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As shown here: https://puzzling.stackexchange.com/questions/97696/consecutive-numbers-which-use-all-digits-a-different-number-of-times/97698#97698, there are arbitrarily long blocks of consecutive integers which, when written all down, use each of the ten digits a different number of times. That block can even begin at 1.

What I am looking for is the smallest N such that writing down all the numbers 1, 2, 3, ...,N will use each of the digits 0 to 9 a different number of times. Moreover, can such numbers be characterized in some nice way?

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  • $\begingroup$ Well if you you them all, and you each of them a different number of times, you could use one of them once, another one twice, ... $\endgroup$
    – saulspatz
    Apr 30, 2020 at 19:26

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Two adjacent digits appear the same number of times unless one of them occurs in $N$. Here $0$ is effectively adjacent to $9$, not to $1$, because numbers don’t have leading zeros, so $0$ appears after $9$, not before $1$.

So we have $9$ pairs of adjacent digits to cover, and each digit only covers $2$ pairs, so we need at least $5$ digits. If this were the whole story, the least admissible $N$ would be $13579$. But the last digit is special: A last digit $d$ only makes $d$ and $d+1$ have different counts, but not $d$ and $d-1$, so $13579$ leads to the same number of $9$s and $8$s. So we actually have to cover $8$ pairs of digits with the remaining $4$ digits, and that only works if we start with $2$.

So the next candidate would be $24689$. But there’s another special feature of the last digit: If it’s a $9$, that means that the penultimate digit $d$ gets the same count as $d-1$, so we lose one distinction that we need. Thus, we can’t have the $9$ in the last digit, so we have to move it up by one.

That leads to $N=24697$, and this is indeed the first number of the kind you’re looking for. Here’s Java code that checks this result.

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