17. On the table, there are 10 number cards from $0 \sim 9$. Three people, A, B, and C, each took three of these cards and summed all the different three-digit numbers that can be formed with their cards. The results for A, B, and C were $1554, 1688, 4662$, respectively. The remaining number card is ( ). (Note: 6 or 9 cannot be flipped to become 9 or 6.)
【Answer】9
【Analysis】The method to find the sum of all three-digit numbers formed by three different digits $\mathrm{a}, \mathrm{b}, \mathrm{c}$:
(1) When $a, b, c$ do not include 0, the sum is $222(a+b+c)$;
(2) When $a, b, c$ do not include 0, set $a=0$, then the sum is $222(a+b+c)-11(b+c)=211(a+b+c)$; so the sum is either 222 times the sum of the digits, or 211 times the sum of the digits.
Jia: $1554=222 \times 7$ ～ Yi: $1688=211 \times 8$; Bing: $4662=222 \times 21$; the sums of the digits are $7,8,21$; the remaining digit is $45-7-8-21=9$.