We write on the board the integers from 1 to 2013. At each step, we erase two of them and write their difference instead. The number of integers thus decreases by 1. Can the last integer obtained at the two thousand and twelfth step be equal to 0?
This is still an argument of parity that allows us to conclude. Since the difference of two integers has the same parity as their sum, the sum of all integers on the board retains the same parity at each step of the process. Initially, it is: $1+2+\ldots+2011=\frac{2011 \times 2012}{2}$, which is even. This sum will therefore remain even, and the last number obtained will be an even number, it cannot be 1.

To prove that it can be 0 (although this is not required), a completely different reasoning would be needed. We would group the numbers as follows: $(1,2,3),(4,5,6,7), \ldots(2008,2009,2010,2011)$. The first triplet can be reduced to 0 by replacing $(2,3)$ with 1 and then $(1,1)$ with 0. Similarly, for each of the following quadruplets, by replacing $(4n, 4n+2)$ with 2, $(4n+1, 4n+3)$ with 2, and then $(2,2)$ with 0. Proving that the answer is "yes" or that the answer is "no" requires completely different reasoning, so it is imperative to guess the correct answer as quickly as possible because a lot of time is lost when one goes in the wrong direction. The technique of invariants is useful for proving an impossibility. To prove that something is possible, usually one shows that it can be constructed explicitly.