## Friday, April 14, 2006

### Move Those Stones. Topic: Logic. Level: AIME/Olympiad.

Problem: (TJ USAMO) Three piles of stones contain $100$, $2000$, and $2004$ stones respectively. If we are allowed to pick two of the piles and move one stone from each to the third pile, is it possible to eventually wind up with three piles of $1368$ stones?

Solution: Let $a$ be the number of times the pile with $100$ is the one that stones are added to. Similarly define $b,c$ for the piles with $2000$ and $2004$, respectively. The number of stones in the piles are

$100+2a-(b+c)$, $2000+2b-(c+a)$, and $2004+2c-(a+b)$.

And these all have to equal $1368$. So from the first two we have

$100+2a-(b+c) = 1368 \Rightarrow 2a-b-c = 1268$ (1)

$2000+2b-(c+a) = 1368 \Rightarrow 2b-c-a = -632$ (2).

Adding (1) and twice of (2), we have

$(2a-b-c)+(4b-2c-2a) = 1268-1264$

$3(b-c) = 4$.

But this clearly is impossible, so the piles cannot end up with $1368$ stones each. QED.

--------------------

Comment: This is basically an invariants problem, but working it out with a system of equations makes it even clearer that there is no solution. Most of the time, problems like this involve straightforward math but some tricky thinking.