Exercise 2. In a game, a strictly positive integer $n$ can be replaced by the integer $a b$ if $n=a+b$, with strictly positive integers $a$ and $b$. Can we obtain the number 2011 starting from $n=5$?
Solution: Yes, we can obtain 2011 starting with $n=5$: indeed, if we proceed with the decomposition $n=(n-2)+2$, the game allows us to replace $n$ with $2n-4$, which (when $n \geq 5$) is strictly greater than $n$. Thus, $5=3+2$ is replaced by $3 \times 2=6$, then $6=4+2$ is replaced by $4 \times 2=8$, then 8 by 12, then 20, 36, 68, 132, 260, 516, 1028, 2052. By this method, we can reach numbers as large as we want. And once we reach 2052, we can go down to 2011, noting that the decomposition $n=(n-1)+1$ allows us to replace $n$ with $(n-1)$, so $2052=2051+1$ is replaced by 2051, then $2050 \ldots$ until 2011.

There are many other ways to reach a number greater than 2011: some found it more natural, for example, to write: $5=3+2$, then $3 \times 2=6=3+3$, then $3 \times 3=9=4+5$, then $4 \times 5=20=10+10$, then $10 \times 10=100=50+50$, and finally $50 \times 50=2500$. It then takes 489 operations to go down to 2011. In reality, it is possible to reach 2011 much more quickly, for example: $5=3+2$, then $3 \times 2=6=4+2$, then $4 \times 2=8=4+4$, then $4 \times 4=16=8+8$, then $8 \times 8=64=36+28$, then $36 \times 28=1008=1006+2$, and finally $1006 \times 2=2012=2011+1$.