Chains of divisibilities

The first set of conditions state that $5,6,7,\ldots,5+m$ must all divide $n-5 \neq 0$ , and we want to determine $m$ such that these conditions always imply that $6+m,7+m$ also divide $n-5$ . In other words, we are asked to determine positive integers $m$ such that $m+6,m+7$ both divide the least common multiple $L(m)$ of $5,6,7,\ldots,5+m$ .

• If $m=1$ , then $L(1) = 30$ is not divisible by $m+6=7$ .
• If $m=2$ then $L(2) = 210$ is not divisible by $m+6=8$ .
• If $m=3$ then $L(3) = 840$ is not divisible by $m+6=9$ .
• If $m=4$ then $L(4) = 2520$ is not divisible by $m+7=11$ .
• If $m=5$ then $L(5) = 2520$ is not divisible by $m+6=11$ .
• If $m=6$ then $L(6) = 27720$ is not divisible by $m+7=13$ .
• If $m=7$ then $L(7) = 27720$ is not divisible by $m+6=13$ .
• $L(8) = 360360 = 25740 \times 14 = 24024 \times 15$ is divisible by both $m+6=14$ and $m+7=15$ .

The answer is $\boxed{8}$ .

Actually calculating n is irrelevant - for any set of consecutive numbers, you can find a set of consecutive numbers which are divisible by the numbers in the 1st set, in the same order - simply find the LCM of the 1st set of numbers, then add the smallest. In this case, it is evident that all of the numbers in the 1st set must divide n-5, and n-5 is at minimum the LCM of {5, 6, ..., 5+m}.

The only important parts are that n is not 5 (which is why n-5 must be non-zero), and that n+m+1 must be divisible by 5+m+1, & n+m+2 is divisible by 5+m+2, in other words, they both divide n-5. This means that these 2 extra numbers (6+m and 7+m) introduce no extra prime factors as the LCM remains unchanged when they are introduced. This means that they cannot be the prime , or a prime to a power p .

Finding the first 2 consecutive integers which fulfill these requirements are 14 and 15, hence n+6 = 14 , n+7 = 15 , and n=8 .