The following problem looks like a riddle. But careful: it's quite difficult and resists all approaches.

Take some prime numbers, e.g. M={11,41,61}, and search for a chain of numbers such that each is divisible by at least one of the chosen prime numbers, e.g. 121,122,123 (11 × 11, 61 × 2, 41 × 3). In this example the chain length is 3. Since the smallest prime number is bigger than the number 3 of chosen primes, the length of the chain is bounded by the number of primes. No prime can be used more than once.

There is a special role for prime number 2 thus it is forbidden. We choose a maximal prime number p und collect all odd primes up to p inluded in a set M, e.g. p=7 and M={3,5,7}. Here yo can find the chain 48,49,50,51 of length 4. Divisibility by 3 was used twice. Thus the chain gets longer than the number of primes: 4>3.

Or choose p=11 and M={3,5,7,11}. We find the chain 515,516,517,518,519,520 with length 6. The prime numbers 3 and 5 are used twice each. Thus the chain is with 6>4 longer than the number of primes.

The set M is determined uniquely by the maximal prime number p.

Question: how long is the maximum length of a chain to a given maximum prime number p (and therefore given set M), with each number in the chain divisible by at least one prime number in M?

For an explicitely given prime number p we can try, because the pattern will repeat after a finite number. This is because of modular arithmetic. We can then make a guess, because there is a pattern. Is it possible to proof the arising conjecture?

WP: Modular arithmetic WP: Chinese remainder theorem WP: Cyclic group WP: Unit (ring theory) WP: Multiplicative group of integers modulo n