14. 3456
14. Such positive integers are all divisible by 9 since each has sum of digits $0+1+2+\cdots+9=45$. Hence we can change the number ' 11111 ' in the question to ' 99999 ' without affecting the answer. We denote such a 10-digit positive integer by $\overline{\text { ABCDEFGHIJ }}$. For this number to be divisible by 99999 , a necessary and sufficient condition is that $\overline{\mathrm{ABCDE}}+\overline{\mathrm{FGHIJ}}$ should be divisible by 99999 (see Remark). Hence $\overline{\mathrm{ABCDE}}+\overline{\mathrm{FGHIJ}}$ must be exactly 99999. In other words, each of $\mathrm{A}+\mathrm{F}, \mathrm{B}+\mathrm{G}, \mathrm{C}+\mathrm{H}, \mathrm{D}+\mathrm{I}$ and $\mathrm{E}+\mathrm{J}$ must be equal to 9 , i.e. we need only choose A, B, C, D, E. There are 9 choices for $\mathrm{A}$ as it cannot be zero (once $\mathrm{A}$ is chosen, $\mathrm{F}$ is also fixed). There are 8 choices for $\mathrm{B}$ as it cannot be equal to $\mathrm{A}$ and $\mathrm{F}$ (once $\mathrm{B}$ is chosen, $\mathrm{G}$ is also fixed). Similarly, there are $6,4,2$ choices for C, D, E respectively, and at the same time H, I, J are fixed. Hence the answer is $9 \times 8 \times 6 \times 4 \times 2=3456$.
Remark. The divisibility test for 99999 is to put the digits into groups of 5 from the right, and then add up the numbers (as positive integers) in each group. The original number is divisible by 99999 if and only if the sum obtained is divisible by 99999 . In the 10 -digit case, this can be explained by the following congruence equation:
$$
\begin{aligned}
\overline{\mathrm{ABCDEFGHIJ}} & =100000 \times \overline{\mathrm{ABCDE}}+\overline{\mathrm{FGHIJ}} \\
& \equiv 1 \times \overline{\mathrm{ABCDE}}+\overline{\mathrm{FGHIJ}} \\
& =\overline{\mathrm{ABCDE}}+\overline{\mathrm{FGHIJ}} \quad(\bmod 99999)
\end{aligned}
$$
14. How many 10-digit positive integers have all 10 digits pairwise distinct and are multiples of 11111 ?
(2 marks)
有多少個 10 位正整數的 10 個數字互不相同, 並且是 11111 的倍數？
(2 分)