Christmas Streak 20/88: Both Even And Odd

$\left\lfloor\dfrac{n^3+m}{n+2}\right\rfloor=\dfrac{n^3+m}{n+2}$ means that $\dfrac{n^3+m}{n+2}$ is a natural number.

Therefore $n+2|n^3+m.$

Knowing that $n^3+8=(n+2)(n^2-2n+4),$ we can figure that $n+2|m-8.$

Then since $m+2>2,$ $m-8$ must have at least 24 distinct divisors, including $1$ and $2.$

Since there are even divisors, the number must contain $2.$

However, if the number is $2^n\cdot k$ where $n\ge2$ and $k$ is odd, then the number of odd divisors is the same with $2k.$

But the number of even divisors are way more than $2k.$

And since the problem asked for at least $11$ odd divisors and $11$ even divisors, we can infer that the number contains only one $2$ and therefore the number has equal amount of odd and even divisors.

So we can infer that $m-8=2k$ where $k$ is an odd number.

$k$ is a minimal number with $12$ distinct divisors. Therefore, $k=3^2\cdot5\cdot7.$

$\therefore~m=2\cdot3^2\cdot5\cdot7+8=638.$