2016 is absolutely awesome 1

For the following derivations, we assume that k doesn't divide 2016.

Let $j=\left\lfloor \frac { 2016 }{ k } \right\rfloor$ . From the first equation we have that: $\frac { 2016 }{ k+1 } <j\le \frac { 2016 }{ k }$ Provided that there is an integer value that $j$ can take in $\left( \frac { 2016 }{ k+1 } ,\frac { 2016 }{ k } \right]$ then eq. 1 will be met. This imples that the first equation must be met if the length of the interval is greater or equal to 1. From this we get:

$\frac { 2016 }{ k } -\frac { 2016 }{ k+1 } =\frac { 2016 }{ k(k+1) } \ge 1$ $0\ge { k }^{ 2 }+k-2016$ $\frac { -1-\sqrt { 8065 } }{ 2 } \le k\le \frac { -1+\sqrt { 8065 } }{ 2 }$ $k \le 44$

Similarly, letting $i=\left\lceil \frac { 2016 }{ k } \right\rceil$ . From the second equation we have that: $\frac { 2016 }{ k } \le i<\frac { 2016 }{ k-1 }$

By similar reasoning to the previous derivation, we get that:

$\frac { 2016 }{ k-1 } -\frac { 2016 }{ k } =\frac { 2016 }{ k(k-1) } \ge 1$ $0\ge { k }^{ 2 }-k-2016$ $\frac { 1-\sqrt { 8065 } }{ 2 } \le k\le \frac { 1+\sqrt { 8065 } }{ 2 }$ $k \le 45$

We no know that all $k \le 44$ satisfy both equations.

Since k doesn't divide 2016, $j+1=i$ .

If both equations are met, then necessarily:

$\left\lfloor \frac { 2016 }{ j } \right\rfloor =\left\lceil \frac { 2016 }{ j+1 } \right\rceil$ $\frac { 2016 }{ j } =\frac { 2016 }{ j+1 } +\alpha$

Where $0<\alpha <2$ . This in turn implies:

$\alpha =\frac { 2016 }{ j(j+1) }$ $j=\frac { -1+\sqrt { 1+8064/\alpha } }{ 2 }$ $j> \frac { -1+\sqrt { 1+8064/2 } }{ 2 }$ $j\ge 32$ $k\le 61$

Therefore, if k doesn't divide 2016, it can only satisfy both equations if it is less than or equal to 61. We now check the interval $44<k\le61$ . In it 45, 46, 48, 50, 51, 54 and 56 satisfy both equations. So far, we found $44+7=51$ cases where both equations are met. The remaining numbers that satisfy the equations are the factors of 2016 that we haven't already counted. These are: 63, 72, 84, 96, 112, 126, 144, 168, 224, 252, 288, 336, 504, 672, 1008 and 2016.

For numbers greater that 2016, eq. one gives $\left\lfloor \frac { 2016 }{ 0 } \right\rfloor \neq k$ , so there are no other solutions.

So, in total there are 67 2016-awesome integers.