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The problem is as such:

Say a natural number $n$ as special if it does not have the digit $0$, has $2021$ as the sum of its digits, and the sum of the digits of $2n$ does not exceed 1202. Let $N$ be the largest number that is special. How many digits does $N$ have?

I thought to let $N = a_k10^k + a_{k-1}10^{k-1}+\dots+a_210^2+a_110+a_0$. This way, the requirements become:

  • $a_i\neq0$ for $i=\{1,2,\dots k\}$.
  • $a_1+a_2+\dots\ a_k=2021$ for $1\leq a_1,a_2,\dots,a_k\leq 9$.
  • $2a_1+2a_2+\dots+2a_k\leq1202$ for $1\leq a_1,a_2,\dots,a_k\leq 9$.

I have tried to meddle with the second and third using stars and bars, but I met a deadend. Help would be appreciated, thanks.

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    $\begingroup$ The sum of the digits of $2N$ is not necessarily $2a_1+ \cdots +2a_k$. For example, $15$ has a digit sum of $6$ and $2(15)=30$ has a digit sum of $3$. $\endgroup$
    – Scene
    Commented Mar 31, 2023 at 3:43
  • $\begingroup$ Ah, I forgot to consider that. Thanks. $\endgroup$
    – ryan.zcd
    Commented Mar 31, 2023 at 3:53
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    $\begingroup$ $316$ fives, followed by $441$ ones, gives digit sum $5\times316+441=2021$, and twice that number has $316$ ones and $441$ twos, sum $316+882=1198$. $\endgroup$ Commented Mar 31, 2023 at 4:55
  • $\begingroup$ How can I be sure that is the largest such $N$? $\endgroup$
    – ryan.zcd
    Commented Mar 31, 2023 at 8:12
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    $\begingroup$ I didn't say that it is. But it does give you something to think about, doesn't it? $\endgroup$ Commented Mar 31, 2023 at 12:34

2 Answers 2

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Assume $M$ is the largest number.


Step $1$:

$M$ cannot contain $2,3,4$ as its digits. For example, if $M=a...b4c ...d$, then $M=a...b1111c...d$ is a greater number which satisfies both of the required conditions.


Step $2$:

$M$ cannot contain $6,7,8,9$ as its digits. For example, if $M=a...b7c ...d$ ($M=a...b8c...d$), then $M=a...b115c...d$ ($M=a...b1115c...d$) is a greater number which satisfies both of the required conditions.


Conclusion:

$M$ only contains $5$ and $1$.


Step $3$:

$M$ is comprised of $m$ fives and $n$ ones ($5m+n=2021$), and the sum of the digits of $2M$ is $m+2n.$


Step $4$:

Our goal is to maximize $m+n=m+(2021-5m)$ where $m+2n=m+2(2021-5m) \leq 1202.$


Step $5$:

From $m+2(2021-5m) \leq 1202$, we conclude that $316 \leq m$, which implies that the maximum of $m+n=m+(2021-5m)$ happens at $m=316$ and $n=441$.

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    $\begingroup$ Ha! I got it right! $\endgroup$ Commented Apr 1, 2023 at 11:59
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Hints towards a solution. If you're stuck, explain what you've tried.

  • Prove that if $n$ has $c_i$ digits that are $i$, then $2n$ has digit sum $2c_1 + 4c_2+6c_3+8c_4 + 1c_5 + 3c_6+5c_7+7c_8 + 9c_9$.
    • This is the crux. Think about why we'd be interested in something like this, and how you could come up with it.
  • Hence, what can you say about $c_1 + 2c_2 + 3c_3 + 4c_4$?
  • What can you say about $5c_5+ 6c_6 + 7c_7 + 8c_8 + 9c_9$?
  • Under these constraints, what is the maximum number of digits ($\sum c_i)$?
  • Hence, what is the maximum $n$?
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