Just For Fun

Just for some (silly) fun, we can actually solve this pretty easy without multiplying out the answer at all, if we assume that the correct answer is listed.

Note that our number is $2016^2 \cdot (1+ 2016) = 2016^2 \cdot 2017.$ The number of digits it has is $\lceil \log_{10} 2016^2 \cdot 2017 \rceil.$ Note that $2016^2 \cdot 2017 = 2.016^2 \times 2.107 \times 10^9 \approx 8 \times 10^9,$ so it has $9+1=10$ digits.

Let's suppose 1 digit $A$ was missing, so one other digit $B$ is there twice. The sum of the digits would then be $(0+1+2+\ldots+9) + B - A = 45 + B - A.$ The number has to be divisible by $9,$ so $B-A$ must be a multiple of 9. This is only possible if $\{A,B\} = \{0,9\}.$ But neither of these digits are choices. So, no digits are missing.

If you put the numbers in your calculator you can see which digits are there. They all are.

If there is no room on your calculator, put this in wolfram alpha or spotlight.(2016^2+2016^3)