9.4. Find the smallest natural number in which each digit occurs exactly once and which is divisible by 990.
Answer: 1234758690.

Solution. The number 990 is the product of coprime numbers 2, 5, 9, and 11. Any ten-digit number composed of different digits, each used once, is divisible by 9, since their sum, which is 45, is divisible by 9. According to the divisibility rule for 10, the desired number must end in 0. It remains to address the divisibility by 11.

The divisibility rule for 11 states: a number is divisible by 11 if and only if the difference between the sum of all its digits in odd positions from left to right and the sum of its digits in even positions is divisible by 11. Let's estimate the value of $\mathrm{S}$, the sum of the digits of the desired number in odd positions: it is no less than $1+2+3+4+5=15$ and no more than $5+6+7+8+9=35$. Therefore, the difference between the sum of all the digits in odd positions and the sum of the digits in even positions, equal to $2S-45$, is an odd number in the interval from -15 to 25, divisible by 11.

There are only two such numbers: -11 and 11, for which $\mathrm{S}$ is 17 and 28, respectively. It is easy to verify that for $S=17$ there are only two options: $S=1+2+3+4+7$ and $S=1+2+3+5+6$. Considering the minimality, this gives the number 1526384970.

For $\mathrm{S}=28$, we will list the smallest possible digits from left to right, as long as it is possible while maintaining the condition that the sum of the digits in odd positions can ultimately be 28, and the sum of the digits in even positions can be 17. This results in 1234, and the sum of the remaining 3 digits in the fifth, seventh, and ninth positions must equal 24, which is only possible if they are 7, 8, and 9, leading to the number in the answer. This number is smaller than the previously found 1526384970 for $S=17$. If a smaller number could be found, it would have $S=28$ and the fifth digit from the left would be less than 7, which, as we have seen, is impossible.

Grading criteria. If the answer is guessed and divisibility is verified: 1 point. Possible values of S are found: 2 points. The answer is written, and then the minimality is proven by full enumeration: 7 points.