Mathematic problem solution

Question

I need help with this question:

When a positive integer $n$ is written in the blackboard, an allowed operation is to substitute it with a number $ab$, where $n=a+b$, and both $a$ and $b$ are positive integers. If $n$ is initially $24$, does there exist a sequence of allowed operations that leads to the number $2021$?

I tried $a^2-24a+2021$ but the result is complex roots.

Thanks.

Answer 1

Start with $n=24=23+1$. Choose $a=23$ and $b=1$. Then $ab=23*1=23$, which is one less than 24. You keep subtracting 1 like that down until 21. If you have $n=21=6+15$ choose $a=6$ and $b=15$. So you have $ab=6*15=90$. Than you can do $n=90=43+47$, so you can again choose $a=43$ and $b=47$, which gives you $43*47=2021$.

The steps are as follows:

$n=24, a=23, b=1$

$n=23, a=22, b=1$

$n=22, a=21, b=1$

$n=21, a=6, b=15$

$n=90, a=43, b=47$

$n=2021$

Answer 2

For any $n > 4$, one can get $2(n - 2)$ from $n$. Note that $2(n - 2)$ is strictly larger than $n$ (and also larger than $4$).

This means that by repeating this procedure, we can get arbitrarily large numbers from $24$.

Finally, note that for any $n > 1$, one can get $n - 1$ from $n$.

Therefore any positive integer can be obtained from $24$.