Problem Post 2-1: Factor Sums, and the Distributive Law

Problem Posts

[Difficulty 1; some parts Difficulty 4]

Who knew power series multiplication could help you at the grocer? (see Question 2)

Just a quick post for younger readers. Often one will find oneself using `brute force' approaches when obvious tricks and shortcuts exist. Most of the time, this is justified -- many tricks are usually too arcane to remember, or too impracticable to execute realistically. This is neither. It's fast, simple, and it could save you in a Do-Or-Die (I've used it before; our team won!)

What is a factor?

First things first, right? Not everybody defines the word “factor” the same way. I will be using the convention used in most local contests: A factor of an integer $n$ is a positive integer $a$ for which there exists an integer $b$  such that $n=ab$. So by our convention $4$ and $-4$ both have exactly three factors: 1, 2, and 4. We do not consider $-2$ as a factor, even if it divides both $4$ and $-4$. On the other hand, a factor of a number is a proper factor iff it is not the number itself. Often 1 is also not considered a proper factor. With this ambiguity, however, the term is not used too often in contests.

Problem 1-1 Kookie's Cookies

Problem Posts

Problem 1-1 (Number Theory) [Difficulty 1] [PDF]

QUESTION
(Sipnayan 2012 High School Final Round: Weightlifting 4) Kookie has a kooky way of eating cookies. He lays them out on a circle. After Kookie eats a cookie, he skips the next (clockwise) remaining cookie in the circle and eats the next (clockwise) remaining cookie after that. Kookie places a batch of $2012$ cookies numbered $1, 2, \ldots 2012$ in that clockwise order, and begins to eat them, cookie $2$ first. Let $k$ be the number on the last of the $2012$ cookies that he eats. Kookie, unsatisfied, arranges another batch of $k$ cookies numbered $2, 4, 6, \ldots 2k$ in that clockwise order, and begins again with the cookie numbered $2$. What is the number on the last cookie Kookie eats from this batch?

Overview
This problem comprises two iterations of the Josephus problem.