Part I

| Comment 2 of 4 |

Part I is trivial.

The numbers 2 though 16 are all divisible by themselves or another prime below 13. This is a sequence of 15 numbers or two sequences of 14. (2 to 15 and 3 to 16)

Posted by Jer on 2023-11-10 07:59:11

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