Vanishing Remainders

Bonus Question 2: Firstly, if the numbers are divisible by $n$ , $n+1$ , $n+2$ , we have two cases:

Case 1: $n$ is even

Thus, $n+2$ is even, so $n$ and $n+2$ have a common factor of 2. The other pairs of numbers are coprime by Euclid's algorithm. so the numbers under 200 are divisible by $\dfrac {n(n+1)(n+2)}{2}$ . We now wish to maximise the value of $\left \lfloor \dfrac{200}{\frac {n(n+1)(n+2)}{2}} \right \rfloor = \left \lfloor \dfrac{400}{n(n+1)(n+2)} \right \rfloor$ . This happens when $n$ is the least value, so $n=2$ . This gives a maximum for $m$ as $m=16$ .

If $m=1$ , $n(n+1)(n+2) \leq 400 < 2n(n+1)(n+2)$ . Solving for $n$ , we get $5\leq n leq 6$ . Thus, $n=6$ in this case for $m=1$ .

Case 2: $n$ is odd

Thus, all the numbers are coprime by Euclid's algorithm, so the numbers under 200 are divisible by $n(n+1)(n+2)$ . Thus, $m=\left \lfloor \dfrac{200}{n(n+1)(n+2)} \right \rfloor$ . Once again, $n$ must be minimum, so $n=1$ , which implies $m=33$ .

If $m=1$ , $n(n+1)(n+2) \leq 200 < 2n(n+1)(n+2)$ , $n=6$ , which is untrue since $n$ is odd.

Thus, the maximum value of $m$ is $33$ and the only satisfying value of $n$ such that $m=1$ is $n=6$ .