Tessellate S.T.E.M.S. (2019) - Mathematics - Category A (School) - Set 3 - Objective Problem 1

The idea is to show that $f(2n) = \frac{10^{2n}}4 + 10^n.$ Then $\sum\limits_{n=1}^{1008} f(2n) = 2525\cdots25 + 111\cdots10,$ where there are $1008$ copies of 25 in the first number and $1008$ 1s in the second number. There are clearly no carries when we sum these two numbers, so the digit sum equals the sum of the digit sums of the two addends, which is $7 \cdot 1008 + 1008 = \fbox{8064}.$

To see that $x_n = \frac{10^{2n}}4 + 10^n$ is not on the board, notice that $\begin{aligned} \left( \frac{10^{2n}}2 + 10^n \right)^2 &= 10^{2n}(x_n+1) \\ \left( \frac{10^{2n}}2 + 10^n - 1 \right)^2 &= 10^{2n} x_n - (2\cdot 10^n-1) \end{aligned}$ so chopping off the rightmost $2n$ digits from these consecutive squares gives the numbers $x_n-1$ and $x_n+1.$

It's a little bit more painful to see why $x_n$ is the smallest number not on the board. There are two steps: first, the squares of any two consecutive numbers less than $10^{2n}/2$ differ by less than $10^{2n},$ so truncating the rightmost $2n$ digits of them will either give the same number or consecutive numbers. Second, it's a straightforward but tedious check that the numbers obtained by truncating the rightmost $2n$ digits of the squares of the $10^n$ consecutive numbers $10^{2n}/2, 10^{2n}/2 + 1, \ldots, 10^{2n}/2 + 10^n - 1$ are consecutive integers. (Indeed, they are the integers $10^{2n}/4, 10^{2n}/4 + 1, \ldots, 10^{2n}/4 + 10^n - 1.$ ) I'll leave that to the reader!