39. Is it possible to form a ten-digit number using each digit exactly once that is divisible by 1980?

Note. To solve this problem, additional information about the divisibility of a number by 11 is required. If the difference between the sums of the digits in the even and odd positions (the alternating sum of the digits) is divisible by 11, then the number is divisible by 11.
39. The desired number must end with the digit 0, then when dividing it by 1980, we can reduce it by 10. The resulting nine-digit number must be divisible by 198. Since $198=2 \cdot 9 \cdot 11$, this nine-digit number must be divisible by 2, 9, and 11. Therefore, it must end with an even digit, have a sum of digits divisible by 9, and an alternating sum of digits divisible by 11. Since $1+2+3+4+5+6+7+8+9=45$ and $45: 9$, such a nine-digit number is indeed divisible by 9. The digits cannot be arranged in such a way that their alternating sum equals zero, as 45 is not divisible by 2. Therefore, the digits must be arranged so that their alternating sum equals 11, $45-11=34$, $34: 2=17$. One such number is 238471596. Therefore, the number 2384715960 is divisible by 1980 without a remainder.