## Task 2 - 150822

Determine all natural numbers $n \geq 1$ for which among the six numbers $n+1, n+2, n+3$, $n+4, n+5, n+6$ a pair can be found where the first number of the pair is a proper divisor of the second number of the pair!

State (for each such $n$) all such pairs!
If twice the smallest of the six numbers $(n+1)$ already exceeds the largest $(n+6)$, i.e., $2(n+1)>(n+6)$ and thus $n>4$, then it is certain that no ordered pair with the required divisibility properties can be found among the six numbers.

Since $n>4$ always implies $2(n+1)>(n+6)$, $n$ can be at most 1, 2, 3, or 4. Similarly, it is found that only for $n=1$ can one of the six numbers be three times another, and the quadruple cannot occur because $4(n+1) \geq (n+7)>(n+6)$. For the same reasons, higher multiples are even less possible.

Let $n=1$.

Among the numbers 2, 3, 4, 5, 6, 7, the divisibility relations $2|4, 2|6$, and $3|6$ hold. No further divisibility relations occur. Consequently, in this case, we obtain exactly the pairs $(2; 4), (2; 6), (3; 6)$.

Let $n=2$.

Among the numbers $3, 4, 5, 6, 7, 8$, exactly the divisibility relations $3|6$ and $4|8$ occur. Thus, we obtain exactly the pairs $(3; 6), (4; 8)$.

Let $n=3$.

Then, from the numbers $4, 5, 6, 7, 8, 9$, we obtain exactly the pair $(4; 8)$.

Let $n=4$.

From the numbers 5, 6, 7, 8, 9, 10, we obtain exactly the pair (5; 10). Thus, all the required pairs have been determined.