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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.

p_square
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Mathuser
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    What is the source of this problem? Questions involving the current year frequently appear in contests or exams. – lulu May 11 '21 at 16:22
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    Voting to close the question, as I assume it is from a current competition. If you can, please edit your post to explain the source of the problem, and to indicate your efforts. – lulu May 11 '21 at 16:35
  • Your approach amounts to trying to get there in one step, so perhaps the answers that reflect making multiple steps will help you. In any case the title should be edited to represent your problem topic in a more apt fashion. – hardmath May 11 '21 at 16:40

2 Answers2

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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:

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

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

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

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

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

$n=2021$

Saha
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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 $24$.

WhatsUp
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