Problem 1792

Solution:
Denote by $M_{n}$ the set of all digits of the numbers $1,2, \ldots, n$. First we find the least positive integer $n=\overline{a_{1} a_{2} \ldots a_{t}}$ such that every two nonzero digits appear different number of times in $M_{n}$. By adding zeros on the left we may assume that all numbers $1,2, \ldots, n-1$ are $t$-digit numbers.

It is clear that every nonzero digit appears the same number of times. Let $B_{i}^{j}, 1 \leq i \leq t, 1 \leq j \leq 9$ be the number of appearances of the digit $j$ in position $i$ among the numbers $1,2, \ldots, n$. Note that for all $i$ and $j \leq 8$, if a number $A$ has $j+1$ in position $i$, then replacing this digit by $j$ we obtain a number which is less than $A$. Therefore $B_{i}^{j} \geq B_{i}^{j+1}$.

Furthermore for a fixed $i$ the inequality $B_{i}^{j} \geq B_{i}^{j+1}$ is fulfilled for at most two pairs of digits $j$ and $j+1$, namely $a_{i-1}$ and $a_{i} ; a_{i}$ and $a_{i+1}$. Moreover, if $i=t$, it is fulfilled only for $a_{t}$ and $a_{t+1}$. Since there are 8 pairs of the form $(j, j+1)$ we have $t \geq 5$. If $n=13578$ then $B_{1}^{1}>B_{1}^{2} ; B_{2}^{2}>B_{2}^{3} ; B_{2}^{3}>B_{2}^{4}$; $B_{3}^{4}>B_{3}^{5} ; B_{3}^{5}>B_{3}^{6} ; \bar{B}_{4}^{6}>B_{4}^{7} ; B_{4}^{7}>B_{4}^{8} ; B_{5}^{8}>B_{5}^{9}$, i.e. $n=13578$ satisfies the condition of the problem.

If $m<13578$ also satisfies the condition then the first digit of $m$ is 1 and the second digit is $0,1,2$ or 3. Since $B_{1}^{j}>B_{1}^{j+1}$ is true only for $j=1$ if the second digit is 0,1 or 2 then at least two consecutive digits appear equal number of times. Therefore the second digit of $m$ is 3. It follows by similar arguments that the third, fourth and the fifth digits of $m$ are respectively 5, 7 and 8. Therefore $n=13578$ is the least positive integer such that every two nonzero digits appear different number of times in $M_{n}$.

Since the number of digits of all numbers $1,2,3, \ldots, 13578$ equals
$$9 \cdot 1 + 90 \cdot 2 + 900 \cdot 3 + 9000 \cdot 4 + 3579 \cdot 5 = 56784$$
we conclude that 56784 has the desired property.

Suppose that there exists $k<56784$ which has the desired property. Then the digits in the sequence are those in $M_{s}$ for some $s<13578$ and some digits of $s+1$. According to the previous observations there exist two consecutive digits that are not digits of $s$ (eventually excluding the last one) appearing equal number of times in $M_{s}$. If the last digit of $s$ is not 9, then the same digits appear equal number of times in the sequence since the digits of $s$ and $s+1$ are the same (except the last one). If the last digit of $s$ is 9 then the last digit of $s+1$ is 0 and therefore $s+1<13578$. Hence we conclude as above that there exist two consecutive digits not among the digits of $s+1$ which appear equal number of times.