Brazilian Mathematical Olympiad

Note that $a_0 + a_1 + \cdots + a_9$ is equal to the sum of number of 0s, 1s, ..., 9s, which is, in turn, equal to total number of digits, 10. In other words, the sum of the digits is 10. In particular, if $i > 5$, then $a_i = 0$ or 1 (because if $a_i = 2$, then we have at least 2 digits bigger than 5 with sum bigger than 10).

Suppose all $a_i \le 5$. Then in particular $a_0 \le 5$, so there are at least 5 non-zero digits, and hence at least 4 of $a_1, a_2, \dots, a_9$ are non-zero. But that means that $a_1 + 2a_2 + 3a_3 + \cdots + 9a_9 \ge 1 + 2 + 3 + 4$ with equality iff $a_1 = a_2 = a_3 = a_4 = 1$ and $a_5 = a_6 = a_7 = a_8 = a_9 = 0$. But we must have equality because $a_1 + 2a_2 + 3a_3 + \cdots + 9a_9$ is the sum of the digits which is 10. However it is not possible to have $a_1 = a_2 = a_3 = a_4 = 1$ and $a_5 = a_6 = a_7 = a_8 = a_9 = 0$ because then there are four 1s, but $a_4 = 1$.

So suppose some $a_i \ge 6$. Then $a_6 \ge 1$. But $a_6 \le 1$, so $a_6 = 1$. Hence $a_1 > 0$. We cannot have $a_1 = 1$, because then there are at least two 1s, so $a_1 \le 2$. We cannot have $a_1 = 6$, because then there are six 1s and one 6, total 12, whereas all the digits only total 10. So $a_1$ must be some other digit $k > 1$. But we now have at least six 0s, at least two 1s, a $k$ and a 6, which is at least
10. So it must be exactly 10. Hence $a_0 = 6$ and $a_1 = 2$. So the number must be 6210001000.