Question

Let $k$ be a positive integer. Bernardo and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernardo starts by writing the smallest perfect square with $k+1$ digits. Every time Bernardo writes a number, Silvia erases the last $k$ digits of it. Bernardo then writes the next perfect square, Silvia erases the last $k$ digits of it, and this process continues until the last two numbers that remain on the board differ by at least $2.$ Let $f(k)$ be the smallest positive integer not written on the board. For example, if $k=1,$ then the numbers that Bernardo writes are $16, 25, 36, 49,$ and $64,$ and the numbers showing on the board after Silvia erases are $1, 2, 3, 4,$ and $6,$ and thus $f(1)=5.$ What is the sum of the digits of $f(2)+f(4)+f(6)+\cdots+f(2016)?$

$7986$

$8002$

$8030$

$8048$

$8064$

Solution:

Take $k=2j.$ The smallest perfect square with $k+1$ digits is $10^{k}=(10^{j})^2,$ and after Silvia erases, the numbers shown are $\left\lfloor n^2/10^{k}\right\rfloor$ for $n=10^{j}, 10^{j}+1,\ldots$ Consecutive terms increase by $0$ or $1$ until the first jump of at least $2.$

That first jump occurs at $n=\dfrac{10^{k}}{2}+m$ with $m=10^{j}-1,$ and one computes that the last number written before the gap gives $f(2j)=\dfrac{10^{2j}}{4}+10^{j}.$

Summing over $j=1,\ldots,1008,$ $\sum_{j=1}^{1008}f(2j)=25\sum_{j=0}^{1007}10^{2j}+10\sum_{j=0}^{1007}10^{j}=\underbrace{2525\cdots25}_{2016\text{ digits}}+\underbrace{111\cdots10}_{1009\text{ digits}}.$ There are no carries, so the digit sum is $1008\cdot(2+5)+1008\cdot 1=1008\cdot 8=8064.$

Thus, the correct answer is E.

2016 AMC 12A Problem 25

Solution:

Problem 25 in Other Years