hatchguy wrote: Alexander and Louise are a pair of burglars. Every morning, Louise steals one third of Alexander's money, but feels remorse later in the afternoon and gives him half of all the money she has. If Louise has no money at the beginning and starts stealing on the first day, what is the least positive integer amount of money Alexander must have so that at the end of the 2012th day they both have an integer amount of money? If $a$ is the starting amount of Alexander's money and $x_n$ is the Louise's amount of money at the end of day $n$ (with $x_0=0$), we get the equation $x_{n+1}=\frac{x_n}3+\frac a6$ And so $x_n=\frac{(3^n-1)a}{4\cdot 3^n}$ If the problem is a pure math problem (considering for example that money is electronic money which can be any rational), we need to find minimal positive integer $a$ such that $\frac{(3^{2012}-1)a}{4\cdot 3^{2012}}\in\mathbb N$ and so trivially $\boxed{a_{min}=3^{2012}}$ If there is some "real life" aspect in this problem and for example money is made of coins and so must be integer number at each step (for example at end of day 1 or at the middle of day $2012$), then $\boxed{a_{min}=2\cdot 3^{2012}}$ And the problem statement, according to me, does not allow to decide between the two values. (I hate the math problems hiding behind pseudo-real life problems )

It is very weird that a problem ask for more than what is written. Since this is a competition problem those aspects were probably considered. And well, if there's even more real life aspect then it is normal to have a non-integer amount of money. Anyways, the answer was $a = 3^{2012}$.

If there is some "real life" aspect in this problem, already $a_{min}=3^{2012}$ makes it out-of-this-world (pray let me possess even $3^{2012}$ eurocents)

This problem is from Alejandro Vargas (Guatemala)

Let $X$ be the total amount of money and let $L_{i}$ be the fraction of money $L$ had the $i-th$ day. See if $L_{k}=t$ then $L_{k+1}=\frac{2t+1}{6}$. So we have $L_{2012}=\frac{2L_{2011}+1}{6}=\frac{2L_{2010}+4}{18}$. Prove by induction that for any $m$ positive integer, with $2012 \ge m > 0$; $L_{2012}=\frac{2L_{2012-m} + N_{m}}{6*3^{m-1}}$, where $N_{m}=3N_{m-1}+1$. Let $N$ be the set of $N_{m}'s$, we have $N=(1, 4, 13, 40, ...)$. Finally we have, $ L_{2012}=\frac{2L_{0} + N_{2012}}{6*3^{2011}}=\frac{N_{2012}}{2*3^{2012}}$ See that $N_{even}$ is even and for any $m$, $N_m$ is $1 mod 3$ so $2$ divides $N_{2012}$, and finally $3^{2012}$ divides $X$ so $X_{min}=3^{2012}$. Notice if the amount of $L$ is integer automatically the amount of $A$ is integer since $L + A = X$ and $X$ is integer.